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  • C# LINQ XML Query with duplicate element names that have attributes

    - by ncain187
    <Party id="Party_1"> <PartyTypeCode tc="1">Person</PartyTypeCode> <FullName>John Doe</FullName> <GovtID>123456789</GovtID> <GovtIDTC tc="1">Social Security Number US</GovtIDTC> <ResidenceState tc="35">New Jersey</ResidenceState> <Person> <FirstName>Frank</FirstName> <MiddleName>Roberts</MiddleName> <LastName>Madison</LastName> <Prefix>Dr.</Prefix> <Suffix>III</Suffix> <Gender tc="1">Male</Gender> <BirthDate>1974-01-01</BirthDate> <Age>35</Age> <Citizenship tc="1">United States of America</Citizenship> </Person> <Address> <AddressTypeCode tc="26">Bill Mailing</AddressTypeCode> <Line1>2400 Meadow Lane</Line1> <Line2></Line2> <Line3></Line3> <Line4></Line4> <City>Somerset</City> <AddressStateTC tc="35">New Jersey</AddressStateTC> <Zip>07457</Zip> <AddressCountryTC tc="1">United States of America</AddressCountryTC> </Address> </Party> <!-- *********************** --> <!-- Insured Information --> <!-- *********************** --> <Party id="Party_2"> <PartyTypeCode tc="1">Person</PartyTypeCode> <FullName>Dollie Robert Madison</FullName> <GovtID>123956239</GovtID> <GovtIDTC tc="1">Social Security Number US</GovtIDTC> <Person> <FirstName>Dollie</FirstName> <MiddleName>R</MiddleName> <LastName>Madison</LastName> <Suffix>III</Suffix> <Gender tc="2">Female</Gender> <BirthDate>1996-10-12</BirthDate> <Citizenship tc="1">United States of America</Citizenship> </Person> <!-- Insured Address --> <Address> <AddressTypeCode tc="26">Bill Mailing</AddressTypeCode> <Line1>2400 Meadow Lane</Line1> <City>Somerset</City> <AddressStateTC tc="35">New Jersey</AddressStateTC> <Zip>07457</Zip> <AddressCountryTC tc="1">United States of America</AddressCountryTC> </Address> <Risk> <!-- Disability Begin Effective Date --> <DisabilityEffectiveStartDate>2006-01-01</DisabilityEffectiveStartDate> <!-- Disability End Effective Date --> <DisabilityEffectiveStopDate>2008-01-01</DisabilityEffectiveStopDate> </Risk> </Party> <!-- ******************************* --> <!-- Company Information --> <!-- ****************************** --> <Party id="Party_3"> <PartyTypeCode tc="2">Organization</PartyTypeCode> <Organization> <DTCCMemberCode>1234</DTCCMemberCode> </Organization> <Carrier> <CarrierCode>105</CarrierCode> </Carrier> </Party> Here is my code which doesn't work because party 3 doesn't contain FullName, I know that partyelements contains 3 parties if I only return the name attribute. Is there a way to loop through each tag seperate? var partyElements = from party in xmlDoc.Descendants("Party") select new { Name = party.Attribute("id").Value, PartyTypeCode = party.Element("PartyTypeCode").Value, FullName = party.Element("FullName").Value, GovtID = party.Element("GovtID").Value, };

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  • How to create Query syntax for multiple DataTable for implementing IN operator of Sql Server

    - by Shantanu Gupta
    I have fetched 3-4 tables by executing my stored procedure. Now they resides on my dataset. I have to maintain this dataset for multiple forms and I am not doing any DML operation on this dataset. Now this dataset contains 4 tables out of which i have to fetch some records to display data. Data stored in tables are in form of one to many relationship. i.e. In case of transactions. N records per record. Then these N records are further mapped to M records of 3rd table. Table 1 MAP_ID GUEST_ID DEPARTMENT_ID PARENT_ID PREFERENCE_ID -------------------- -------------------- -------------------- -------------------- -------------------- 19 61 1 1 5 14 61 1 5 15 15 61 2 4 10 18 61 2 13 23 17 61 2 20 26 16 61 40 40 41 20 62 1 5 14 21 62 1 5 15 22 62 1 6 16 24 62 2 3 4 23 62 2 4 9 27 62 2 13 23 25 62 2 20 24 26 62 2 20 25 28 63 1 1 5 29 63 1 1 8 34 63 1 5 15 30 63 2 4 10 33 63 2 4 11 31 63 2 13 23 32 63 40 40 41 35 65 1 NULL 1 36 65 1 NULL 1 38 68 2 13 22 37 68 2 20 25 39 68 2 23 27 40 92 1 NULL 1 Table 2 Department_ID Department_Name Parent_Id Parent_Name -------------------- ----------------------- --------------- ---------------------------------------------------------------------------------- 1 Food 1, 5, 6 Food, North Indian, South Indian 2 Lodging 3, 4, 13, 20, 23 Room, Floor, Non Air Conditioned, With Balcony, Without Balcony 40 New 40 SubNew TABLE 3 Parent_Id Parent_Name Preference_ID Preference_Name -------------------- ----------------------------------------------- ----------------------- ------------------- NULL NULL NULL NULL 1 Food 5, 8 North Indian, Italian 3 Room 4 Floor 4 Floor 9, 10, 11 First, Second, Third 5 North Indian 14, 15 X, Y 6 South Indian 16 Dosa 13 Non Air Conditioned 22, 23 With Balcony, Without Balcony 20 With Balcony 24, 25, 26 Mountain View, Ocean View, Garden View 23 Without Balcony 27 Mountain View 40 New 41 SubNew I have these 3 tables that are related in some fashion like this. Table 1 will be the master for these 2 tables i.e. table 2 and table 3. I need to query on them as SELECT Department_Id, Department_Name, Parent_Name FROM Table2 WHERE Department_Id in ( SELECT Department_Id FROM Table1 WHERE guest_id=65 ) SELECT Parent_Id, Parent_Name, Preference_Name FROM Table3 WHERE PARENT_ID in ( SELECT parent_id FROM Table1 WHERE guest_id=65 ) Now I need to use these queries on DataTables. So I am using Query Syntax for this and reached up to this point. var dept_list= from dept in DtMapGuestDepartment.AsEnumerable() where dept.Field<long>("PK_GUEST_ID")==long.Parse(63) select dept; This should give me the list of all departments that has guest id =63 Now I want to select all departments_name and parent_name from Table 2 where guest_id=63 i.e. departments that i fetched above. This same case will be followed for Table3. Please suggest how to do this. Thanks for keeping up patience for reading my question.

