week number - Page 145 - Developer IT

Best practice - logging events (general) and changes (database)

- by b0x0rz

need help with logging all activities on a site as well as database changes. requirements: * should be in database * should be easily searchable by initiator (user name / session id), event (activity type) and event parameters i can think of a database design but either it involves a lot of tables (one per event) so i can log each of the parameters of an event in a separate field OR it involves one table with generic fields (7 int numeric and 7 text types) and log everything in one table with event type field determining what parameter got written where (and hoping that i don't need more than 7 fields of a certain type, or 8 or 9 or whatever number i choose)... example of entries (the usual things): [username] login failed @datetime [username] login successful @datetime [username] changed password @datetime, estimated security of password [low/ok/high/perfect] @datetime [username] clicked result [result number] [result id] after searching for [search string] and got [number of results] @datetime [username] clicked result [result number] [result id] after searching for [search string] and got [number of results] @datetime [username] changed profile name from [old name] to [new name] @datetime [username] verified name with [credit card type] credit card @datetime datbase table [table name] purged of old entries @datetime via automated process etc... so anyone dealt with this before? any best practices / links you can share? i've seen it done with the generic solution mentioned above, but somehow that goes against what i learned from database design, but as you can see the sheer number of events that need to be trackable (each user will be able to see this info) is giving me headaches, BUT i do LOVE the one event per table solution more than the generic one. any thoughts? edit: also, is there maybe an authoritative list of such (likely) events somewhere? thnx stack overflow says: the question you're asking appears subjective and is likely to be closed. my answer: probably is subjective, but it is directly related to my issue i have with designing a database / writing my code, so i'd welcome any help. also i tried narrowing down the ideas to 2 so hopefully one of these will prevail, unless there already is an established solution for these kinds of things.

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How to properly translate the "var" result of a lambda expression to a concrete type?

- by CrimsonX

So I'm trying to learn more about lambda expressions. I read this question on stackoverflow, concurred with the chosen answer, and have attempted to implement the algorithm using a console app in C# using a simple LINQ expression. My question is: how do I translate the "var result" of the lambda expression into a usable object that I can then print the result? I would also appreciate an in-depth explanation of what is happening when I declare the outer => outer.Value.Frequency (I've read numerous explanations of lambda expressions but additional clarification would help) C# //Input : {5, 13, 6, 5, 13, 7, 8, 6, 5} //Output : {5, 5, 5, 13, 13, 6, 6, 7, 8} //The question is to arrange the numbers in the array in decreasing order of their frequency, preserving the order of their occurrence. //If there is a tie, like in this example between 13 and 6, then the number occurring first in the input array would come first in the output array. List<int> input = new List<int>(); input.Add(5); input.Add(13); input.Add(6); input.Add(5); input.Add(13); input.Add(7); input.Add(8); input.Add(6); input.Add(5); Dictionary<int, FrequencyAndValue> dictionary = new Dictionary<int, FrequencyAndValue>(); foreach (int number in input) { if (!dictionary.ContainsKey(number)) { dictionary.Add(number, new FrequencyAndValue(1, number) ); } else { dictionary[number].Frequency++; } } var result = dictionary.OrderByDescending(outer => outer.Value.Frequency); // How to translate the result into something I can print??

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Naive Bayesian classification (spam filtering) - Doubt in one calculation? Which one is right? Plz c

- by Microkernel

Hi guys, I am implementing Naive Bayesian classifier for spam filtering. I have doubt on some calculation. Please clarify me what to do. Here is my question. In this method, you have to calculate P(S|W) - Probability that Message is spam given word W occurs in it. P(W|S) - Probability that word W occurs in a spam message. P(W|H) - Probability that word W occurs in a Ham message. So to calculate P(W|S), should I do (1) (Number of times W occuring in spam)/(total number of times W occurs in all the messages) OR (2) (Number of times word W occurs in Spam)/(Total number of words in the spam message) So, to calculate P(W|S), should I do (1) or (2)? (I thought it to be (2), but I am not sure, so plz clarify me) I am refering http://en.wikipedia.org/wiki/Bayesian_spam_filtering for the info by the way. I got to complete the implementation by this weekend :( Thanks and regards, MicroKernel :) @sth: Hmm... Shouldn't repeated occurrence of word 'W' increase a message's spam score? In the your approach it wouldn't, right?. Lets take a scenario and discuss... Lets say, we have 100 training messages, out of which 50 are spam and 50 are Ham. and say word_count of each message = 100. And lets say, in spam messages word W occurs 5 times in each message and word W occurs 1 time in Ham message. So total number of times W occuring in all the spam message = 5*50 = 250 times. And total number of times W occuring in all Ham messages = 1*50 = 50 times. Total occurance of W in all of the training messages = (250+50) = 300 times. So, in this scenario, how do u calculate P(W|S) and P(W|H) ? Naturally we should expect, P(W|S) P(W|H)??? right. Please share your thought...

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Array: Recursive problem cracked me up

- by VaioIsBorn

An array of integers A[i] (i 1) is defined in the following way: an element A[k] ( k 1) is the smallest number greater than A[k-1] such that the sum of its digits is equal to the sum of the digits of the number 4* A[k-1] . You need to write a program that calculates the N th number in this array based on the given first element A[1] . INPUT: In one line of standard input there are two numbers seperated with a single space: A[1] (1 <= A[1] <= 100) and N (1 <= N <= 10000). OUTPUT: The standard output should only contain a single integer A[N] , the Nth number of the defined sequence. Input: 7 4 Output: 79 Explanation: Elements of the array are as follows: 7, 19, 49, 79... and the 4th element is solution. I tried solving this by coding a separate function that for a given number A[k] calculates the sum of it's digits and finds the smallest number greater than A[k-1] as it says in the problem, but with no success. The first testing failed because of a memory limit, the second testing failed because of a time limit, and now i don't have any possible idea how to solve this. One friend suggested recursion, but i don't know how to set that. Anyone who can help me in any way please write, also suggest some ideas about using recursion/DP for solving this problem. Thanks.

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Why can't decimal numbers be represented exactly in binary?

