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• how to print data in a table from mysql

  - by robertdd

  hello, i want to extract the last eight entries from my database and print them into a 2 columns table!like this: |1|2| |3|4| |5|6| |7|8| is that possible? this is my code: $db = new Database(DB_SERVER, DB_USER, DB_PASS, DB_DATABASE); $db->connect(); $sql = "SELECT ID, movieno FROM movies ORDER BY ID DESC LIMIT 8 "; $rows = $db->query($sql); print '<table width="307" border="0" cellspacing="5" cellpadding="4">'; while ($record = $db->fetch_array($rows)) { $vidaidi = $record['movieno']; print <<<END <tr> <td> <a href="http://www.youtube.com/watch?v=$vidaidi" target="_blank"> <img src="http://img.youtube.com/vi/$vidaidi/1.jpg" width="123" height="80"></a> </td> </tr> END; } print '</table>';

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• How to retrive user input data entered tab bar View Controller when application start in iphone app

  - by TechFusion

  Hello, I have created window based application. Tab bar controller as root controller and it has three tabs. One Tab has Labels and TextFiled inputs like Name, Username and Password. I am looking to store this text filed inputs when user enters and able retrieve in other tabs. Previously I have set key for different text fields and setobject:withkey task and able to retrive text filed values in same view Controller [[NSUserDefaults standardUserDefaults] stringForKey:key] task. Now I am looking to create database which has different objects and each objects has data values of different Text Field inputs that I can access in whole application. like DatabaseName - Object1 - Name, Username & Password - Object2 - Name, Username & Password Something like structure in Normal C so it would be easy to retrieve data. I am looking NSUserDefaults Class and User Defaults Programming Topics in Cocoa(http://developer.apple.com/iPhone/library/documentation/Cocoa/Conceptual/UserDefaults/UserDefaults.html#//apple_ref/doc/uid/10000059-BCIDJFHD). Also Referring Archives and Serialization Programming guide(http://developer.apple.com/iphone/library/documentation/Cocoa/Conceptual/Archiving/Archiving.html#//apple_ref/doc/uid/10000047i). Which method i need to use to create such type of database ? Thanks,

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• Logic in the db for maintaining a points system relationship?

  - by MarcusBooster

  I'm making a little web based game and need to determine where to put logic that checks the integrity of some underlying data in the sql database. Each user keeps track of points assigned to him, and points are awarded by various tasks. I keep a record of each task transaction to make sure they're not repeated, and to keep track of the value of the task at the time of completion, since an individual award level my fluctuate over time. My schema looks like this so far: create table player ( player_ID serial primary key, player_Points int not null default 0 ); create table task ( task_ID serial primary key, task_PointsAwarded int not null ); create table task_list ( player_ID int references player(player_ID), task_ID int references task(task_ID), when_completed timestamp default current_timestamp, point_value int not null, --not fk because task value may change later constraint pk_player_task_id primary key (player_ID, task_ID) ); So, the player.player_Points should be the total of all his cumulative task points in the task_list. Now where do I put the logic to enforce this? Should I do away with player.player_Points altogether and do queries every time I want to know the total score? Which seems wasteful since I'll be doing that query a lot over the course of a game. Or, put a trigger in the task_list that automatically updates the player.player_Points? Is that too much logic to have in the database and should just maintain this relationship in the application? Thanks.

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• SQL -- How is DISTINCT so fast without an index?

  - by Jonathan

  Hi, I have a database with a table called 'links' with 600 million rows in it in SQLite. There are 2 columns in the database - a "src" column and a "dest" column. At present there are no indices. There are a fair number of common values between src and dest, but also a fair number of duplicated rows. The first thing I'm trying to do is remove all the duplicate rows, and then perform some additional processing on the results, however I've been encountering some weird issues. Firstly, SELECT * FROM links WHERE src=434923 AND dest=5010182. Now this returns one result fairly quickly and then takes quite a long time to run as I assume it's performing a tablescan on the rest of the 600m rows. However, if I do SELECT DISTINCT * FROM links, then it immediately starts returning rows really quickly. The question is: how is this possible?? Surely for each row, the row must be compared against all of the other rows in the table, but this would require a tablescan of the remaining rows in the table which SHOULD takes ages! Any ideas why SELECT DISTINCT is so much quicker than a standard SELECT?

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• How to store and synchronize a big list of strings

  - by Joel

  I have a large database table in SQLExpress on Windows, with a particular field of interest 'code'. I have an Apache web server with MySQL on Linux. The web application on the Linux box needs access to the list of all codes. The only thing it will use the list for is checking for the existence of a given code. Having the Linux server call out to the Windows server is impractical as the Windows server is behind a NAT'ed office internet connection, and it may not always be accessible. I have set it so the Windows server will push the list of codes to the web server by means of a simple HTTP POST request. However, at this point I have not implemented the storage of the codes on the Linux box. Should I store them in a MySQL table with a single field 'code'? Then I get fast indexed lookups O(1), however I think synchronization will be an issue - given an updated list of codes, pushed from the Windows box, how would I optimally synchronize the list with the database? TRUNCATE, followed by INSERT? Should I instead store them in a flat file? Then I have O(n) look up time rather than O(1). Additionally an extra constant-time overhead too, as I will be processing the file in Ruby. However, synchronization is easy - simply replace the file.

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• find the difference between two very large list

  - by user157195

  I have two large list(could be a hundred million items), the source of each list can be either from a database table or a flat file. both lists are of comparable sizes, both unsorted. I need to find the difference between them. so I have 3 scenarios: 1. List1 is a database table(assume each row simply have one item(key) that is a string), List2 is a large file. 2. Both lists are from 2 db tables. 3. both lists are from two files. in case 2, I plan to use: select a.item from MyTable a where a.item not in (select b.item form MyTable b) this clearly is inefficient, is there a better way? Another approach is: I plan to sort each list, and then walk down both of them to find the diff. If the list is from a file, I have to read it into a db table first, then use db sorting to output the list. Is the run time complexity still O(nlogn) in db sorting? either approach is a pain and seems would be very slow when the list involved has hundreds of millions of items. any suggestions?