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  • C++ string array binary search

    - by Jose Vega
    string Haystack[] = { "Alabama", "Alaska", "American Samoa", "Arizona", "Arkansas", "California", "Colorado", "Connecticut", "Delaware", "District of Columbia", "Florida", "Georgia", "Guam", "Hawaii", "Idaho", "Illinois", "Indiana", "Iowa", "Kansas", "Kentucky", "Louisiana", "Maine", "Maryland", "Massachusetts", "Michigan", "Minnesota", "Mississippi", "Missouri", "Montana", "Nebraska", "Nevada", "New Hampshire", "New Jersey", "New Mexico", "New York", "North Carolina", "North Dakota", "Northern Mariana Islands", "Ohio", "Oklahoma", "Oregon", "Pennsylvania", "Puerto Rico", "Rhode Island", "South Carolina", "South Dakota", "Tennessee", "Texas", "US Virgin Islands", "Utah", "Vermont", "Virginia", "Washington", "West Virginia", "Wisconsin", "Wyoming"}; string Needle = "Virginia"; if(std::binary_search(Haystack, Haystack+56, Needle)) cout<<"Found"; If I also wanted to find the location of the needle in the string array, is there an "easy" way to find out?

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  • facing outsourced wages, can i still eat and survive as a computing science major ?

    - by wefwgeweg
    offshore outsourced programmers charge fraction of what costs a North American developer. should I still pursue my major as computing science ? Why would companies spend more on North American/local developers where they can get the same quality if not better job done offshore ? I am just concerned for the development labor market, the free market wants the lowest cost provider. not just programming but many high skilled labor such as engineering, scientists, artists and etc. perhaps i should become a lawyer ?

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  • problem in adding image to JFrame

    - by firestruq
    Hi, I'm having problems in adding a picture into JFrame, something is missing probebly or written wrong. here are the classes: main class: public class Tester { public static void main(String args[]) { BorderLayoutFrame borderLayoutFrame = new BorderLayoutFrame(); borderLayoutFrame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); borderLayoutFrame.setSize(600,600); borderLayoutFrame.setVisible(true); } } public class BorderLayoutFrame extends JFrame implements ActionListener { private JButton buttons[]; // array of buttons to hide portions private final String names[] = { "North", "South", "East", "West", "Center" }; private BorderLayout layout; // borderlayout object private PicPanel picture = new PicPanel(); // set up GUI and event handling public BorderLayoutFrame() { super( "Philosofic Problem" ); layout = new BorderLayout( 5, 5 ); // 5 pixel gaps setLayout( layout ); // set frame layout buttons = new JButton[ names.length ]; // set size of array // create JButtons and register listeners for them for ( int count = 0; count < names.length; count++ ) { buttons[ count ] = new JButton( names[ count ] ); buttons[ count ].addActionListener( this ); } add( buttons[ 0 ], BorderLayout.NORTH ); // add button to north add( buttons[ 1 ], BorderLayout.SOUTH ); // add button to south add( buttons[ 2 ], BorderLayout.EAST ); // add button to east add( buttons[ 3 ], BorderLayout.WEST ); // add button to west add( picture, BorderLayout.CENTER ); // add button to center } // handle button events public void actionPerformed( ActionEvent event ) { } } I'v tried to add the image into the center of layout. here is the image class: public class PicPanel extends JPanel { Image img; private int width = 0; private int height = 0; public PicPanel() { super(); img = Toolkit.getDefaultToolkit().getImage("table.jpg"); } public void paintComponent(Graphics g) { super.paintComponents(g); if ((width <= 0) || (height <= 0)) { width = img.getWidth(this); height = img.getHeight(this); } g.drawImage(img,0,0,width,height,this); } } Please your help, what is the problem? thanks BTW: i'm using eclipse, which directory the image suppose to be in?

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  • What Conferences would you recommend for a UI / Frontend Web Developer in the next 6 months?

    - by rsturim
    Hello, I'm looking for a strong conference(s) to attend in the next 6 months. I may be able to attend one or two. I'm looking for something surrounding Frontend Web Development -- web standards, CSS3, html5, javascript, UX, and usability are strong interests of mine. I'm also starting to consider diving deep into designing for Mobile devices. I've discovered these 2 conferences so far -- they look very good -- but am I missing anything HUGE and/or obvious? An Event Apart - Wash DC (http://aneventapart.com/2010/dc/) Web Directions North - Altanta (http://north.webdirections.org/) Thoughts?