- by Barry Brown

There have been several questions posted to SO about floating-point representation. For example, the decimal number 0.1 doesn't have an exact binary representation, so it's dangerous to use the == operator to compare it to another floating-point number. I understand the principles behind floating-point representation. What I don't understand is why, from a mathematical perspective, are the numbers to the right of the decimal point any more "special" that the ones to the left? For example, the number 61.0 has an exact binary representation because the integral portion of any number is always exact. But the number 6.10 is not exact. All I did was move the decimal one place and suddenly I've gone from Exactopia to Inexactville. Mathematically, there should be no intrinsic difference between the two numbers -- they're just numbers. By contrast, if I move the decimal one place in the other direction to produce the number 610, I'm still in Exactopia. I can keep going in that direction (6100, 610000000, 610000000000000) and they're still exact, exact, exact. But as soon as the decimal crosses some threshold, the numbers are no longer exact. What's going on? Edit: to clarify, I want to stay away from discussion about industry-standard representations, such as IEEE, and stick with what I believe is the mathematically "pure" way. In base 10, the positional values are: ... 1000 100 10 1 1/10 1/100 ... In binary, they would be: ... 8 4 2 1 1/2 1/4 1/8 ... There are also no arbitrary limits placed on these numbers. The positions increase indefinitely to the left and to the right.

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Insert and delete SIM contacs working but it needs to be phone restart to update the changes

- by girishgm08

Hi All, I am able to insert the contacts into SIM card and delete from it. But it needs to be phone restart to update the changes. The below is the code woks for delete the conatcs, Uri simUri = Uri.parse("content://icc/adn"); Cursor cur = context.getContentResolver().query(simUri, null, null, null, null); prn("Number of SIM Contacts are.."+cur.getCount()); int row =0; while(cur.moveToNext()){ String name = cur.getString(cur.getColumnIndex("name")); prn("Name..."+name); String data = cur.getString(cur.getColumnIndex("number")); if(!data.equals("")) prn("Number.."+data); String where = null; if(!name.equals("") && !data.equals("")){ where = "tag =" + name + "AND" + "number =" +data; } else if(name.equals("") && !data.equals("")){ where = "number ="+data; } else { where = "tag ="+name+ "AND" +"number="+null; } context.getContentResolver().delete(simUri, where, null); row++; } prn(row+" are deleted"); cur.close(); cur = null; Please look into this issue and give suggestions on this. Thanks, Girish G M

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Java ternary operator and boxing Integer/int?

- by Markus

I tripped across a really strange NullPointerException the other day caused by an unexpected type-cast in the ternary operator. Given this (useless exemplary) function: Integer getNumber() { return null; } I was expecting the following two code segments to be exactly identical after compilation: Integer number; if (condition) { number = getNumber(); } else { number = 0; } vs. Integer number = (condition) ? getNumber() : 0; . Turns out, if condition is true, the if-statement works fine, while the ternary opration in the second code segment throws a NullPointerException. It seems as though the ternary operation has decided to type-cast both choices to int before auto-boxing the result back into an Integer!?! In fact, if I explicitly cast the 0 to Integer, the exception goes away. In other words: Integer number = (condition) ? getNumber() : 0; is not the same as: Integer number = (condition) ? getNumber() : (Integer) 0; . So, it seems that there is a byte-code difference between the ternary operator and an equivalent if-else-statement (something I didn't expect). Which raises three questions: Why is there a difference? Is this a bug in the ternary implementation or is there a reason for the type cast? Given there is a difference, is the ternary operation more or less performant than an equivalent if-statement (I know, the difference can't be huge, but still)?

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MySQL Paritioning performance

- by Imran Pathan

Measured performance on key partitioned tables and normal tables separately. But we couldn't find any performance improvement with partitioning. Queries are pruned. Using MySQL 5.1.47 on RHEL 4. Table details: UserUsage - Will have entries for user mobile number and data usage for each date. Mobile number and Date as PRI KEY. UserProfile - Queries prev table and stores summary for each mobile number. Mobile number PRI KEY. CREATE TABLE `UserUsage` ( `Msisdn` decimal(20,0) NOT NULL, `Date` date NOT NULL, . . PRIMARY KEY USING BTREE (`Msisdn`,`Date`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 PARTITION BY KEY(Msisdn) PARTITIONS 50; CREATE TABLE `UserProfile` ( `Msisdn` decimal(20,0) NOT NULL, . . PRIMARY KEY (`Msisdn`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 PARTITION BY KEY(Msisdn) PARTITIONS 50; Second table is updated by query select and order by date in first table in a perl program, query is select * from UserUsage where Msisdn=number order by Date desc limit 7 [Process data in perl] update UserProfile values(....) where Msisdn=number explain partition for select, shows row being scanned in a particular partition only. Is something wrong with partition design or queries as partitioning is taking almost same or more time compared to normal tables?

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Animate and form rows, arrays, AS3

- by VideoDnd

Question How can I animate and form rows together? Explanation One 'for loop' is for animation, the other 'for loop' is for making rows. I want to understand how to use arrays and create a row of sprite animations. 'for loop' for animation //FRAMES ARRAY //THIS SETS UP MY ANIMATION FOR TIMER EVENT var frames:Array = [ new Frame1(), new Frame2(), new Frame3(), new Frame4(), new Frame5(), new Frame6(), new Frame7(), new Frame8(), new Frame9(), new Frame0(), ]; for each (var frame:Sprite in frames) { addChild(frame); } 'for loop' for rows //THIS MAKES A ROW OF DISPLAY OBJECTS var numberOfClips:Number = 11; var xStart:Number = 0; var yStart:Number = 0; var xVal:Number = xStart; var xOffset:Number = 2; for (var $:Number=0; $<numberOfClips; $++) { //DUDE ARRAY var dude:Array = frames; dude.y = yStart +11; dude.x = xVal +55; xVal = dude.x + dude.width + this.xOffset; } timer var timer:Timer = new Timer(100); timer.addEventListener(TimerEvent.TIMER, countdown); function countdown(event:TimerEvent) { var currentFrame:int = timer.currentCount % frames.length; for (var i:int = 0; i < frames.length; ++i) { frames[i].visible = (i == currentFrame); } } timer.start(); counter experiment My new class I'm working on loops through 10 different display objects that are numbers. For those following, I'm trying to make something like NumbersView.

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Knowing which annotation is selected and accessing properties of it.

