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  • Java date long value changed after insert, select query

    - by StackExploded
    AFAIK, java Date type is independent from Timezone which means that it represents specific moment of time as long typed value. I found really weird thing here. This is the original value i tried to insert. (http-0.0.0.0-9080-4) 1352955600000 <-- long integer. (http-0.0.0.0-9080-4) Thu Nov 15 00:00:00 EST 2012 <-- User Friendly Format. After i finished inserting into Oracle 11g database, the value has changed! (http-0.0.0.0-9080-4) 1352952000000 (http-0.0.0.0-9080-4) Wed Nov 14 23:00:00 EST 2012 How could this happen?? The more weird thing is it only happens specific environments such as Jboss. I'm currently using below environments. java 1.6 ibatis 2.34 jboss-5.1 (server) tomcat 6.0 (local) oracle 11g Is there anybody who can give me a clue or link to be helpful? it really bugging me!!

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  • convert string array to cell in MATLAB

    - by Maddy
    I need to output a cell to an excel file. Before this I need to convert a date column of integers to datestrings. I know how to do this, but I am not able to put this new string array back into the cell - mycell = { 'AIR' [780] [1] [734472] [0.01] ; ... 'ABC' [780] [1] [734472] [0.02]} I did this -- dates = datestr(cell2mat(mycell(:,4))) ; What I need as an answer is: {'AIR' [780] [1] '14-Dec-2010' [0.01] ; 'ABC' [780] [1] '23-Dec-2010' [0.03] ; } so that I can now send it to an excel file using xlswrite.m

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  • recursive enumeration of integer subsets?

    - by KDaker
    I have an NSArray of NSNumbers with integer values such as [1,10,3]. I want to get the sum of all the possible subsets of these numbers. For example for 1,10 and 3 i would get: 1, 10, 3, 1+10=11, 1+3=4, 10+3=13, 1+10+3=14 there are 2^n possible combinations. I understand the math of it but im having difficulties putting this into code. so how can i put this into a method that would take the initial array of numbers and return an array with all the sums of the subsets? e.g -(NSArray *) getSums:(NSArray *)numbers; I understand that the results grow exponentially but im going to be using it for small sets of numbers.

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  • It is the arranging game in which

    - by bachchan
    1 2 3 13 5 4 7 10 6 14 11 9 8 15 12 1.Every time when we refresh the page the numbers in the cells will change but the These numbers will remain unique n from 1 to 15 2.Whenever we double click the number in the cell which is surrounded the empty cell then it will replace the empty cell with that number n that number cell become empty. 3.If we double click the cell which is not surrounded the empty cell then it will not replace the empty cell. 4.e.g. if we click 8 then it will not move to empty cell But if we click either 13, 7 , or 11 then it will move to empty cell 5.And every time when we click the cell it’s num color will change for a moment

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  • Where to store users visited pages?

    - by kofto4ka
    Hi there. I have a project, where I have posts for example. The task is next: I must show to user his last posts visit. This is my solution: every time user visits new (for him) topic, I create a new record in table visits. Table visits has next structure: id, user_id, post_id, last_visit. Now my tables visits has ~14,000,000 records and its still growing every day.. May be my solution isnt optimal and exists another way how to store users visits? Its important to save every visit as standalone record, because I also have feature to select and use users visits. And I cant purge this table, because data could be needed later month, year. How I could optimize this situation?

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  • Need an efficient algorithm solve this kind of complex structure

    - by Rizvan
    Problem Statement is : Given 2 Dimensional array, print output for example If 4 rows and 6 columns, output would be: 1 2 3 4 5 6 16 17 18 19 20 7 15 24 23 22 21 8 14 13 12 11 10 9 I tried it is looking like square within square but when I attempted this problem, I put so many while and if loops but didn't got exact answer. If row and columns increases how to handle it? This is not homework. I was learning solving complex structure so I need to understand it by some guidance.

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  • Uploading Files to AWS S3 from an Android App

    - by Abhishek Kaushik
    Edited 7th June,14 My Android app needs to have a feature where clients can upload their files. I want AWS S3 as my storage. Moreover i dont want to use SECRET_KEY and ACCESS_KEY_ID on client side. What is the the best way to do this. Can someone provide the working code too ? I read that i can request to AWS for a signed URL and then make client directly upload to that URL. How to achieve this ?

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  • How to roeder the rows of one matrix with respect to the other matrix?

    - by user2806363
    I have two big matrices A and B with diffrent dimensions.I want to order the rows of matrix B with respect to rows of the matrix A. and add the rows with values 0 to matrix B, if that row is not exist in B but in A Here is the reproduceable example and expected output: A<-matrix(c(1:40), ncol=8) rownames(A)<-c("B", "A", "C", "D", "E") > A [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] B 1 6 11 16 21 26 31 36 A 2 7 12 17 22 27 32 37 C 3 8 13 18 23 28 33 38 D 4 9 14 19 24 29 34 39 E 5 10 15 20 25 30 35 40 > B<-matrix(c(100:108),ncol=3) rownames(B)<-c("A", "E", "C") > B [,1] [,2] [,3] A 100 103 106 E 101 104 107 C 102 105 108 Here is the Expected output : >B [,1] [,2] [,3] B 0 0 0 A 100 103 106 C 102 105 108 D 0 0 0 E 101 104 107 > Would someone help me to implement this in R ?