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  • Partitioning requests in code among several servers

    - by Jacques René Mesrine
    I have several forum servers (what they are is irrelevant) which stores posts from users and I want to be able to partition requests among these servers. I'm currently leaning towards partitioning them by geographic location. To improve the locality of data, users will be separated into regions e.g. North America, South America and so on. Is there any design pattern on how to implement the function that maps the partioning property to the server, so that this piece of code has high availability and would not become a single point of failure ? f( Region ) -> Server IP

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  • Table not Echoing out if another Table has a Zero value

    - by John
    Hello, The table below with mysql_query($sqlStr3) (the one with the word "Joined" in its row) does not echo if the result associated with mysql_query($sqlStr1) has a value of zero. This happens even if mysql_query($sqlStr3) returns a result. In other words, if a given loginid has an entry in the table "login", but not one in the table "submission", then the table associated with mysql_query($sqlStr3) does not echo. I don't understand why the "submission" table would have any effect on mysql_query($sqlStr3), since the $sqlStr3 only deals with another table, called "login", as seen below. Any ideas why this is happening? Thanks in advance, John W. <?php echo '<div class="profilename">User Profile for </div>'; echo '<div class="profilename2">'.$profile.'</div>'; $tzFrom = new DateTimeZone('America/New_York'); $tzTo = new DateTimeZone('America/Phoenix'); $profile = mysql_real_escape_string($_GET['profile']); $sqlStr = "SELECT l.username, l.loginid, s.loginid, s.submissionid, s.title, s.url, s.datesubmitted, s.displayurl FROM submission AS s INNER JOIN login AS l ON s.loginid = l.loginid WHERE l.username = '$profile' ORDER BY s.datesubmitted DESC"; $result = mysql_query($sqlStr); $arr = array(); echo "<table class=\"samplesrec1\">"; while ($row = mysql_fetch_array($result)) { $dt = new DateTime($row["datesubmitted"], $tzFrom); $dt->setTimezone($tzTo); echo '<tr>'; echo '<td class="sitename3">'.$dt->format('F j, Y &\nb\sp &\nb\sp g:i a').'</a></td>'; echo '<td class="sitename1"><a href="http://www.'.$row["url"].'">'.$row["title"].'</a></td>'; echo '</tr>'; } echo "</table>"; $sqlStr1 = "SELECT l.username, l.loginid, s.loginid, s.submissionid, s.title, s.url, s.datesubmitted, s.displayurl, l.created, count(s.submissionid) countSubmissions FROM submission AS s INNER JOIN login AS l ON s.loginid = l.loginid WHERE l.username = '$profile'"; $result1 = mysql_query($sqlStr1); $arr1 = array(); echo "<table class=\"samplesrec2\">"; while ($row1 = mysql_fetch_array($result1)) { echo '<tr>'; echo '<td class="sitename5">Submissions: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$row1["countSubmissions"].'</td>'; echo '</tr>'; } echo "</table>"; $sqlStr2 = "SELECT l.username, l.loginid, c.loginid, c.commentid, c.submissionid, c.comment, c.datecommented, l.created, count(c.commentid) countComments FROM comment AS c INNER JOIN login AS l ON c.loginid = l.loginid WHERE l.username = '$profile'"; $result2 = mysql_query($sqlStr2); $arr2 = array(); echo "<table class=\"samplesrec3\">"; while ($row2 = mysql_fetch_array($result2)) { echo '<tr>'; echo '<td class="sitename5">Comments: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$row2["countComments"].'</td>'; echo '</tr>'; } echo "</table>"; $tzFrom3 = new DateTimeZone('America/New_York'); $tzTo3 = new DateTimeZone('America/Phoenix'); $sqlStr3 = "SELECT created, username FROM login WHERE username = '$profile'"; $result3 = mysql_query($sqlStr3); $arr3 = array(); echo "<table class=\"samplesrec4\">"; while ($row3 = mysql_fetch_array($result3)) { $dt3 = new DateTime($row3["created"], $tzFrom3); $dt3->setTimezone($tzTo3); echo '<tr>'; echo '<td class="sitename5">Joined: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$dt->format('F j, Y').'</td>'; echo '</tr>'; } echo "</table>"; ?> </body> </html>

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  • Finding if a path between 2 sides of a game board exists

    - by Meny
    Hi, i'm currently working on a game as an assignment for school in java. the game cuurently is designed for Console. the game is for 2 players, one attacking from north to south, and the other from west to east. the purpose of the game is to build a "bridge"/"path" between the 2 of your sides before your opponent does. for example: A B C D E F 1 _ _ X _ _ _ 1 2 O X X _ _ _ 2 3 O X O O O O 3 4 O X O _ _ _ 4 5 X X _ _ _ _ 5 6 X O _ _ _ _ 6 A B C D E F player that attacks from north to south won (path/bridge from C to A) my problem is, what algorithm would be good to check if the user have managed to create a path (will be checked at the end of each turn). you're help would be very appreciated.

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  • How can I find the nearest intersection via the Google Maps API?