- by kevin Mendoza

So far my program can display a database of custom annotation views. Eventually I want my program to be able to display extra information after a button on the annotation bubble is clicked. Each element in the database has a unique entry Number, so I thought it would be a good idea to add this entry number as a property of the custom annotation. The problem I am having is that after the button is clicked and the program switches to a new view I am unable to retrieve the entry number of the annotation I selected. Below is the code that assigns the entry Number property to the annotation: for (id mine in mines) { workingCoordinate.latitude = [[mine latitudeInitial] doubleValue]; workingCoordinate.longitude = [[mine longitudeInitial] doubleValue]; iProspectAnnotation *tempMine = [[iProspectAnnotation alloc] initWithCoordinate:workingCoordinate]; [tempMine setTitle:[mine mineName]]; [tempMine setAnnotationEntryNumber:[mine entryNumber]]; } [mines dealloc]; When the button on an annotation is selected, this is the code that initializes the new view: - (void)mapView:(MKMapView *)mapView annotationView:(MKAnnotationView *)view calloutAccessoryControlTapped:(UIControl *)control { mineInformationController *controller = [[mineInformationController alloc] initWithNibName:@"mineInformationController" bundle:nil]; controller.modalTransitionStyle = UIModalTransitionStyleCrossDissolve; [self presentModalViewController:controller animated:YES]; [controller release]; } and lastly is my attempt at retrieving the entryNumber property from the new view so that I can compare it to the mines database and retrieve more information on the array element. iProspectFresno_LiteAppDelegate *appDelegate = (iProspectFresno_LiteAppDelegate *)[[UIApplication sharedApplication] delegate]; NSMutableArray* mines = [[NSMutableArray alloc] initWithArray:(NSMutableArray *)appDelegate.mines]; for(id mine in mines) { if ([[mine entryNumber] isEqualToNumber: /*the entry Number of the selected annotation*/]) { /* display the information in the mine object */ } } So how do I access this entry number property in this new view controller?

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how do I get the program to Know which annotation is selected and be able to access properties of it

- by kevin Mendoza

So far my program can display a database of custom annotation views. Eventually I want my program to be able to display extra information after a button on the annotation bubble is clicked. Each element in the database has a unique entry Number, so I thought it would be a good idea to add this entry number as a property of the custom annotation. The problem I am having is that after the button is clicked and the program switches to a new view I am unable to retrieve the entry number of the annotation I selected. Below is the code that assigns the entry Number property to the annotation: for (id mine in mines) { workingCoordinate.latitude = [[mine latitudeInitial] doubleValue]; workingCoordinate.longitude = [[mine longitudeInitial] doubleValue]; iProspectAnnotation *tempMine = [[iProspectAnnotation alloc] initWithCoordinate:workingCoordinate]; [tempMine setTitle:[mine mineName]]; [tempMine setAnnotationEntryNumber:[mine entryNumber]]; } [mines dealloc]; When the button on an annotation is selected, this is the code that initializes the new view: - (void)mapView:(MKMapView *)mapView annotationView:(MKAnnotationView *)view calloutAccessoryControlTapped:(UIControl *)control { mineInformationController *controller = [[mineInformationController alloc] initWithNibName:@"mineInformationController" bundle:nil]; controller.modalTransitionStyle = UIModalTransitionStyleCrossDissolve; [self presentModalViewController:controller animated:YES]; [controller release]; } and lastly is my attempt at retrieving the entryNumber property from the new view so that I can compare it to the mines database and retrieve more information on the array element. iProspectFresno_LiteAppDelegate *appDelegate = (iProspectFresno_LiteAppDelegate *)[[UIApplication sharedApplication] delegate]; NSMutableArray* mines = [[NSMutableArray alloc] initWithArray:(NSMutableArray *)appDelegate.mines]; for(id mine in mines) { if ([[mine entryNumber] isEqualToNumber: /*the entry Number of the selected annotation*/]) { /* display the information in the mine object */ } } So how do I access this entry number property in this new view controller?

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How do I do a count on that meet a specific condition, dependent on several has_many relationships i

- by Angela

I have a Model Campaign. A Campaign has many Events. Each Event has an attribute :days. A Campaign also has_many Contacts. Each Contact as a :date_entered attribute. The from_today(contact,event) method returns a number, which is the number of days from the contact's :date_entered till today minus the event's :days. In other words, a positive number shows the number of days from today till the :days of the event is elapsed. If it is negative, if means that the number of days that has elapsed since the :date_entered is greater than the :days attribute of an event. In other words, the event is overdue. What I would like to be able to do is do campaign.overdue and this would result in a total number of contacts that have an overdue event. It shouldn't count multiple events for a single contact, just one contact. How do I do that? It seems like I would need to cycle through all the events for every contact and keep a counter but I'm assuming that there is a better way.

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What is a more "ruby way" to write this code?

- by steadfastbuck

This was a homework assignment for my students (I am a teaching assistant) in c and I am trying to learn Ruby, so I thought I would code it up. The goal is to read integers from a redirected file and print some simple information. The first line in the file is the number of elements, and then each integer resides on its own line. This code works (although perhaps inefficiently), but how can I make the code more Ruby-like? #!/usr/bin/ruby -w # first line is number of inputs (Don't need it) num_inputs = STDIN.gets.to_i # read inputs as ints h = Hash.new STDIN.each do |n| n = n.to_i h[n] = 1 unless h[n] and h[n] += 1 end # find smallest mode h.sort.each do |k,v| break puts "Mode is: #{k}", "\n" if v == h.values.max end # mode unique? v = h.values.sort print "Mode is unique: " puts v.pop == v.pop, "\n" # print number of singleton odds, # odd elems repeated odd number times in desc order # even singletons in desc order odd_once = 0 odd = Array.new even = Array.new h.each_pair do |k, v| odd_once += 1 if v == 1 and k.odd? odd << k if v.odd? even << k if v == 1 and k.even? end puts "Number of elements with an odd value that appear only once: #{odd_once}", "\n" puts "Elements repeated an odd number of times:" puts odd.sort.reverse, "\n" puts "Elements with an even value that appear exactly once:" puts even.sort.reverse, "\n" # print fib numbers in the hash class Fixnum def is_fib? l, h = 0, 1 while h <= self return true if h == self l, h = h, l+h end end end puts "Fibonacci numbers:" h.keys.sort.each do |n| puts n if n.is_fib? end

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dividing two array elements

- by pradeep

main() { int prime_array[2339],prime1_count=0,mul1_count=0; int i, prime, lim_up, lim_low, n,j=0; int mul,count=0; int mul_count[65026]={0},number[7096]; printf("\n ENTER THE LOWER LIMIT…: "); scanf("%d", &lim_low); printf("\n ENTER THE UPPER LIMIT…: "); scanf("%d", &lim_up); for(n=lim_low+1; n<lim_up; n++) { prime = 1; for(i=2; i<n; i++) if(n%i == 0) { prime = 0; break; } if(prime) { prime_array[j]=n; j++; } } for(i=1;i<=255;i++) { for(j=1;j<=255;j++) { mul = j*i; mul_count[mul]++; } } for(i=1;i<=65025;i++) if( mul_count[i]!=2 && mul_count[i]!=0 ) { number[count]=i; count++; } for(prime1_count=0;prime1_count<2339;prime1_count++) { printf("\nprime number used is:%d",prime_array[prime1_count]); for(mul1_count=0;mul1_count<7096;mul1_count++) { printf("\n%d\t",number[mul1_count] % prime_array[prime1_count]); } } } I want to find the modulus of (number[mul1_count] % prime_array[prime1_count] ), but the output which I get is wrong. What is the mistake here. The prime number should be in the range 40000 to 65025. What changes should i make here?