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• How to store generated eigen faces for future face recognition?

  - by user3237134

  My code works in the following manner: 1.First, it obtains several images from the training set 2.After loading these images, we find the normalized faces,mean face and perform several calculation. 3.Next, we ask for the name of an image we want to recognize 4.We then project the input image into the eigenspace, and based on the difference from the eigenfaces we make a decision. 5.Depending on eigen weight vector for each input image we make clusters using kmeans command. Source code i tried: clear all close all clc % number of images on your training set. M=1200; %Chosen std and mean. %It can be any number that it is close to the std and mean of most of the images. um=60; ustd=32; %read and show images(bmp); S=[]; %img matrix for i=1:M str=strcat(int2str(i),'.jpg'); %concatenates two strings that form the name of the image eval('img=imread(str);'); [irow icol d]=size(img); % get the number of rows (N1) and columns (N2) temp=reshape(permute(img,[2,1,3]),[irow*icol,d]); %creates a (N1*N2)x1 matrix S=[S temp]; %X is a N1*N2xM matrix after finishing the sequence %this is our S end %Here we change the mean and std of all images. We normalize all images. %This is done to reduce the error due to lighting conditions. for i=1:size(S,2) temp=double(S(:,i)); m=mean(temp); st=std(temp); S(:,i)=(temp-m)*ustd/st+um; end %show normalized images for i=1:M str=strcat(int2str(i),'.jpg'); img=reshape(S(:,i),icol,irow); img=img'; end %mean image; m=mean(S,2); %obtains the mean of each row instead of each column tmimg=uint8(m); %converts to unsigned 8-bit integer. Values range from 0 to 255 img=reshape(tmimg,icol,irow); %takes the N1*N2x1 vector and creates a N2xN1 matrix img=img'; %creates a N1xN2 matrix by transposing the image. % Change image for manipulation dbx=[]; % A matrix for i=1:M temp=double(S(:,i)); dbx=[dbx temp]; end %Covariance matrix C=A'A, L=AA' A=dbx'; L=A*A'; % vv are the eigenvector for L % dd are the eigenvalue for both L=dbx'*dbx and C=dbx*dbx'; [vv dd]=eig(L); % Sort and eliminate those whose eigenvalue is zero v=[]; d=[]; for i=1:size(vv,2) if(dd(i,i)>1e-4) v=[v vv(:,i)]; d=[d dd(i,i)]; end end %sort, will return an ascending sequence [B index]=sort(d); ind=zeros(size(index)); dtemp=zeros(size(index)); vtemp=zeros(size(v)); len=length(index); for i=1:len dtemp(i)=B(len+1-i); ind(i)=len+1-index(i); vtemp(:,ind(i))=v(:,i); end d=dtemp; v=vtemp; %Normalization of eigenvectors for i=1:size(v,2) %access each column kk=v(:,i); temp=sqrt(sum(kk.^2)); v(:,i)=v(:,i)./temp; end %Eigenvectors of C matrix u=[]; for i=1:size(v,2) temp=sqrt(d(i)); u=[u (dbx*v(:,i))./temp]; end %Normalization of eigenvectors for i=1:size(u,2) kk=u(:,i); temp=sqrt(sum(kk.^2)); u(:,i)=u(:,i)./temp; end % show eigenfaces; for i=1:size(u,2) img=reshape(u(:,i),icol,irow); img=img'; img=histeq(img,255); end % Find the weight of each face in the training set. omega = []; for h=1:size(dbx,2) WW=[]; for i=1:size(u,2) t = u(:,i)'; WeightOfImage = dot(t,dbx(:,h)'); WW = [WW; WeightOfImage]; end omega = [omega WW]; end % Acquire new image % Note: the input image must have a bmp or jpg extension. % It should have the same size as the ones in your training set. % It should be placed on your desktop ed_min=[]; srcFiles = dir('G:\newdatabase\*.jpg'); % the folder in which ur images exists for b = 1 : length(srcFiles) filename = strcat('G:\newdatabase\',srcFiles(b).name); Imgdata = imread(filename); InputImage=Imgdata; InImage=reshape(permute((double(InputImage)),[2,1,3]),[irow*icol,1]); temp=InImage; me=mean(temp); st=std(temp); temp=(temp-me)*ustd/st+um; NormImage = temp; Difference = temp-m; p = []; aa=size(u,2); for i = 1:aa pare = dot(NormImage,u(:,i)); p = [p; pare]; end InImWeight = []; for i=1:size(u,2) t = u(:,i)'; WeightOfInputImage = dot(t,Difference'); InImWeight = [InImWeight; WeightOfInputImage]; end noe=numel(InImWeight); % Find Euclidean distance e=[]; for i=1:size(omega,2) q = omega(:,i); DiffWeight = InImWeight-q; mag = norm(DiffWeight); e = [e mag]; end ed_min=[ed_min MinimumValue]; theta=6.0e+03; %disp(e) z(b,:)=InImWeight; end IDX = kmeans(z,5); clustercount=accumarray(IDX, ones(size(IDX))); disp(clustercount); QUESTIONS: 1.It is working fine for M=50(i.e Training set contains 50 images) but not for M=1200(i.e Training set contains 1200 images).It is not showing any error.There is no output.I waited for 10 min still there is no output. I think it is going infinite loop.What is the problem?Where i was wrong? 2.Instead of running the training set everytime how eigen faces generated are stored so that stored eigen faces are used for future face recoginition for a new input image.So it reduces wastage of time.