    - by dusoft
    How can I find the closest intersection of the street I have coordinates of? For instance, say I have street A running from south to north that is crossed by street X on the north and by street Y on the south. Does the Google Maps API allow for finding coordinates of the nearest crossroad (either X or Y) of street A? I couldn't find it mentioned anywhere. PS: The only solution I am aware of is to guess the lowest number and the highest number of building on the street A and to draw polyline between them. I am not sure about this though.

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  • How to get global access to enum types in C#?

    - by lala
    This is probably a stupid question, but I can't seem to do it. I want to set up some enums in one class like this: public enum Direction { north, east, south, west }; Then have that enum type accessible to all classes so that some other class could for instance have: Direction dir = north; and be able to pass the enum type between classes: public void changeDirection(Direction direction) { dir = direction; } I thought that setting the enum to public would make this automatically possible, but it doesn't seem to be visible outside of the class I declared the enum in.

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  • Why does Java return a date in GMT-4.5 when choosing Co-ordinated Universal Time time zone in Window

    - by Simon Nickerson
    We have seen a strange issue on some Windows XP machines involving the "Co-ordinated Universal Time" time zone. Not all Windows XP machines seem to have it, but on those that do, the following simple Java program public class TimeTest { public static void main(String[] args) { System.out.println(java.util.TimeZone.getDefault()); System.out.println(new java.util.Date()); } } on JDK 1.6.0_06 prints: sun.util.calendar.ZoneInfo[id="America/Caracas",offset=-16200000,dstSavings=0,useDaylight=false,transitions=5,lastRule=null] Fri Nov 13 05:34:14 VET 2009 (i.e. 4 and a half hours behind GMT). I should add that I am based in London, and have never been to South America. :-) My questions are: Where does Java get this time zone from? I thought Co-ordinated Universal Time was supposed to be the new name for GMT. Why do some Windows machines have this time zone but not others?

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  • Access is refusing to run an query with linked table?

    - by Mahmoud
    Hey all i have 3 tables each as follow cash_credit Bank_Name-------in_date-------Com_Id---Amount America Bank 15/05/2010 1 200 HSBC 17/05/2010 3 500 Cheque_credit Bank_Name-----Cheque_Number-----in_date-------Com_Id---Amount America Bank 74835435-5435 15/05/2010 2 600 HSBC 41415454-2851 17/05/2010 5 100 Companies com_id----Com_Name 1 Ebay 2 Google 3 Facebook 4 Amazon Companies table is a linked table when i tried to create an query as follow SELECT cash_credit.Amount, Companies.Com_Name, cheque_credit.Amount FROM cheque_credit INNER JOIN (cash_credit INNER JOIN Companies ON cash_credit.com_id = Companies.com_id) ON cheque_credit.com_id = Companies.com_id; I get an error saying that my inner Join is incorrectly, this query was created using Access 2007 query design the error is Type mismatch in expression then i thought it might be the inner join so i tried Left Join and i get an error that this method is not used JOIN expression is not supported I am confused on where is the problem that is causing all this

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  • Codeigniter view file looping query

    - by user2505513
    Right, I'm unsure about how to code my view file to generate following query results WITHOUT compromising the principles of mvc. Query in model: SELECT * FROM events GROUP BY country, area ORDER BY country, area View: <?php if (isset($query)):?> <?php foreach ($query as $row):?> <h2><?=$row->country?></h2> <h3><?=$row->area?></h3> <?php endforeach;?> <?php endif;?> I want the results to display: England North South West - utilising the GROUP BY parameter As opposed to: England North England South England West Has anybody any advice as to how to achieve this?

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  • html source does not show all visible data

    - by every_answer_gets_a_point
    if you go here: http://whois.domaintools.com/iconplc.com and view the source why can't you see the registrant data in the HTML source? is it at all possible to get this data through the html source? this stuff is not in the html source: Registrant: ICON Clinical Research 212 Church Road North Wales, PA 19454 US Domain Name: ICONPLC.COM Administrative Contact, Technical Contact: ICON Clinical Research 212 Church Road North Wales, PA 19454 US 215-616-3359 fax: 123 123 1234 Record expires on 08-Sep-2019. Record created on 12-Dec-2007. Domain servers in listed order: UDNS1.ULTRADNS.NET UDNS2.ULTRADNS.NET

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  • How to return result set based on other rows

    - by understack
    I've 2 tables - packages and items. Items table contains all items belonging to the packages along with location information. Like this: Packages table id, name, type(enum{general,special}) 1, name1, general 2, name2, special Items table id, package_id, location 1, 1, America 2, 1, Africa 3, 1, Europe 4, 2, Europe Question: I want to find all 'special' packages belonging to a location and if no special package is found then it should return 'general' packages belonging to same location. So, for 'Europe' : package 2 should be returned since it is special package (Though package 1 also belongs to Europe but not required since its a general package) for 'America' : package 1 should be returned since there are no special packages

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  • Alfresco: Restrict categories that can be selected

    - by vegemite4me
    I have a custom content type in Alfresco (3.3 Enterprise, if it matters) and I can assign one or more categories to that content. So far so good. But can I restrict the set of possible categories to only a subset of all categories? If, for example, categories looked like below, how can I restrict the user to only selecting a region subcategory (e.g. Europe, South America, etc). Categories + Software Document Classification <- I do not want these to be picked. | + Utilisation Documents | + Software Descriptions | + ... | + Regions <- I want to restrict the | + Latin America user to this subset of categories. | + Europe | + ... + ... Is this possible?