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PHP's page generation time takes 0.01s. 1/0.01 = 100; however i'm having problems reaching that number of request per seconds. Why?

- by cedivad

On average, my PHP page generation time is 10ms. So i should be able to execute 100 requests one after the other one (using a single core on the server, since that php is not multithreaded). However, i'm having problems reaching 50 pages per seconds. As of now i do 25 on avg., with a medium load. The application is really light, it consist in a read (<5KB) from a pool of SSDs, some read queries solved by indexes. Where should i look to solve this bottleneck?

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How to account for non-prime numbers 0 and 1 in java?

- by shady

I'm not sure if this is the right place to be asking this, but I've been searching for a solution for this on my own for quite some time, so hopefully I've come to the right place. When calculating prime numbers, the starting number that each number has to be divisible by is 2 to be a non-prime number. In my java program, I want to include all the non-prime numbers in the range from 0 to a certain number, so how do I include 0 and 1? Should I just have separate if and else-if statements for 0 and 1 that state that they are not prime numbers? I think that maybe 0 and 1 should be included in the java for loop, but I don't know how to go about doing that. for (int i = 2; i < num; i++){ if (num % i == 0){ System.out.println(i + " is not a prime number. "); } else{ System.out.println(i + " is a prime number. "); } }

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selected option not clearing from memory android

- by user2980560

I have a small random number spinner that when you click gives a random number. I am having two problems. The first is when the main activity loads it displays a random number on the screen without the random number spinner being clicked. I am unsure what to set to false to keep it from opening with the main activity. The second problem is that when you select an option from the spinner it does not clear. Meaning that If you click on option D6 or D20 then you can not click on the same option again until selecting the other option first. Essentially the selection does not clear out of memory after the random number is selected. Here is the random number code public void onItemSelected(AdapterView<?> parent, View view, int pos, long id) { Random rand = new Random(); int roll; // An item was selected. if (spinner1.getSelectedItemPosition()==0) { roll = rand.nextInt(6)+1; } else { roll = rand.nextInt(20)+1; } // Put the result into a string. String text = "You rolled a " + roll; // Build a dialog box and with the result string and a single button AlertDialog.Builder builder = new AlertDialog.Builder(this); builder.setMessage(text).setCancelable(false) .setPositiveButton("OK", new DialogInterface.OnClickListener() { public void onClick(DialogInterface dialog, int id) { // do things when the user clicks ok. } }); AlertDialog alert = builder.create(); // Show the dialog box. alert.show(); }

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IASA Sessions on Social Networking Note Influence of Millennial Generation on Insurance Technology

- by [email protected]

Helen Pitts, senior product marketing manager for Oracle Insurance is blogging from the 2010 IASA Annual Conference and Business Show this week. Social networking continues to be a buzzword for many in the industry. Erin Esurance, the Geico Gecko and even Nationwide's "The World's Greatest Spokesperson in the World" all have a prominent presence in the social media world. Sessions at the 2010 IASA Annual Conference and Business Show this week in Grapevine, Texas, highlighted how the millennial generation's exploding use of social media is spurring more carriers to leverage tools like Facebook, LinkedIn and other social networks to engage prospect and customers. While panelists encouraged carriers to leverage social networking tools for marketing and communications, they expressed the need for caution and corporate governance when it comes to using the tools as a part of claims, underwriting, and human resources recruitment business practices, and interactions with producers. (A previous Oracle Insurance blog entry by my colleague Susan Keuer noted that social networking and its impact on the underwriting process was also a hot topic at the recent AHOU conference.) Speaking of the millennial generation, IASA announced a new scholarship program and awarded three scholarships during the association's conference this week. The IASA Insurance Industry Collegiate Scholarship Program awards $2,000 scholarships to students in their second or third year of college who are studying an insurance-related field at a four-year college or university. The IASA scholarship committee is co-chaired by Wendy Gibson, vice president of business development for Oracle Insurance. Gibson, a long time IASA volunteer, is completing her second term on IASA's volunteer management team as vice president of industry relations. Helen Pitts is senior product marketing manager for Oracle Insurance.

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Ask How-To Geek: Diagnosing DSL Hang Ups, Extracting Media from PowerPoint, Restricting IE to a Single Web Page

- by Jason Fitzpatrick

This week we take a look at flaky DSL connections, extracting media from PowerPoint presentations, and how to lock down IE to a single website without any additional software or network configuration hacking necessary. Once a week we dip into our reader mailbag and help readers solve their problems, sharing the useful solutions with you in the process. Read on to see our fixes for this week’s reader dilemmas. Latest Features How-To Geek ETC How to Get Amazing Color from Photos in Photoshop, GIMP, and Paint.NET Learn To Adjust Contrast Like a Pro in Photoshop, GIMP, and Paint.NET Have You Ever Wondered How Your Operating System Got Its Name? Should You Delete Windows 7 Service Pack Backup Files to Save Space? What Can Super Mario Teach Us About Graphics Technology? Windows 7 Service Pack 1 is Released: But Should You Install It? Page Zipper Unpacks Multi-Page Articles for Single-Page Display Minty Bug: Build an FM Bug Inside a Mint Container Get the MakeUseOf eBook Guide to Hacker Proofing Your PC Sync Your Windows Computer with Your Ubuntu One Account [Desktop Client] Awesome 10 Meter Curved Touchscreen at the University of Groningen [Video] TV Antenna Helper Makes HDTV Antenna Calibration a Snap

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SRs @ Oracle: How do I License Thee?