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• Vectorization of matlab code for faster execution

  - by user3237134

  My code works in the following manner: 1.First, it obtains several images from the training set 2.After loading these images, we find the normalized faces,mean face and perform several calculation. 3.Next, we ask for the name of an image we want to recognize 4.We then project the input image into the eigenspace, and based on the difference from the eigenfaces we make a decision. 5.Depending on eigen weight vector for each input image we make clusters using kmeans command. Source code i tried: clear all close all clc % number of images on your training set. M=1200; %Chosen std and mean. %It can be any number that it is close to the std and mean of most of the images. um=60; ustd=32; %read and show images(bmp); S=[]; %img matrix for i=1:M str=strcat(int2str(i),'.jpg'); %concatenates two strings that form the name of the image eval('img=imread(str);'); [irow icol d]=size(img); % get the number of rows (N1) and columns (N2) temp=reshape(permute(img,[2,1,3]),[irow*icol,d]); %creates a (N1*N2)x1 matrix S=[S temp]; %X is a N1*N2xM matrix after finishing the sequence %this is our S end %Here we change the mean and std of all images. We normalize all images. %This is done to reduce the error due to lighting conditions. for i=1:size(S,2) temp=double(S(:,i)); m=mean(temp); st=std(temp); S(:,i)=(temp-m)*ustd/st+um; end %show normalized images for i=1:M str=strcat(int2str(i),'.jpg'); img=reshape(S(:,i),icol,irow); img=img'; end %mean image; m=mean(S,2); %obtains the mean of each row instead of each column tmimg=uint8(m); %converts to unsigned 8-bit integer. Values range from 0 to 255 img=reshape(tmimg,icol,irow); %takes the N1*N2x1 vector and creates a N2xN1 matrix img=img'; %creates a N1xN2 matrix by transposing the image. % Change image for manipulation dbx=[]; % A matrix for i=1:M temp=double(S(:,i)); dbx=[dbx temp]; end %Covariance matrix C=A'A, L=AA' A=dbx'; L=A*A'; % vv are the eigenvector for L % dd are the eigenvalue for both L=dbx'*dbx and C=dbx*dbx'; [vv dd]=eig(L); % Sort and eliminate those whose eigenvalue is zero v=[]; d=[]; for i=1:size(vv,2) if(dd(i,i)>1e-4) v=[v vv(:,i)]; d=[d dd(i,i)]; end end %sort, will return an ascending sequence [B index]=sort(d); ind=zeros(size(index)); dtemp=zeros(size(index)); vtemp=zeros(size(v)); len=length(index); for i=1:len dtemp(i)=B(len+1-i); ind(i)=len+1-index(i); vtemp(:,ind(i))=v(:,i); end d=dtemp; v=vtemp; %Normalization of eigenvectors for i=1:size(v,2) %access each column kk=v(:,i); temp=sqrt(sum(kk.^2)); v(:,i)=v(:,i)./temp; end %Eigenvectors of C matrix u=[]; for i=1:size(v,2) temp=sqrt(d(i)); u=[u (dbx*v(:,i))./temp]; end %Normalization of eigenvectors for i=1:size(u,2) kk=u(:,i); temp=sqrt(sum(kk.^2)); u(:,i)=u(:,i)./temp; end % show eigenfaces; for i=1:size(u,2) img=reshape(u(:,i),icol,irow); img=img'; img=histeq(img,255); end % Find the weight of each face in the training set. omega = []; for h=1:size(dbx,2) WW=[]; for i=1:size(u,2) t = u(:,i)'; WeightOfImage = dot(t,dbx(:,h)'); WW = [WW; WeightOfImage]; end omega = [omega WW]; end % Acquire new image % Note: the input image must have a bmp or jpg extension. % It should have the same size as the ones in your training set. % It should be placed on your desktop ed_min=[]; srcFiles = dir('G:\newdatabase\*.jpg'); % the folder in which ur images exists for b = 1 : length(srcFiles) filename = strcat('G:\newdatabase\',srcFiles(b).name); Imgdata = imread(filename); InputImage=Imgdata; InImage=reshape(permute((double(InputImage)),[2,1,3]),[irow*icol,1]); temp=InImage; me=mean(temp); st=std(temp); temp=(temp-me)*ustd/st+um; NormImage = temp; Difference = temp-m; p = []; aa=size(u,2); for i = 1:aa pare = dot(NormImage,u(:,i)); p = [p; pare]; end InImWeight = []; for i=1:size(u,2) t = u(:,i)'; WeightOfInputImage = dot(t,Difference'); InImWeight = [InImWeight; WeightOfInputImage]; end noe=numel(InImWeight); % Find Euclidean distance e=[]; for i=1:size(omega,2) q = omega(:,i); DiffWeight = InImWeight-q; mag = norm(DiffWeight); e = [e mag]; end ed_min=[ed_min MinimumValue]; theta=6.0e+03; %disp(e) z(b,:)=InImWeight; end IDX = kmeans(z,5); clustercount=accumarray(IDX, ones(size(IDX))); disp(clustercount); Running time for 50 images:Elapsed time is 103.947573 seconds. QUESTIONS: 1.It is working fine for M=50(i.e Training set contains 50 images) but not for M=1200(i.e Training set contains 1200 images).It is not showing any error.There is no output.I waited for 10 min still there is no output. I think it is going infinite loop.What is the problem?Where i was wrong?

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• How can I map a Windows group login to the dbo schema in a database?

  - by Christian Hayter

  I have a database for which I want to restrict access to 3 named individuals. I thought I could do the following: Create a local Windows group on the database server and add the named individuals to it. Create a Windows login in SQL Server mapped to the local Windows group. Map the login to the "dbo" schema in the database, so that the users can access all objects without having to qualify them with the schema name. When I try to do step 3, I get the following error: Msg 15353, Level 16, State 1, Line 1 An entity of type database cannot be owned by a role, a group, an approle, or by principals mapped to certificates or asymmetric keys. I have tried to do this via the IDE, the sp_changedbowner sproc, and the ALTER AUTHORIZATION command, and I get the same error each time. After searching MSDN and Google, I find that this restriction is by design. Great, that's useful. Can anyone tell me: Why this restriction exists? It seems very arbitrary. More importantly, can I accomplish my requirement some other way? Other info that might be pertinent: The server is fully up to date with service packs and hotfixes. All objects in the database are owned by the "dbo" schema, and it's not feasible to change that. The database is running in compatibility level 80, and it's not feasible to change that to 90 yet. I am free to make any other changes (within reason, depending on what they are).