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  • JD Edwards World Reporting Made Easy with Real Time Reporting Tools from The GL Company

    Fred talks to Paul Yarwood, US Operations General Manager and Richard Crotty, North America Business Development Manager for The GL Company, an Oracle Certified Partner, and Denise Grills, Senior Director of Marketing and Product Strategy for Oracle's JD Edwards World products. They discuss how the finance department of JD Edwards World customers can have complete control over their management reporting with a true inquiry, consolidation, and reporting solution from The GL Company, freeing up the finance team from being dependent upon IT time and resources.

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  • C# XNA Normals Question

    - by Wade
    Hello all! I have been working on some simple XNA proof of concept for a game idea I have as well as just to further my learning in XNA. However, i seem to be stuck on these dreaded normals, and using the BasicEffect with default lighting i can't seem to tell if my normals are being calculated correctly, hence the question. I'm mainly drawing cubes at the moment, I'm using a triangle list and a VertexBuffer to get the job done. The north face of my cube has two polygons and 6 vectors: Vector3 startPosition = new Vector3(0,0,0); corners[0] = startPosition; // This is the start position. Block size is 5. corners[1] = new Vector3(startPosition.X, startPosition.Y + BLOCK_SIZE, startPosition.Z); corners[2] = new Vector3(startPosition.X + BLOCK_SIZE, startPosition.Y, startPosition.Z); corners[3] = new Vector3(startPosition.X + BLOCK_SIZE, startPosition.Y + BLOCK_SIZE, startPosition.Z); verts[0] = new VertexPositionNormalTexture(corners[0], normals[0], textCoordBR); verts[1] = new VertexPositionNormalTexture(corners[1], normals[0], textCoordTR); verts[2] = new VertexPositionNormalTexture(corners[2], normals[0], textCoordBL); verts[3] = new VertexPositionNormalTexture(corners[3], normals[0], textCoordTL); verts[4] = new VertexPositionNormalTexture(corners[2], normals[0], textCoordBL); verts[5] = new VertexPositionNormalTexture(corners[1], normals[0], textCoordTR); Using those coordinates I want to generate the normal for the north face, I have no clue how to get the average of all those vectors and create a normal for the two polygons that it makes. Here is what i tried: normals[0] = Vector3.Cross(corners[1], corners[2]); normals[0].Normalize(); It seems like its correct, but then using the same thing for other sides of the cube the lighting effect seems weird, and not cohesive with where i think the light source is coming from, not really sure with the BasicEffect. Am I doing this right? Can anyone explain in lay mans terms how normals are calculated. Any help is much appreciated. Note: I tried going through Riemers and such to figure it out with no luck, it seems no one really goes over the math well enough. Thanks!

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  • links for 2010-05-21

    - by Bob Rhubart
    Stewart Bryson: Rittman Mead America is Recruiting "We don’t employ any junior consultants," writes Bryson, "so you would have to be highly experienced with some or all of the Oracle BI Stack (OBIEE, OWB, ODI, the Oracle Database, Hyperion), preferably have consulting experience, and excellent client-facing skills. Our consultants also provide all of our training services, and most of us write and speak at conferences, so being articulate and passionate about Oracle BI is another requirement. (tags: jobs employment consultants oracle businessintelligence)

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  • Einladung zum Oracle Partner Day 2012

    - by A&C Redaktion
    EINLADUNG: ORACLE PARTNER DAY GERMANY - 29. Oktober 2012: COMMERZBANK ARENA, Frankfurt Sehr geehrter Oracle Partner, volle Kraft voraus! Unsere Neuausrichtung und Zusammenfassung der Ressourcen, die Sie sicherlich aufmerksam verfolgt haben, ist nun abgeschlossen. Wir sind sehr überzeugt davon, dass wir alle gemeinsam von diesen Veränderungen nachhaltig profitieren werden. Nur ein Stichwort dazu: „One Oracle Red Stack“. Oracle Alliances & Channels und seine Partner können ab sofort das komplette Oracle Produktportfolio bestehend aus Oracle Software Technology, Oracle Applications und Oracle Hardware anbieten – bei bestmöglichem Support aus unserer neuen Organisation. Mein Name ist Christian Werner. Ich bin seit einigen Wochen verantwortlich für alle Alliances & Channels Bereiche in Deutschland. Frau Silvia Kaske leitet jetzt den kompletten A&C Bereich in Europe North in der Funktion als Senior Director Europe North Alliances & Channel Sales. Willkommen an Bord! Was Sie als Oracle Partner davon erwarten können? Wir haben alle unsere Kräfte konzentriert. Aus drei eigenen Bereichen wird ein großes Ganzes. Für Sie bedeutet das: fokussiertes Wachstum, geballte Kompetenz und Supportinfrastruktur. Abgestimmte Prozesse und kürzere Zeiten für Approvals. Wir meinen: Die Zeit ist reif, um jetzt gemeinsam den Hebel auf „Volle Kraft voraus“ zu legen. Und die Bedingungen für Sie sind besser als jemals zuvor, wenn Sie Ihre Kundenbasis erweitern oder neue Marktbereiche erschließen wollen. Leinen los: Kommen Sie zum Oracle Partner Day, wenn Sie wissen wollen, was sich für Sie verbessern wird: am 29. Oktober 2012 in der Commerzbank Arena in Frankfurt. Treffen Sie Ihr neu aufgestelltes A&C-Team. Erleben Sie die Produktneuheiten von der Oracle Open World in San Francisco (30. September bis 4. Oktober 2012) aus erster Hand. Nutzen Sie das neue Speed Dating-Format mit ausgewählten Oracle Experten vor Ort – für konkrete Fragen zu Vertrieb und Produkten. Ganz besonders freue ich mich, dass dieses Jahr Jürgen Kunz, SVP Technology Northern Europe & Country Leader Germany, als Keynote Speaker dabei sein wird. Ich lade Sie ganz herzlich ein und freue mich, Sie in Frankfurt zu begrüßen! Die Teilnahme ist für Sie als Oracle Partner selbstverständlich kostenfrei. Hier finden Sie weitere Informationen zum Oracle Partner Day sowie die unkomplizierte Anmeldemöglichkeit. Ich freue mich auf Sie! Ihr Christian WernerSenior Director Alliances & Channels Germany Übrigens: Direkt nach dem Oracle Partner Day findet der Oracle Day für Endkunden statt. Sie als Partner können natürlich gemeinsam mit Ihren Kunden an dieser Veranstaltung teilnehmen. Bitte melden Sie sich schnell an, da die Plätze limitiert sind. Hier finden Sie weitere Infos zum Oracle Day.