- by [email protected]

With the release of the new Sun Ray product last week comes the advent of a different software licensing model. Where Sun had initially taken the approach of '1 desktop device = one license', we later changed things to be '1 concurrent connection to the server software = one license', and while there were ways to tell how many connections there were at a time, it wasn't the easiest thing to do. And, when should you measure concurrency? At your busiest time, of course... but when might that be? 9:00 Monday morning this week might yield a different result than 9:00 Monday morning last week.In the acquisition of this desktop virtualization product suite Oracle has changed things to be, in typical Oracle fashion, simpler. There are now two choices for customers around licensing: Named User licenses and Per Device licenses.Here's how they work, and some examples:The Rules1) A Sun Ray device, and PC running the Desktop Access Client (DAC), are both considered unique devices.OR, 2) Any user running a session on either a Sun Ray or an DAC is still just one user.So, you have a choice of path to go down.Some Examples:Here are 6 use cases I can think of right now that will help you choose the Oracle server software licensing model that is right for your business:Case 1If I have 100 Sun Rays for 100 users, and 20 of them use DAC at home that is 100 user licenses.If I have 100 Sun Rays for 100 users, and 20 of them use DAC at home that is 120 device licenses.Two cases using the same metrics - different licensing models and therefore different results.Case 2If I have 100 Sun Rays for 200 users, and 20 of them use DAC at home that is 200 user licenses.If I have 100 Sun Rays for 200 users, and 20 of them use DAC at home that is 120 device licenses.Same metrics - very different results.Case 3If I have 100 Sun Rays for 50 users, and 20 of them use DAC at home that is 50 user licenses.If I have 100 Sun Rays for 50 users, and 20 of them use DAC at home that is 120 device licenses.Same metrics - but again - very different results.Based on the way your business operates you should be able to see which of the two licensing models is most advantageous to you.Got questions? I'll try to help.(Thanks to Brad Lackey for the clarifications!)

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SQL SERVER – Question to You – When to use Function and When to use Stored Procedure

- by pinaldave

This week has been very interesting week. I have asked few questions to users and have received remarkable participation on the subject. Q1) SQL SERVER – Puzzle – SELECT * vs SELECT COUNT(*) Q2) SQL SERVER – Puzzle – Statistics are not Updated but are Created Once Keeping the same spirit up, I am asking the third question over here. Q3) When to use User Defined Function and when to use Stored Procedure in your development? Personally, I believe that they are both different things - they cannot be compared. I can say, it will be like comparing apples and oranges. Each has its own unique use. However, they can be used interchangeably at many times and in real life (i.e., production environment). I have personally seen both of these being used interchangeably many times. This is the precise reason for asking this question. When do you use Function and when do you use Stored Procedure? What are Pros and Cons of each of them when used instead of each other? If you are going to answer that ‘To avoid repeating code, you use Function’ - please think harder! Stored procedure can do the same. In SQL Server Denali, even the stored procedure can return the result just like Function in SELECT statement; so if you are going to answer with ‘Function can be used in SELECT, whereas Stored Procedure cannot be used’ - again think harder! (link). Now, what do you say? I will post the answers of all the three questions with due credit next week. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, Readers Question, SQL, SQL Authority, SQL Function, SQL Puzzle, SQL Query, SQL Server, SQL Stored Procedure, SQL Tips and Tricks, SQLServer, T SQL, Technology

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Know more about Cache Buffer Handle

- by Liu Maclean(???)