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• How to know the root device size of some public AMI?

  - by red23jordan

  Since I do not want to pay money for my testing, the free limit size is 10G. I can know the root device root for some default AMI such as Amazon Linux AMI 2012.03 The Amazon Linux AMI 2012.03 is an EBS-backed, PV-GRUB image. It includes Linux 3.2, AWS tools, and repository access to multiple versions of MySQL, PostgreSQL, Python, Ruby, and Tomcat. Root Device Size: 8 GB And the last row displayed 8GB. However, if I find AMI in Community Page, it does not show the root device size. Can anyone know how to use the instance such as centOS that is not provided by default but it is under 10GB so that I can still free use?

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• MaxClients in apache. How to know the size of my proccess?

  - by Larry

  From http://httpd.apache.org/docs/2.2/misc/perf-tuning.html The single biggest hardware issue affecting webserver performance is RAM. A webserver should never ever have to swap, as swapping increases the latency of each request beyond a point that users consider "fast enough". This causes users to hit stop and reload, further increasing the load. You can, and should, control the MaxClients setting so that your server does not spawn so many children it starts swapping. This procedure for doing this is simple: determine the size of your average Apache process, by looking at your process list via a tool such as top, and divide this into your total available memory, leaving some room for other processes. The main issue is that I can't understand how to know the size, because, well i have the size of httpd on no more of 3888 But, if we need to determine the number for MaxClients, and I have 4GB of RAM, so I get: 972, so I should use like 900 in the MaxClients?

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• SQL Server 2008 Bring Database Online trying to open a file from a drive that doesn't exist

  - by Nai

  This is my error I am facing TITLE: Microsoft.SqlServer.Smo Set offline failed for Database 'Go3D_Retailer ------------------------------ ADDITIONAL INFORMATION: An exception occurred while executing a Transact-SQL statement or batch. (Microsoft.SqlServer.ConnectionInfo) Unable to open the physical file "E:\Program Files\Microsoft SQL Server\MSSQL10.MSSQLSERVER\MSSQL\DATA\ftrow_Go3D_catalog.ndf". Operating system error 2: "2(failed to retrieve text for this error. Reason: 15105)". Database 'Go3D_Retailer' cannot be opened due to inaccessible files or insufficient memory or disk space. See the SQL Server errorlog for details. ALTER DATABASE statement failed. (Microsoft SQL Server, Error: 5120) Background to this error I've been trying to move my destination logshipping database to another physical server for analysis purposes. Because I do not have domain keys and active directory set up, I had to hack my process by using the same username/password for both the source and destination servers to get the process to work. Following that, I used this guy's solution to move the destination database to another server. However, this error occurs when I try to bring the database back online. I don't have an E drive on my server and I have no idea why it's trying to open a file from E drive. I have over a 100gb left on my hard disk so it's definitely not a space issue. This sounds like a bug... Any ideas?

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• Maximum disk size for a Debian 6 VMware installation?

  - by Mohamed Abobakr

  I'm planning to install a variety of Linux distributions on VMware Player to get more experience with Linux (I've only used Fedora back in university). I'm starting with Debian 6.0.5 Squeeze, and have already downloaded the small CD i386 netinst image. In the new virtual machine wizard, it's asking me for the maximum disk size of the image. It says that the recommended size for Debian 6 is 8 GB. However, I'm trying to figure out what is the absolute minimal size that I can get away with, since I'm only testing it out with very few software installations (mainly testing web server applications).

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• How to de-dupe identical photos that have a slightly different file size?

  - by GJ.

  I imported many photos using the new "camera import" feature of Dropbox. Many of those were duplicates of photos previously imported by direct copying from the camera. Strangely, the Dropbox import appears to slightly reduce the file size. E.g. here on the right is the file imported through Dropbox: Comparison of the two files using pdiff returns "Images are binary identical", but tools such as fdupes or even the Picasa "show duplicate files" feature, consider them as unique. What can be the cause of this file size change? Is there any way to undo it? Most importantly: how can I de-dupe efficiently without regard to file size comparison? (running pdiff comparison over all photo pairs in my library is obviously impractical...) A solution for either OS X or Windows would do.

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• SQL Server 2008: Getting Login failed for user "Domain\User". Failed to open the explicitly specified database [CLIENT: IP.ADD.RR.ESS]

  - by GodEater

  This is a very similar issue to " SQL Server 2008 login problem with ASP.NET application: Failed to open the explicitly specified database " which unfortunately seems to have gone unsolved. My issue here is subtly different. Firstly the account failing login is not 'NT AUTHORITY\NETWORK SERVICE' - it's an actual domain account. Secondly, there are two machines involved - I gathered from the first question it was a single machine running both the IIS and SQL instances. The application which is trying to connect to the database is an ASP.NET one running on another server (if that makes any different, I'm not sure it does.) The ConnectionString being used in the web.config for the application is : data source=MySQLServer;initial catalog=MyDatabase;integrated security=sspi; And the Application Pool is set to NetworkService for Identity. So - in the web app, I get the following error : Cannot open database "MyDatabase" requested by the login. The login failed. Login failed for user 'MyDomain\WebServerMachineName$' In the SQL Server logs I see : Login failed for user 'MyDomain\WebServerMachineName$'. Reason: Failed to open the explicitly specified database. [CLIENT: Web.Server.IP.Address] Running this bit of SQL against the database in question : USE [MyDatabase] GO SELECT SDP.name AS [User Name], SDP.type_desc AS [User Type], UPPER(SDPS.name) AS [Database Role] FROM sys.database_principals SDP INNER JOIN sys.database_role_members SDRM ON SDP.principal_id=SDRM.member_principal_id INNER JOIN sys.database_principals SDPS ON SDRM.role_principal_id = SDPS.principal_id Gets me this result : MyDomain\WebServerMachineName$ WINDOWS_USER DB_DDLADMIN MyDomain\WebServerMachineName$ WINDOWS_USER DB_DATAREADER MyDomain\WebServerMachineName$ WINDOWS_USER DB_DATAWRITER Which appears to me to indicate I've got the permissions right. Anyone have any idea why it's not working, or how I can narrow the issue down some more?