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  • Einladung zum Oracle Partner Day 2012

    - by A&C Redaktion
    EINLADUNG: ORACLE PARTNER DAY GERMANY - 29. Oktober 2012: COMMERZBANK ARENA, Frankfurt Sehr geehrter Oracle Partner, volle Kraft voraus! Unsere Neuausrichtung und Zusammenfassung der Ressourcen, die Sie sicherlich aufmerksam verfolgt haben, ist nun abgeschlossen. Wir sind sehr überzeugt davon, dass wir alle gemeinsam von diesen Veränderungen nachhaltig profitieren werden. Nur ein Stichwort dazu: „One Oracle Red Stack“. Oracle Alliances & Channels und seine Partner können ab sofort das komplette Oracle Produktportfolio bestehend aus Oracle Software Technology, Oracle Applications und Oracle Hardware anbieten – bei bestmöglichem Support aus unserer neuen Organisation. Mein Name ist Christian Werner. Ich bin seit einigen Wochen verantwortlich für alle Alliances & Channels Bereiche in Deutschland. Frau Silvia Kaske leitet jetzt den kompletten A&C Bereich in Europe North in der Funktion als Senior Director Europe North Alliances & Channel Sales. Willkommen an Bord! Was Sie als Oracle Partner davon erwarten können? Wir haben alle unsere Kräfte konzentriert. Aus drei eigenen Bereichen wird ein großes Ganzes. Für Sie bedeutet das: fokussiertes Wachstum, geballte Kompetenz und Supportinfrastruktur. Abgestimmte Prozesse und kürzere Zeiten für Approvals. Wir meinen: Die Zeit ist reif, um jetzt gemeinsam den Hebel auf „Volle Kraft voraus“ zu legen. Und die Bedingungen für Sie sind besser als jemals zuvor, wenn Sie Ihre Kundenbasis erweitern oder neue Marktbereiche erschließen wollen. Leinen los: Kommen Sie zum Oracle Partner Day, wenn Sie wissen wollen, was sich für Sie verbessern wird: am 29. Oktober 2012 in der Commerzbank Arena in Frankfurt. Treffen Sie Ihr neu aufgestelltes A&C-Team. Erleben Sie die Produktneuheiten von der Oracle Open World in San Francisco (30. September bis 4. Oktober 2012) aus erster Hand. Nutzen Sie das neue Speed Dating-Format mit ausgewählten Oracle Experten vor Ort – für konkrete Fragen zu Vertrieb und Produkten. Ganz besonders freue ich mich, dass dieses Jahr Jürgen Kunz, SVP Technology Northern Europe & Country Leader Germany, als Keynote Speaker dabei sein wird. Ich lade Sie ganz herzlich ein und freue mich, Sie in Frankfurt zu begrüßen! Die Teilnahme ist für Sie als Oracle Partner selbstverständlich kostenfrei. Hier finden Sie weitere Informationen zum Oracle Partner Day sowie die unkomplizierte Anmeldemöglichkeit. Ich freue mich auf Sie! Ihr Christian WernerSenior Director Alliances & Channels Germany Übrigens: Direkt nach dem Oracle Partner Day findet der Oracle Day für Endkunden statt. Sie als Partner können natürlich gemeinsam mit Ihren Kunden an dieser Veranstaltung teilnehmen. Bitte melden Sie sich schnell an, da die Plätze limitiert sind. Hier finden Sie weitere Infos zum Oracle Day.