??????«latch free:cache buffer handles???SQL????»?????cache buffer handle latch?????,?????????: “?????pin?buffer header???????buffer handle,??buffer handle?????????cache buffer handles?,??????cache buffer handles??????,???????cache???buffer handles,?????(reserved set)?????????????_db_handles_cached(???5)???,?????????????????SQL??????????????????????,????pin??????,????????handle,?????????5?cached buffer handles???handle????????????????,Oracle?????????????????pin?”????“?buffer,????????????????handle???db_block_buffers/processes,????_cursor_db_buffers_pinned???????cache buffer handles?????,??????,????????????SQL,????cache?buffer handles?????????,??????????????,???????????/?????” ????T.ASKMACLEAN.COM????,??????cache Buffer handle?????: cache buffer handle ??: ------------------------------ | Buffer state object | ------------------------------ | Place to hang the buffer | ------------------------------ | Consistent Get? | ------------------------------ | Proc Owning SO | ------------------------------ | Flags(RIR) | ------------------------------ ???? cache buffer handle SO: 70000046fdfe530, type: 24, owner: 70000041b018630, flag: INIT/-/-/0×00(buffer) (CR) PR: 70000048e92d148 FLG: 0×500000lock rls: 0, class bit: 0kcbbfbp: [BH: 7000001c7f069b0, LINK: 70000046fdfe570]where: kdswh02: kdsgrp, why: 0BH (7000001c7f069b0) file#: 12 rdba: 0×03061612 (12/398866) class: 1 ba: 7000001c70ee000set: 75 blksize: 8192 bsi: 0 set-flg: 0 pwbcnt: 0dbwrid: 2 obj: 66209 objn: 48710 tsn: 6 afn: 12hash: [700000485f12138,700000485f12138] lru: [70000025af67790,700000132f69ee0]lru-flags: hot_bufferckptq: [NULL] fileq: [NULL] objq: [700000114f5dd10,70000028bf5d620]use: [70000046fdfe570,70000046fdfe570] wait: [NULL]st: SCURRENT md: SHR tch: 0flags: affinity_lockLRBA: [0x0.0.0] HSCN: [0xffff.ffffffff] HSUB: [65535]where: kdswh02: kdsgrp, why: 0 # Example:# (buffer) (CR) PR: 37290 FLG: 0# kcbbfbp : [BH: befd8, LINK: 7836c] (WAITING) Buffer handle (X$KCBBF) kernel cache, buffer buffer_handles Query x$kcbbf – lists all the buffer handles ???? _db_handles System-wide simultaneous buffer operations ,no of buffer handles_db_handles_cached Buffer handles cached each process , no of processes default 5_cursor_db_buffers_pinned additional number of buffers a cursor can pin at once_session_kept_cursor_pins Number of cursors pins to keep in a session When a buffer is pinned it is attached to buffer state object. ??? ???????? cache buffer handles latch ? buffer pin???: SESSION A : SQL> select * from v$version; BANNER ---------------------------------------------------------------- Oracle Database 10g Enterprise Edition Release 10.2.0.5.0 - 64bi PL/SQL Release 10.2.0.5.0 - Production CORE 10.2.0.5.0 Production TNS for Linux: Version 10.2.0.5.0 - Production NLSRTL Version 10.2.0.5.0 - Production SQL> create table test_cbc_handle(t1 int); Table created. SQL> insert into test_cbc_handle values(1); 1 row created. SQL> commit; Commit complete. SQL> select rowid from test_cbc_handle; ROWID ------------------ AAANO6AABAAAQZSAAA SQL> select * from test_cbc_handle where rowid='AAANO6AABAAAQZSAAA'; T1 ---------- 1 SQL> select addr,name from v$latch_parent where name='cache buffer handles'; ADDR NAME ---------------- -------------------------------------------------- 00000000600140A8 cache buffer handles SQL> select to_number('00000000600140A8','xxxxxxxxxxxxxxxxxxxx') from dual; TO_NUMBER('00000000600140A8','XXXXXXXXXXXXXXXXXXXX') ---------------------------------------------------- 1610694824 ??cache buffer handles????parent latch ??? child latch ???SESSION A hold ??????cache buffer handles parent latch ???? oradebug call kslgetl ??, kslgetl?oracle??get latch??? SQL> oradebug setmypid; Statement processed. SQL> oradebug call kslgetl 1610694824 1; Function returned 1 ?????SESSION B ???: SQL> select * from v$latchholder; PID SID LADDR NAME GETS ---------- ---------- ---------------- ---------------------------------------------------------------- ---------- 15 141 00000000600140A8 cache buffer handles 119 cache buffer handles latch ???session A hold??,????????acquire cache buffer handle latch SQL> select * from test_cbc_handle where rowid='AAANO6AABAAAQZSAAA'; T1 ---------- 1 ?????Server Process?????? read buffer, ????????"_db_handles_cached", ??process?cache 5? cache buffer handle ??"_db_handles_cached"=0,?process????5????cache buffer handle , ???? process ???pin buffer,???hold cache buffer handle latch??????cache buffer handle SQL> alter system set "_db_handles_cached"=0 scope=spfile; System altered. ????? shutdown immediate; startup; session A: SQL> oradebug setmypid; Statement processed. SQL> oradebug call kslgetl 1610694824 1; Function returned 1 session B: select * from test_cbc_handle where rowid='AAANO6AABAAAQZSAAA'; session B hang!! WHY? SQL> oradebug setmypid; Statement processed. SQL> oradebug dump systemstate 266; Statement processed. SO: 0x11b30b7b0, type: 2, owner: (nil), flag: INIT/-/-/0x00 (process) Oracle pid=22, calls cur/top: (nil)/0x11b453c38, flag: (0) - int error: 0, call error: 0, sess error: 0, txn error 0 (post info) last post received: 0 0 0 last post received-location: No post last process to post me: none last post sent: 0 0 0 last post sent-location: No post last process posted by me: none (latch info) wait_event=0 bits=8 holding (efd=4) 600140a8 cache buffer handles level=3 SO: 0x11b305810, type: 2, owner: (nil), flag: INIT/-/-/0x00 (process) Oracle pid=10, calls cur/top: 0x11b455ac0/0x11b450a58, flag: (0) - int error: 0, call error: 0, sess error: 0, txn error 0 (post info) last post received: 0 0 0 last post received-location: No post last process to post me: none last post sent: 0 0 0 last post sent-location: No post last process posted by me: none (latch info) wait_event=0 bits=2 Location from where call was made: kcbzgs: waiting for 600140a8 cache buffer handles level=3 FBD93353:000019F0 10 162 10005 1 KSL WAIT BEG [latch: cache buffer handles] 1610694824/0x600140a8 125/0x7d 0/0x0 FF936584:00002761 10 144 10005 1 KSL WAIT BEG [latch: cache buffer handles] 1610694824/0x600140a8 125/0x7d 0/0x0 PID=22 holding ??cache buffer handles latch PID=10 ?? cache buffer handles latch, ????"_db_handles_cached"=0 ?? process??????cache buffer handles ??systemstate???? kcbbfbp cache buffer handle??, ?? "_db_handles_cached"=0 ? cache buffer handles latch?hold ?? ????cache buffer handles latch , ??? buffer?pin?????????? session A exit session B: SQL> select * from v$latchholder; no rows selected SQL> insert into test_cbc_handle values(2); 1 row created. SQL> commit; Commit complete. SQL> SQL> select t1,rowid from test_cbc_handle; T1 ROWID ---------- ------------------ 1 AAANPAAABAAAQZSAAA 2 AAANPAAABAAAQZSAAB SQL> select spid,pid from v$process where addr = ( select paddr from v$session where sid=(select distinct sid from v$mystat)); SPID PID ------------ ---------- 19251 10 ? GDB ? SPID=19215 ?debug , ?? kcbrls ????breakpoint ??? ????release buffer [oracle@vrh8 ~]$ gdb $ORACLE_HOME/bin/oracle 19251 GNU gdb (GDB) Red Hat Enterprise Linux (7.0.1-37.el5) Copyright (C) 2009 Free Software Foundation, Inc. License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html> This is free software: you are free to change and redistribute it. There is NO WARRANTY, to the extent permitted by law. Type "show copying" and "show warranty" for details. This GDB was configured as "x86_64-redhat-linux-gnu". For bug reporting instructions, please see: <http://www.gnu.org/software/gdb/bugs/>... Reading symbols from /s01/oracle/product/10.2.0.5/db_1/bin/oracle...