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• May the file size returned by stat be compromised?

  - by codeholic

  I want to make sure that nobody changed a file. In order to accomplish that, I want not only to check MD5 sum of the file, but also check its size, since as far as I understand this additional simple check can sophisticate falsification by several digits. May I trust the size that stat returns? I don't mean if changes were made to stat itself. I don't go that deep. But, for instance, may one compromise the file size that stat returns by hacking the directory file? Or by similar means, that do not require superuser privileges? It's Linux.

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• File property information (last write time and file size) in explorer out of date by hours over netw

  - by David L Morris

  An application is running on a windows XP prof machine picking up file from a network share from another windows machine. It detects that the file has been updated (by date and time or optionally file size) and reads it for any new data. Most of the time the last write time and file size, seems to be up to date. Occasionally, this information stops being updated, even though the file is growing (intermittently during the day) with appended content, so that the last write time and file size remain fixed at some arbitrary moment. This is visible in explorer, where it shows a fixed last write time on the reading machine. Just opening the file to edit it in notepad, immediately refreshes the file properties, and the other application picks up where it left of. The file location can't be changed, nor the location of the relevant applications. Any solutions to resolve this problem?

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• How can I get the size of an Amazon S3 bucket?

  - by Garret Heaton

  I'd like to graph the size (in bytes, and # of items) of an Amazon S3 bucket and am looking for an efficient way to get the data. The s3cmd tools provide a way to get the total file size using s3cmd du s3://bucket_name, but I'm worried about its ability to scale since it looks like it fetches data about every file and calculates its own sum. Since Amazon charges users in GB-Months it seems odd that they don't expose this value directly. Although Amazon's REST API returns the number of items in a bucket, [s3cmd] doesn't seem to expose it. I could do s3cmd ls -r s3://bucket_name | wc -l but that seems like a hack. The Ruby AWS::S3 library looked promising, but only provides the # of bucket items, not the total bucket size. Is anyone aware of any other command line tools or libraries (prefer Perl, PHP, Python, or Ruby) which provide ways of getting this data?

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• Ubuntu 12.04 3.2 Kernel showing incorrect "cache size" in cpuinfo?

  - by Tom G

  2x Xeon E5620 . 16 Cores altogether. /proc/cpuinfo shows cache is only @ 4096kb According to intel this should have 12MB of "smart cache". Doing searched for E5620 and CPUinfo shows the correct number: cache size : 12288 KB However mine shows this: processor : 15 v endor_id : GenuineIntel cpu family : 6 model : 44 model name : Intel(R) Xeon(R) CPU E5620 @ 2.40GHz stepping : 2 microcode : 0x1 cpu MHz : 2400.104 cache size : 4096 KB fpu : yes fpu_exception : yes cpuid level : 11 wp : yes flags : fpu vme de pse tsc msr pae mce cx8 apic mtrr pge mca cmov pat pse36 clflush mmx bogomips : 4800.20 clflush size : 64 cache_alignment : 64 address sizes : 40 bits physical, 48 bits virtual Any idea on why it shows like it's missing 8MB in CPU cache? .

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• How can I increase the size of the hard disk used by my Virtual Machine Manager (VMM) vm?

  - by Scott Langham

  In Virtual Machine Manager admin console, I used the 'New virtual machine' wizard and created a machine from an existing template. When I got to the Configure Hardware step, I selected the primary hard disk (which is a dynamic) disk and tried to increase it's maximum size, but this control is disabled so I can't increase it. It's not big enough for my purposes. how do I increase the maximum size of the hard-disk? Thanks. Edit: I am able to add a second hard disk of a size I specify. That's not useful though as I'm trying to install some software, some of which insists on being on drive C.

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• How can I print legible text in a font size <1.5?

  - by user330372

  For biological research, I need to print characters so tiny that 2 of them fit in less than 0.5 mm, which I will read under a microscope. I am currently printing from Excel at font size of 1.5, using a HP LaserJet 400M. The result is slightly larger than what I need it to be, but printing at size 1 produces unreadable results. How can I print a smaller font size but still get readable results? Are there specialized printers for that? Where could I find one?

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• Re-encoding a video file and increasing the size -- does this improve the quality?

  - by Josh

  If I have a video file at 320x240 resolution which I want to re-encode (because I don't like the encoding it's in now) and I also want to play it at double size (640x480), will I get higher quality if I scale it up to 640x480 when I convert it to a new format, verses keeping it at 320x240 in the new format and playing it at double size? This probably depends on the program used to convert, and if so, please let me know any program which might increase the quality. Here's my thinking. If I play a 320x240 file at double size, the system has to scale up each frame in real time, whereas if I scale up while recompressing the system may be able to use a more intensive algorythm like Bicubic interpolation . However I am not sure if this is true or not.