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  • Oracle Database 12 c New Partition Maintenance Features by Gwen Lazenby

    - by hamsun
    One of my favourite new features in Oracle Database 12c is the ability to perform partition maintenance operations on multiple partitions. This means we can now add, drop, truncate and merge multiple partitions in one operation, and can split a single partition into more than two partitions also in just one command. This would certainly have made my life slightly easier had it been available when I administered a data warehouse at Oracle 9i. To demonstrate this new functionality and syntax, I am going to create two tables, ORDERS and ORDERS_ITEMS which have a parent-child relationship. ORDERS is to be partitioned using range partitioning on the ORDER_DATE column, and ORDER_ITEMS is going to partitioned using reference partitioning and its foreign key relationship with the ORDERS table. This form of partitioning was a new feature in 11g and means that any partition maintenance operations performed on the ORDERS table will also take place on the ORDER_ITEMS table as well. First create the ORDERS table - SQL CREATE TABLE orders ( order_id NUMBER(12), order_date TIMESTAMP, order_mode VARCHAR2(8), customer_id NUMBER(6), order_status NUMBER(2), order_total NUMBER(8,2), sales_rep_id NUMBER(6), promotion_id NUMBER(6), CONSTRAINT orders_pk PRIMARY KEY(order_id) ) PARTITION BY RANGE(order_date) (PARTITION Q1_2007 VALUES LESS THAN (TO_DATE('01-APR-2007','DD-MON-YYYY')), PARTITION Q2_2007 VALUES LESS THAN (TO_DATE('01-JUL-2007','DD-MON-YYYY')), PARTITION Q3_2007 VALUES LESS THAN (TO_DATE('01-OCT-2007','DD-MON-YYYY')), PARTITION Q4_2007 VALUES LESS THAN (TO_DATE('01-JAN-2008','DD-MON-YYYY')) ); Table created. Now the ORDER_ITEMS table SQL CREATE TABLE order_items ( order_id NUMBER(12) NOT NULL, line_item_id NUMBER(3) NOT NULL, product_id NUMBER(6) NOT NULL, unit_price NUMBER(8,2), quantity NUMBER(8), CONSTRAINT order_items_fk FOREIGN KEY(order_id) REFERENCES orders(order_id) on delete cascade) PARTITION BY REFERENCE(order_items_fk) tablespace example; Table created. Now look at DBA_TAB_PARTITIONS to get details of what partitions we have in the two tables – SQL select table_name,partition_name, partition_position position, high_value from dba_tab_partitions where table_owner='SH' and table_name like 'ORDER_%' order by partition_position, table_name; TABLE_NAME PARTITION_NAME POSITION HIGH_VALUE -------------- --------------- -------- ------------------------- ORDERS Q1_2007 1 TIMESTAMP' 2007-04-01 00:00:00' ORDER_ITEMS Q1_2007 1 ORDERS Q2_2007 2 TIMESTAMP' 2007-07-01 00:00:00' ORDER_ITEMS Q2_2007 2 ORDERS Q3_2007 3 TIMESTAMP' 2007-10-01 00:00:00' ORDER_ITEMS Q3_2007 3 ORDERS Q4_2007 4 TIMESTAMP' 2008-01-01 00:00:00' ORDER_ITEMS Q4_2007 4 Just as an aside it is also now possible in 12c to use interval partitioning on reference partitioned tables. In 11g it was not possible to combine these two new partitioning features. For our first example of the new 12cfunctionality, let us add all the partitions necessary for 2008 to the tables using one command. Notice that the partition specification part of the add command is identical in format to the partition specification part of the create command as shown above - SQL alter table orders add PARTITION Q1_2008 VALUES LESS THAN (TO_DATE('01-APR-2008','DD-MON-YYYY')), PARTITION Q2_2008 VALUES LESS THAN (TO_DATE('01-JUL-2008','DD-MON-YYYY')), PARTITION Q3_2008 VALUES LESS THAN (TO_DATE('01-OCT-2008','DD-MON-YYYY')), PARTITION Q4_2008 VALUES LESS THAN (TO_DATE('01-JAN-2009','DD-MON-YYYY')); Table altered. Now look at DBA_TAB_PARTITIONS and we can see that the 4 new partitions have been added to both tables – SQL select table_name,partition_name, partition_position position, high_value from dba_tab_partitions where table_owner='SH' and table_name like 'ORDER_%' order by partition_position, table_name; TABLE_NAME PARTITION_NAME POSITION HIGH_VALUE -------------- --------------- -------- ------------------------- ORDERS Q1_2007 1 TIMESTAMP' 2007-04-01 00:00:00' ORDER_ITEMS Q1_2007 1 ORDERS Q2_2007 2 TIMESTAMP' 2007-07-01 00:00:00' ORDER_ITEMS Q2_2007 2 ORDERS Q3_2007 3 TIMESTAMP' 2007-10-01 00:00:00' ORDER_ITEMS Q3_2007 3 ORDERS Q4_2007 4 TIMESTAMP' 2008-01-01 00:00:00' ORDER_ITEMS Q4_2007 4 ORDERS Q1_2008 5 TIMESTAMP' 2008-04-01 00:00:00' ORDER_ITEMS Q1_2008 5 ORDERS Q2_2008 6 TIMESTAMP' 2008-07-01 00:00:00' ORDER_ITEM Q2_2008 6 ORDERS Q3_2008 7 TIMESTAMP' 2008-10-01 00:00:00' ORDER_ITEMS Q3_2008 7 ORDERS Q4_2008 8 TIMESTAMP' 2009-01-01 00:00:00' ORDER_ITEMS Q4_2008 8 Next, we can drop or truncate multiple partitions by giving a comma separated list in the alter table command. Note the use of the plural ‘partitions’ in the command as opposed to the singular ‘partition’ prior to 