(no debugging symbols found)...done. Attaching to program: /s01/oracle/product/10.2.0.5/db_1/bin/oracle, process 19251 Reading symbols from /s01/oracle/product/10.2.0.5/db_1/lib/libskgxp10.so...(no debugging symbols found)...done. Loaded symbols for /s01/oracle/product/10.2.0.5/db_1/lib/libskgxp10.so Reading symbols from /s01/oracle/product/10.2.0.5/db_1/lib/libhasgen10.so...(no debugging symbols found)...done. Loaded symbols for /s01/oracle/product/10.2.0.5/db_1/lib/libhasgen10.so Reading symbols from /s01/oracle/product/10.2.0.5/db_1/lib/libskgxn2.so...(no debugging symbols found)...done. Loaded symbols for /s01/oracle/product/10.2.0.5/db_1/lib/libskgxn2.so Reading symbols from /s01/oracle/product/10.2.0.5/db_1/lib/libocr10.so...(no debugging symbols found)...done. Loaded symbols for /s01/oracle/product/10.2.0.5/db_1/lib/libocr10.so Reading symbols from /s01/oracle/product/10.2.0.5/db_1/lib/libocrb10.so...(no debugging symbols found)...done. Loaded symbols for /s01/oracle/product/10.2.0.5/db_1/lib/libocrb10.so Reading symbols from /s01/oracle/product/10.2.0.5/db_1/lib/libocrutl10.so...(no debugging symbols found)...done. Loaded symbols for /s01/oracle/product/10.2.0.5/db_1/lib/libocrutl10.so Reading symbols from /s01/oracle/product/10.2.0.5/db_1/lib/libjox10.so...(no debugging symbols found)...done. Loaded symbols for /s01/oracle/product/10.2.0.5/db_1/lib/libjox10.so Reading symbols from /s01/oracle/product/10.2.0.5/db_1/lib/libclsra10.so...(no debugging symbols found)...done. Loaded symbols for /s01/oracle/product/10.2.0.5/db_1/lib/libclsra10.so Reading symbols from /s01/oracle/product/10.2.0.5/db_1/lib/libdbcfg10.so...(no debugging symbols found)...done. Loaded symbols for /s01/oracle/product/10.2.0.5/db_1/lib/libdbcfg10.so Reading symbols from /s01/oracle/product/10.2.0.5/db_1/lib/libnnz10.so...(no debugging symbols found)...done. Loaded symbols for /s01/oracle/product/10.2.0.5/db_1/lib/libnnz10.so Reading symbols from /usr/lib64/libaio.so.1...(no debugging symbols found)...done. Loaded symbols for /usr/lib64/libaio.so.1 Reading symbols from /lib64/libdl.so.2...(no debugging symbols found)...done. Loaded symbols for /lib64/libdl.so.2 Reading symbols from /lib64/libm.so.6...(no debugging symbols found)...done. Loaded symbols for /lib64/libm.so.6 Reading symbols from /lib64/libpthread.so.0...(no debugging symbols found)...done. [Thread debugging using libthread_db enabled] Loaded symbols for /lib64/libpthread.so.0 Reading symbols from /lib64/libnsl.so.1...(no debugging symbols found)...done. Loaded symbols for /lib64/libnsl.so.1 Reading symbols from /lib64/libc.so.6...(no debugging symbols found)...done. Loaded symbols for /lib64/libc.so.6 Reading symbols from /lib64/ld-linux-x86-64.so.2...(no debugging symbols found)...done. Loaded symbols for /lib64/ld-linux-x86-64.so.2 Reading symbols from /lib64/libnss_files.so.2...(no debugging symbols found)...done. Loaded symbols for /lib64/libnss_files.so.2 0x00000035c000d940 in __read_nocancel () from /lib64/libpthread.so.0 (gdb) break kcbrls Breakpoint 1 at 0x10e5d24 session B: select * from test_cbc_handle where rowid='AAANPAAABAAAQZSAAA'; select hang !! GDB (gdb) c Continuing. Breakpoint 1, 0x00000000010e5d24 in kcbrls () (gdb) bt #0 0x00000000010e5d24 in kcbrls () #1 0x0000000002e87d25 in qertbFetchByUserRowID () #2 0x00000000030c62b8 in opifch2 () #3 0x00000000032327f0 in kpoal8 () #4 0x00000000013b7c10 in opiodr () #5 0x0000000003c3c9da in ttcpip () #6 0x00000000013b3144 in opitsk () #7 0x00000000013b60ec in opiino () #8 0x00000000013b7c10 in opiodr () #9 0x00000000013a92f8 in opidrv () #10 0x0000000001fa3936 in sou2o () #11 0x000000000072d40b in opimai_real () #12 0x000000000072d35c in main () SQL> oradebug setmypid; Statement processed. SQL> oradebug dump systemstate 266; Statement processed. ?????? kcbbfbp buffer cache handle ? SO state object ? BH BUFFER HEADER link??? ---------------------------------------- SO: 0x11b452348, type: 3, owner: 0x11b305810, flag: INIT/-/-/0x00 (call) sess: cur 11b41bd18, rec 0, usr 11b41bd18; depth: 0 ---------------------------------------- SO: 0x1182dc750, type: 24, owner: 0x11b452348, flag: INIT/-/-/0x00 (buffer) (CR) PR: 0x11b305810 FLG: 0x108000 class bit: (nil) kcbbfbp: [BH: 0xf2fc69f8, LINK: 0x1182dc790] where: kdswh05: kdsgrp, why: 0 BH (0xf2fc69f8) file#: 1 rdba: 0x00410652 (1/67154) class: 1 ba: 0xf297c000 set: 3 blksize: 8192 bsi: 0 set-flg: 2 pwbcnt: 272 dbwrid: 0 obj: 54208 objn: 54202 tsn: 0 afn: 1 hash: [f2fc47f8,1181f3038] lru: [f2fc6b88,f2fc6968] obj-flags: object_ckpt_list ckptq: [1182ecf38,1182ecf38] fileq: [1182ecf58,1182ecf58] objq: [108712a28,108712a28] use: [1182dc790,1182dc790] wait: [NULL] st: XCURRENT md: SHR tch: 12 flags: buffer_dirty gotten_in_current_mode block_written_once redo_since_read LRBA: [0xc7.73b.0] HSCN: [0x0.1cbe52] HSUB: [1] Using State Objects ---------------------------------------- SO: 0x1182dc750, type: 24, owner: 0x11b452348, flag: INIT/-/-/0x00 (buffer) (CR) PR: 0x11b305810 FLG: 0x108000 class bit: (nil) kcbbfbp: [BH: 0xf2fc69f8, LINK: 0x1182dc790] where: kdswh05: kdsgrp, why: 0 buffer tsn: 0 rdba: 0x00410652 (1/67154) scn: 0x0000.001cbe52 seq: 0x01 flg: 0x02 tail: 0xbe520601 frmt: 0x02 chkval: 0x0000 type: 0x06=trans data tab 0, row 0, @0x1f9a tl: 6 fb: --H-FL-- lb: 0x0 cc: 1 col 0: [ 2] c1 02 tab 0, row 1, @0x1f94 tl: 6 fb: --H-FL-- lb: 0x2 cc: 1 col 0: [ 2] c1 15 end_of_block_dump (buffer) (CR) PR: 0x11b305810 FLG: 0x108000 st: XCURRENT md: SHR tch: 12 ? buffer header?status= XCURRENT mode=KCBMSHARE KCBMSHR current share ????? x$kcbbf ????? cache buffer handle SQL> select distinct KCBBPBH from x$kcbbf ; KCBBPBH ---------------- 00 00000000F2FC69F8 ==>0xf2fc69f8 SQL> select * from x$kcbbf where kcbbpbh='00000000F2FC69F8'; ADDR INDX INST_ID KCBBFSO_TYP KCBBFSO_FLG KCBBFSO_OWN ---------------- ---------- ---------- ----------- ----------- ---------------- KCBBFFLG KCBBFCR KCBBFCM KCBBFMBR KCBBPBH ---------- ---------- ---------- ---------------- ---------------- KCBBPBF X0KCBBPBH X0KCBBPBF X1KCBBPBH ---------------- ---------------- ---------------- ---------------- X1KCBBPBF KCBBFBH KCBBFWHR KCBBFWHY ---------------- ---------------- ---------- ---------- 00000001182DC750 748 1 24 1 000000011B452348 1081344 1 0 00 00000000F2FC69F8 00000001182DC750 00 00000001182DC750 00 00000001182DC7F8 00 583 0 SQL> desc x$kcbbf; Name Null? Type ----------------------------------------- -------- ---------------------------- ADDR RAW(8) INDX NUMBER INST_ID NUMBER KCBBFSO_TYP NUMBER KCBBFSO_FLG NUMBER KCBBFSO_OWN RAW(8) KCBBFFLG NUMBER KCBBFCR NUMBER KCBBFCM NUMBER KCBBFMBR RAW(8) KCBBPBH RAW(8) KCBBPBF RAW(8) X0KCBBPBH RAW(8) X0KCBBPBF RAW(8) X1KCBBPBH RAW(8) X1KCBBPBF RAW(8) KCBBFBH RAW(8) KCBBFWHR NUMBER KCBBFWHY NUMBER gdb ?? ?process??????kcbrls release buffer? ???cache buffer handle??? SQL> select distinct KCBBPBH from x$kcbbf ; KCBBPBH ---------------- 00