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• SQL SERVER – Guest Posts – Feodor Georgiev – The Context of Our Database Environment – Going Beyond the Internal SQL Server Waits – Wait Type – Day 21 of 28

  - by pinaldave

  This guest post is submitted by Feodor. Feodor Georgiev is a SQL Server database specialist with extensive experience of thinking both within and outside the box. He has wide experience of different systems and solutions in the fields of architecture, scalability, performance, etc. Feodor has experience with SQL Server 2000 and later versions, and is certified in SQL Server 2008. In this article Feodor explains the server-client-server process, and concentrated on the mutual waits between client and SQL Server. This is essential in grasping the concept of waits in a ‘global’ application plan. Recently I was asked to write a blog post about the wait statistics in SQL Server and since I had been thinking about writing it for quite some time now, here it is. It is a wide-spread idea that the wait statistics in SQL Server will tell you everything about your performance. Well, almost. Or should I say – barely. The reason for this is that SQL Server is always a part of a bigger system – there are always other players in the game: whether it is a client application, web service, any other kind of data import/export process and so on. In short, the SQL Server surroundings look like this: This means that SQL Server, aside from its internal waits, also depends on external waits and settings. As we can see in the picture above, SQL Server needs to have an interface in order to communicate with the surrounding clients over the network. For this communication, SQL Server uses protocol interfaces. I will not go into detail about which protocols are best, but you can read this article. Also, review the information about the TDS (Tabular data stream). As we all know, our system is only as fast as its slowest component. This means that when we look at our environment as a whole, the SQL Server might be a victim of external pressure, no matter how well we have tuned our database server performance. Let’s dive into an example: let’s say that we have a web server, hosting a web application which is using data from our SQL Server, hosted on another server. The network card of the web server for some reason is malfunctioning (think of a hardware failure, driver failure, or just improper setup) and does not send/receive data faster than 10Mbs. On the other end, our SQL Server will not be able to send/receive data at a faster rate either. This means that the application users will notify the support team and will say: “My data is coming very slow.” Now, let’s move on to a bit more exciting example: imagine that there is a similar setup as the example above – one web server and one database server, and the application is not using any stored procedure calls, but instead for every user request the application is sending 80kb query over the network to the SQL Server. (I really thought this does not happen in real life until I saw it one day.) So, what happens in this case? To make things worse, let’s say that the 80kb query text is submitted from the application to the SQL Server at least 100 times per minute, and as often as 300 times per minute in peak times. Here is what happens: in order for this query to reach the SQL Server, it will have to be broken into a of number network packets (according to the packet size settings) – and will travel over the network. On the other side, our SQL Server network card will receive the packets, will pass them to our network layer, the packets will get assembled, and eventually SQL Server will start processing the query – parsing, allegorizing, generating the query execution plan and so on. So far, we have already had a serious network overhead by waiting for the packets to reach our Database Engine. There will certainly be some processing overhead – until the database engine deals with the 80kb query and its 20 subqueries. The waits you see in the DMVs are actually collected from the point the query reaches the SQL Server and the packets are assembled. Let’s say that our query is processed and it finally returns 15000 rows. These rows have a certain size as well, depending on the data types returned. This means that the data will have converted to packages (depending on the network size package settings) and will have to reach the application server. There will also be waits, however, this time you will be able to see a wait type in the DMVs called ASYNC_NETWORK_IO. What this wait type indicates is that the client is not consuming the data fast enough and the network buffers are filling up. Recently Pinal Dave posted a blog on Client Statistics. What Client Statistics does is captures the physical flow characteristics of the query between the client(Management Studio, in this case) and the server and back to the client. As you see in the image, there are three categories: Query Profile Statistics, Network Statistics and Time Statistics. Number of server roundtrips–a roundtrip consists of a request sent to the server and a reply from the server to the client. For example, if your query has three select statements, and they are separated by ‘GO’ command, then there will be three different roundtrips. TDS Packets sent from the client – TDS (tabular data stream) is the language which SQL Server speaks, and in order for applications to communicate with SQL Server, they need to pack the requests in TDS packets. TDS Packets sent from the client is the number of packets sent from the client; in case the request is large, then it may need more buffers, and eventually might even need more server roundtrips. TDS packets received from server –is the TDS packets sent by the server to the client during the query execution. Bytes sent from client – is the volume of the data set to our SQL Server, measured in bytes; i.e. how big of a query we have sent to the SQL Server. This is why it is best to use stored procedures, since the reusable code (which already exists as an object in the SQL Server) will only be called as a name of procedure + parameters, and this will minimize the network pressure. Bytes received from server – is the amount of data the SQL Server has sent to the client, measured in bytes. Depending on the number of rows and the datatypes involved, this number will vary. But still, think about the network load when you request data from SQL Server. Client processing time – is the amount of time spent in milliseconds between the first received response packet and the last received response packet by the client. Wait time on server replies – is the time in milliseconds between the last request packet which left the client and the first response packet which came back from the server to the client. Total execution time – is the sum of client processing time and wait time on server replies (the SQL Server internal processing time) Here is an illustration of the Client-server communication model which should help you understand the mutual waits in a client-server environment. Keep in mind that a query with a large ‘wait time on server replies’ means the server took a long time to produce the very first row. This is usual on queries that have operators that need the entire sub-query to evaluate before they proceed (for example, sort and top operators). However, a query with a very short ‘wait time on server replies’ means that the query was able to return the first row fast. However a long ‘client processing time’ does not necessarily imply the client spent a lot of time processing and the server was blocked waiting on the client. It can simply mean that the server continued to return rows from the result and this is how long it took until the very last row was returned. The bottom line is that developers and DBAs should work together and think carefully of the resource utilization in the client-server environment. From experience I can say that so far I have seen only cases when the application developers and the Database developers are on their own and do not ask questions about the other party’s world. I would recommend using the Client Statistics tool during new development to track the performance of the queries, and also to find a synchronous way of utilizing resources between the client – server – client. Here is another example: think about similar setup as above, but add another server to the game. Let’s say that we keep our media on a separate server, and together with the data from our SQL Server we need to display some images on the webpage requested by our user. No matter how simple or complicated the logic to get the images is, if the images are 500kb each our users will get the page slowly and they will still think that there is something wrong with our data. Anyway, I don’t mean to get carried away too far from SQL Server. Instead, what I would like to say is that DBAs should also be aware of ‘the big picture’. I wrote a blog post a while back on this topic, and if you are interested, you can read it here about the big picture. And finally, here are some guidelines for monitoring the network performance and improving it: Run a trace and outline all queries that return more than 1000 rows (in Profiler you can actually filter and sort the captured trace by number of returned rows). This is not a set number; it is more of a guideline. The general thought is that no application user can consume that many rows at once. Ask yourself and your fellow-developers: ‘why?’. Monitor your network counters in Perfmon: Network Interface:Output queue length, Redirector:Network errors/sec, TCPv4: Segments retransmitted/sec and so on. Make sure to establish a good friendship with your network administrator (buy them coffee, for example J ) and get into a conversation about the network settings. Have them explain to you how the network cards are setup – are they standalone, are they ‘teamed’, what are the settings – full duplex and so on. Find some time to read a bit about networking. In this short blog post I hope I have turned your attention to ‘the big picture’ and the fact that there are other factors affecting our SQL Server, aside from its internal workings. As a further reading I would still highly recommend the Wait Stats series on this blog, also I would recommend you have the coffee break conversation with your network admin as soon as possible. This guest post is written by Feodor Georgiev. Read all the post in the Wait Types and Queue series. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, Readers Contribution, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, SQL Wait Stats, SQL Wait Types, T SQL