12c– SQL alter table orders drop partitions Q3_2008,Q2_2008,Q1_2008; Table altered. Now look at DBA_TAB_PARTITIONS and we can see that the 3 partitions have been dropped in both the two tables – TABLE_NAME PARTITION_NAME POSITION HIGH_VALUE -------------- --------------- -------- ------------------------- ORDERS Q1_2007 1 TIMESTAMP' 2007-04-01 00:00:00' ORDER_ITEMS Q1_2007 1 ORDERS Q2_2007 2 TIMESTAMP' 2007-07-01 00:00:00' ORDER_ITEMS Q2_2007 2 ORDERS Q3_2007 3 TIMESTAMP' 2007-10-01 00:00:00' ORDER_ITEMS Q3_2007 3 ORDERS Q4_2007 4 TIMESTAMP' 2008-01-01 00:00:00' ORDER_ITEMS Q4_2007 4 ORDERS Q4_2008 5 TIMESTAMP' 2009-01-01 00:00:00' ORDER_ITEMS Q4_2008 5 Now let us merge all the 2007 partitions together to form one single partition – SQL alter table orders merge partitions Q1_2005, Q2_2005, Q3_2005, Q4_2005 into partition Y_2007; Table altered. TABLE_NAME PARTITION_NAME POSITION HIGH_VALUE -------------- --------------- -------- ------------------------- ORDERS Y_2007 1 TIMESTAMP' 2008-01-01 00:00:00' ORDER_ITEMS Y_2007 1 ORDERS Q4_2008 2 TIMESTAMP' 2009-01-01 00:00:00' ORDER_ITEMS Q4_2008 2 Splitting partitions is a slightly more involved. In the case of range partitioning one of the new partitions must have no high value defined, and in list partitioning one of the new partitions must have no list of values defined. I call these partitions the ‘everything else’ partitions, and will contain any rows contained in the original partition that are not contained in the any of the other new partitions. For example, let us split the Y_2007 partition back into 4 quarterly partitions – SQL alter table orders split partition Y_2007 into (PARTITION Q1_2007 VALUES LESS THAN (TO_DATE('01-APR-2007','DD-MON-YYYY')), PARTITION Q2_2007 VALUES LESS THAN (TO_DATE('01-JUL-2007','DD-MON-YYYY')), PARTITION Q3_2007 VALUES LESS THAN (TO_DATE('01-OCT-2007','DD-MON-YYYY')), PARTITION Q4_2007); Now look at DBA_TAB_PARTITIONS to get details of the new partitions – TABLE_NAME PARTITION_NAME POSITION HIGH_VALUE -------------- --------------- -------- ------------------------- ORDERS Q1_2007 1 TIMESTAMP' 2007-04-01 00:00:00' ORDER_ITEMS Q1_2007 1 ORDERS Q2_2007 2 TIMESTAMP' 2007-07-01 00:00:00' ORDER_ITEMS Q2_2007 2 ORDERS Q3_2007 3 TIMESTAMP' 2007-10-01 00:00:00' ORDER_ITEMS Q3_2007 3 ORDERS Q4_2007 4 TIMESTAMP' 2008-01-01 00:00:00' ORDER_ITEMS Q4_2007 4 ORDERS Q4_2008 5 TIMESTAMP' 2009-01-01 00:00:00' ORDER_ITEMS Q4_2008 5 Partition Q4_2007 has a high value equal to the high value of the original Y_2007 partition, and so has inherited its upper boundary from the partition that was split. As for a list partitioning example let look at the following another table, SALES_PAR_LIST, which has 2 partitions, Americas and Europe and a partitioning key of country_name. SQL select table_name,partition_name, high_value from dba_tab_partitions where table_owner='SH' and table_name = 'SALES_PAR_LIST'; TABLE_NAME PARTITION_NAME HIGH_VALUE -------------- --------------- ----------------------------- SALES_PAR_LIST AMERICAS 'Argentina', 'Canada', 'Peru', 'USA', 'Honduras', 'Brazil', 'Nicaragua' SALES_PAR_LIST EUROPE 'France', 'Spain', 'Ireland', 'Germany', 'Belgium', 'Portugal', 'Denmark' Now split the Americas partition into 3 partitions – SQL alter table sales_par_list split partition americas into (partition south_america values ('Argentina','Peru','Brazil'), partition north_america values('Canada','USA'), partition central_america); Table altered. Note that no list of values was given for the ‘Central America’ partition. However it should have inherited any values in the original ‘Americas’ partition that were not assigned to either the ‘North America’ or ‘South America’ partitions. We can confirm this by looking at the DBA_TAB_PARTITIONS view. SQL select table_name,partition_name, high_value from dba_tab_partitions where table_owner='SH' and table_name = 'SALES_PAR_LIST'; TABLE_NAME PARTITION_NAME HIGH_VALUE --------------- --------------- -------------------------------- SALES_PAR_LIST SOUTH_AMERICA 'Argentina', 'Peru', 'Brazil' SALES_PAR_LIST NORTH_AMERICA 'Canada', 'USA' SALES_PAR_LIST CENTRAL_AMERICA 'Honduras', 'Nicaragua' SALES_PAR_LIST EUROPE 'France', 'Spain', 'Ireland', 'Germany', 'Belgium', 'Portugal', 'Denmark' In conclusion, I hope that DBA’s whose work involves maintaining partitions will find the operations a bit more straight forward to carry out once they have upgraded to Oracle Database 12c. Gwen Lazenby is a Principal Training Consultant at Oracle. She is part of Oracle University's Core Technology delivery team based in the UK, teaching Database Administration and Linux courses. Her specialist topics include using Oracle Partitioning and Parallelism in Data Warehouse environments, as well as Oracle Spatial and RMAN.

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