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My Expert F# Book has now arrived!

- by MarkPearl

So it has finally arrived from Amazon. Expert F# by Don Syme, Adam Granicz & Antonio Cisternino. I got a note from the post office yesterday that I needed to collect a package from their offices. After paying a 10% customs fee (that I wasn’t expecting) I had my new Yellow & Black F# Book… it’s so shinny. Trust my luck though – I have a few university assignments due this week as well as a crazy week of work so it has been sitting on my desk for a day and I haven’t managed to get into it. Eventually I managed to take a few minutes this evening to page through it and it looks really good. I can’t wait! So my goal this week is to cover Chapter 2 (by the end of the weekend) and put the appropriate posts up. F# is slowly working on me but I am keen to get a deeper understanding of the language which I am hoping this book will help me achieve.

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Google+ Platform Office Hours for April 11, 2012: Recent Activity jQuery Plugin

Google+ Platform Office Hours for April 11, 2012: Recent Activity jQuery Plugin Here is the edited video from last week's Google+ platform office hours. Discuss this video on Google+: goo.gl This week we spent the first half of the show live coding a jQuery plugin that fetched recent public activity from a Google+ profile or page for inclusion on your website. Get the source code: goo.gl 1:15 - A demo of the implemented plugin 2:04 - The design of the plugin 2:57 - The coding begins! - Use the Google+ badge config tool to discover your userId (goo.gl or follow these instructions for pages: goo.gl Q&A 17:10 - Is there any kind of beta group that I can join for Google+? - Sign up for for the publisher preview group - goo.gl 19:03 - When will the API be available? 20:11 - When will there be more moderation tools for hangouts? 21:36 - How do I get Hangouts on air? 22:18 - An update on last week's report of Google Analytics social actions not agreeing with the +1 button count 25:53 - How do I join the hangout for these office hours? 26:44 - Using the activities search API is there any way to only see new activities? 28:18 - An announcement about our future office hours schedule From: GoogleDevelopers Views: 3387 47 ratings Time: 29:29 More in Science & Technology

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Oracle Linux Partner Pavilion Spotlight

- by Ted Davis

With the first day of Oracle OpenWorld starting in less than a week, we wanted to showcase some of our premier partners exhibiting in the Oracle Linux Partner Pavilion ( Booth #1033) this year. We have Independent Hardware Vendors, Independent Software Vendors and Systems Integrators that show the breadth of support in the Oracle Linux and Oracle VM ecosystem. We'll be highlighting partners all week so feel free to come back check us out. Centrify delivers integrated software and cloud-based solutions that centrally control, secure and audit access to cross-platform systems, mobile devices and applications by leveraging the infrastructure organizations already own. From the data center and into the cloud, more than 4,500 organizations, including 40 percent of the Fortune 50 and more than 60 Federal agencies, rely on Centrify's identity consolidation and privilege management solutions to reduce IT expenses, strengthen security and meet compliance requirements. Visit Centrify at Oracle OpenWorld 2102 for a look at Centrify Suite and see how you can streamline security management on Oracle Linux. Unify identities across the enterprise and remove the pain and security issues associated with managing local user accounts by leveraging Active Directory Implement a least-privilege security model with flexible, role-based controls that protect privileged operations while still granting users the privileges they need to perform their job Get a central, global view of audited user sessions across your Oracle Linux environment "Data Intensity's cloud infrastructure leverages Oracle VM and Oracle Linux to provide highly available enterprise application management solutions. Engineers will be available to answer questions about and demonstrate the technology, including management tools, configuration do's and don'ts, high availability, live migration, integrating the technology with Oracle software, and how the integrated support process works." Mellanox’s end-to-end InfiniBand and Ethernet server and storage interconnect solutions deliver the highest performance, efficiency and scalability for enterprise, high-performance cloud and web 2.0 applications. Mellanox’s interconnect solutions accelerate Oracle RAC query throughput performance to reach 50Gb/s compared to TCP/IP based competing solutions that cap off at less than 12Gb/s. Mellanox solutions help Oracle’s Exadata to deliver 10X performance boost at 50% Hardware cost making it the world’s leading database appliance. Thanks for reviewing today's Partner spotlight. We will highlight new partners each day this week leading up to Oracle OpenWorld.

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