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• Streaming desktop with avconv - severe sound issues

  - by Tommy Brunn

  I'm trying to do some live streaming in Ubuntu 12.10, but I'm having some problems with audio. More specifically, the quality is complete garbage and it's at least 10 seconds out of sync with the video. I'm using an excellent guide found here to set up my loopback devices so that I can combine the desktop audio with the microphone input. It seems to work, as I'm able to stream both audio and video to Twitch.tv. But, as I said, the audio quality is terrible. The microphone audio is very, very low, but if I increase it, I get a horrible garbled sound that is absolutely unbearable. Nothing like that is present during VoIP calls or when recording sound alone with the sound recorder, so it's not an issue with the microphone itself. The entire audio stream is also delayed about 10-15 seconds compared to the video stream. I put together an imgur album of my settings. Here is some example output from when I'm streaming: avconv version 0.8.4-6:0.8.4-0ubuntu0.12.10.1, Copyright (c) 2000-2012 the Libav developers built on Nov 6 2012 16:51:11 with gcc 4.7.2 [x11grab @ 0x162fd80] device: :0.0+570,262 -> display: :0.0 x: 570 y: 262 width: 1280 height: 720 [x11grab @ 0x162fd80] shared memory extension found [x11grab @ 0x162fd80] Estimating duration from bitrate, this may be inaccurate Input #0, x11grab, from ':0.0+570,262': Duration: N/A, start: 1353181686.735113, bitrate: 884736 kb/s Stream #0.0: Video: rawvideo, bgra, 1280x720, 884736 kb/s, 30 tbr, 1000k tbn, 30 tbc [alsa @ 0x163fce0] capture with some ALSA plugins, especially dsnoop, may hang. [alsa @ 0x163fce0] Estimating duration from bitrate, this may be inaccurate Input #1, alsa, from 'pulse': Duration: N/A, start: 1353181686.773841, bitrate: N/A Stream #1.0: Audio: pcm_s16le, 48000 Hz, 2 channels, s16, 1536 kb/s Incompatible pixel format 'bgra' for codec 'libx264', auto-selecting format 'yuv420p' [buffer @ 0x1641ec0] w:1280 h:720 pixfmt:bgra [scale @ 0x1642480] w:1280 h:720 fmt:bgra -> w:852 h:480 fmt:yuv420p flags:0x4 [libx264 @ 0x165ae80] VBV maxrate unspecified, assuming CBR [libx264 @ 0x165ae80] using cpu capabilities: MMX2 SSE2Fast SSSE3 FastShuffle SSE4.2 [libx264 @ 0x165ae80] profile Main, level 3.1 [libx264 @ 0x165ae80] 264 - core 123 r2189 35cf912 - H.264/MPEG-4 AVC codec - Copyleft 2003-2012 - http://www.videolan.org/x264.html - options: cabac=1 ref=2 deblock=1:0:0 analyse=0x1:0x111 me=hex subme=6 psy=1 psy_rd=1.00:0.00 mixed_ref=0 me_range=16 chroma_me=1 trellis=1 8x8dct=0 cqm=0 deadzone=21,11 fast_pskip=1 chroma_qp_offset=-2 threads=4 sliced_threads=0 nr=0 decimate=1 interlaced=0 bluray_compat=0 constrained_intra=0 bframes=3 b_pyramid=0 b_adapt=1 b_bias=0 direct=1 weightb=0 open_gop=1 weightp=1 keyint=250 keyint_min=25 scenecut=40 intra_refresh=0 rc_lookahead=30 rc=cbr mbtree=1 bitrate=712 ratetol=1.0 qcomp=0.60 qpmin=0 qpmax=69 qpstep=4 vbv_maxrate=712 vbv_bufsize=512 nal_hrd=none ip_ratio=1.25 aq=1:1.00 Output #0, flv, to 'rtmp://live.justin.tv/app/live_23011330_Pt1plSRM0z5WVNJ0QmCHvTPmpUnfC4': Metadata: encoder : Lavf53.21.0 Stream #0.0: Video: libx264, yuv420p, 852x480, q=-1--1, 712 kb/s, 1k tbn, 30 tbc Stream #0.1: Audio: libmp3lame, 44100 Hz, 2 channels, s16, 712 kb/s Stream mapping: Stream #0:0 -> #0:0 (rawvideo -> libx264) Stream #1:0 -> #0:1 (pcm_s16le -> libmp3lame) Press ctrl-c to stop encoding frame= 17 fps= 0 q=0.0 size= 0kB time=10000000000.00 bitrate= 0.0kbitframe= 32 fps= 31 q=0.0 size= 0kB time=10000000000.00 bitrate= 0.0kbitframe= 40 fps= 23 q=29.0 size= 44kB time=0.03 bitrate=13786.2kbits/s dup=frame= 47 fps= 21 q=31.0 size= 93kB time=2.73 bitrate= 277.7kbits/s dup=0frame= 62 fps= 23 q=29.0 size= 160kB time=3.23 bitrate= 406.2kbits/s dup=0frame= 77 fps= 24 q=23.0 size= 209kB time=3.71 bitrate= 462.5kbits/s dup=0frame= 92 fps= 25 q=20.0 size= 267kB time=4.91 bitrate= 445.2kbits/s dup=0frame= 107 fps= 25 q=20.0 size= 318kB time=5.41 bitrate= 482.1kbits/s dup=0frame= 123 fps= 26 q=18.0 size= 368kB time=5.96 bitrate= 505.7kbits/s dup=0frame= 139 fps= 26 q=16.0 size= 419kB time=6.48 bitrate= 529.7kbits/s dup=0frame= 155 fps= 27 q=15.0 size= 473kB time=7.00 bitrate= 553.6kbits/s dup=0frame= 170 fps= 27 q=14.0 size= 525kB time=7.52 bitrate= 571.7kbits/s dup=0 frame= 180 fps= 25 q=-1.0 Lsize= 652kB time=7.97 bitrate= 670.0kbits/s dup=0 drop=32 //Here I stop the streaming video:531kB audio:112kB global headers:0kB muxing overhead 1.345945% [libx264 @ 0x165ae80] frame I:1 Avg QP:30.43 size: 39748 [libx264 @ 0x165ae80] frame P:45 Avg QP:11.37 size: 11110 [libx264 @ 0x165ae80] frame B:134 Avg QP:15.93 size: 27 [libx264 @ 0x165ae80] consecutive B-frames: 0.6% 0.0% 1.7% 97.8% [libx264 @ 0x165ae80] mb I I16..4: 7.3% 0.0% 92.7% [libx264 @ 0x165ae80] mb P I16..4: 0.1% 0.0% 0.1% P16..4: 49.1% 1.2% 2.1% 0.0% 0.0% skip:47.4% [libx264 @ 0x165ae80] mb B I16..4: 0.0% 0.0% 0.0% B16..8: 0.1% 0.0% 0.0% direct: 0.0% skip:99.9% L0:42.5% L1:56.9% BI: 0.6% [libx264 @ 0x165ae80] coded y,uvDC,uvAC intra: 82.3% 87.4% 71.9% inter: 7.1% 8.4% 7.0% [libx264 @ 0x165ae80] i16 v,h,dc,p: 27% 29% 16% 28% [libx264 @ 0x165ae80] i4 v,h,dc,ddl,ddr,vr,hd,vl,hu: 22% 21% 14% 8% 8% 8% 7% 5% 7% [libx264 @ 0x165ae80] i8c dc,h,v,p: 47% 22% 20% 11% [libx264 @ 0x165ae80] Weighted P-Frames: Y:0.0% UV:0.0% [libx264 @ 0x165ae80] ref P L0: 96.4% 3.6% [libx264 @ 0x165ae80] kb/s:474.19 Received signal 2: terminating. Any ideas on how I can resolve this? The video delay is perfectly acceptable, so I wouldn't think that it's a network issue that's causing the delay in the audio. Any help would be appreciated.

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• Coherence Warnings in WLS

  - by john.graves(at)oracle.com

  With 11g (10.3.4 WLS), coherence is now built into many applications.  I’ve been noticing errors in my OSB logs like these:####<10/03/2011 10:45:40 AM EST> <Warning> <Coherence> <osb-jeos> <osb_server1> <Logger@324239121 3.6.0.4> <<anonymous>> <> <583c1 0bfdbd326ba:-8c38159:12e9d02c829:-8000-0000000000000003> <1299714340643> <BEA-000000> <Oracle Coherence 3.6.0.4 (member=n/a): Unic astUdpSocket failed to set receive buffer size to 714 packets (1023KB); actual size is 12%, 89 packets (127KB). Consult your OS do cumentation regarding increasing the maximum socket buffer size. Proceeding with the actual value may cause sub-optimal performanc e.> ####<10/03/2011 10:45:40 AM EST> <Warning> <Coherence> <osb-jeos> <osb_server1> <Logger@324239121 3.6.0.4> <<anonymous>> <> <583c1 0bfdbd326ba:-8c38159:12e9d02c829:-8000-0000000000000003> <1299714340650> <BEA-000000> <Oracle Coherence 3.6.0.4 (member=n/a): Pref erredUnicastUdpSocket failed to set receive buffer size to 1428 packets (1.99MB); actual size is 6%, 89 packets (127KB). Consult y our OS documentation regarding increasing the maximum socket buffer size. Proceeding with the actual value may cause sub-optimal p erformance.> ####<10/03/2011 10:45:40 AM EST> <Warning> <Coherence> <osb-jeos> <osb_server1> <Logger@324239121 3.6.0.4> <<anonymous>> <> <583c1 0bfdbd326ba:-8c38159:12e9d02c829:-8000-0000000000000003> <1299714340659> <BEA-000000> <Oracle Coherence 3.6.0.4 (member=n/a): Mult icastUdpSocket failed to set receive buffer size to 714 packets (1023KB); actual size is 12%, 89 packets (127KB). Consult your OS documentation regarding increasing the maximum socket buffer size. Proceeding with the actual value may cause sub-optimal performa nce.> I was able to “fix” this on my ubuntu system by adding the following lines to the /etc/sysctl.conf file:# Setup networking for coherence # maximum receive socket buffer size, default 131071 net.core.rmem_max = 2000000 # maximum send socket buffer size, default 131071 net.core.wmem_max = 1000000 # default receive socket buffer size, default 65535 net.core.rmem_default = 2524287 # default send socket buffer size, default 65535 net.core.wmem_default = 2524287 .csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; }   .csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; }

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