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  • Objective-C FontAwesome

    - by sdover102
    I'm Attempting to get FontAwesome up and running on an iOS app and could use a little assistance. I have the following code for iOS: UIBarButtonItem * viewDeckButton = [[UIBarButtonItem alloc] initWithTitle:@"\uf0c9" style:UIBarButtonItemStyleBordered target:self.viewDeckController action:@selector(toggleLeftView)]; NSDictionary * customTextAttrs = [NSDictionary dictionaryWithObjectsAndKeys: [UIFont fontWithName:@"FontAwesome" size:14.0], UITextAttributeFont, nil]; [viewDeckButton setTitleTextAttributes:customTextAttrs forState:UIControlStateNormal]; @"\uf0c9" corresponds to the css class, icon-reorder. The font appears to be installed on my system (see http://cl.ly/image/2F1x1z2H0i2N). I'm getting the standard box character, as if the font is not loaded (see http://madmonkdev.com/badchar.png). Any help would be appreciated.

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  • java - reduce external jar file size

    - by joe_shmoe
    Hi all, still learning, so be patient :) I've developed a module for a Java project. The module depends on external library (fastutil). the problem is, the fastutil.jar file is a couple of times heavier than the whole project itself (14 MB). I only use a tiny subset of the classes from the library. the module is now finished, and no-one is likely to extend it in future. is there a way I could extract only the relevant class to some fastutil_small.jar so that others don't have to download all this extra weight? there's probably a simple answer to this, but as I said, I still consider myself a noob. Thanks a lot

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  • Is it good to create a usercontrol for Recursive code in xaml?

    - by user281947
    <Border BorderBrush="#C4C8CC" BorderThickness="0,0,0,1"> <TextBlock x:Name="SectionTitle" FontFamily="Trebuchet MS" FontSize="14" FontWeight="Bold" Foreground="#3D3D3D" /> </Border> I have to use the same above format at many places in a single xaml page, so for this i created a usercontrol and defined the above code inside it. So my question is, What i am doing is it right approach? Will it make the page to load slower then the above code used as it is without defining it in a new user control?

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  • What does the R function `poly` really do?

    - by merlin2011
    I have read through the manual page ?poly (which I admit I did not completely comphrehend) and also read the description of the function in book Introduction to Statistical Learning. My current understanding is that a call to poly(horsepower, 2) should be equivalent to writing horsepower + I(horsepower^2). However, this seems to be contradicted by the output of the following code. library(ISLR) summary(lm(mpg~poly(horsepower,2), data=Auto))$coef summary(lm(mpg~horsepower+I(horsepower^2), data=Auto))$coef Output: Estimate Std. Error t value Pr(>|t|) (Intercept) 23.44592 0.2209163 106.13030 2.752212e-289 poly(horsepower, 2)1 -120.13774 4.3739206 -27.46683 4.169400e-93 poly(horsepower, 2)2 44.08953 4.3739206 10.08009 2.196340e-21 Estimate Std. Error t value Pr(>|t|) (Intercept) 56.900099702 1.8004268063 31.60367 1.740911e-109 horsepower -0.466189630 0.0311246171 -14.97816 2.289429e-40 I(horsepower^2) 0.001230536 0.0001220759 10.08009 2.196340e-21 My question is, why does the output not match, and what is poly really doing?

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  • Row for each hour even if there is no record

    - by peku33
    I've got a trouble with creating mysql Query. My PHP script executes this query on each run: INSERT INTO Executes SET UserIp='%s' (%s is user IP) Executes table is: ExecuteId UNSIGNED BIGINT AI PRIMARY Date TIMESTAMP DEFAULT CURRENT_TIMESTAMP INDEX UserIp CHAR(24) ... | Some Columns I want to retrive number of Executes in each hour. The most obvious solution would be: SELECT COUNT(*) as ExecutesNum, DATE(Date) as D, HOUR(Date) as H GROUP BY D, H And it works, BUT it does not create rows for hours where there were no executes. What should I modify to get result like: 1 | 2012-09-01 | 14 **0 | 2012-09-01 | 15** 11 | 2012-09-01 | 16 1 | 2012-09-01 | 17

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  • Is it good to create a usercontrol for Recurrsive code in xaml?

    - by user281947
    <Border BorderBrush="#C4C8CC" BorderThickness="0,0,0,1"> <TextBlock x:Name="SectionTitle" FontFamily="Trebuchet MS" FontSize="14" FontWeight="Bold" Foreground="#3D3D3D" /> </Border> I have to use the same above format at many places in a single xaml page, so for this i created a usercontrol and defined the above code inside it. So my question is, what i am doing is it right approach ? Will it make the page to load slower then the above code used as it is without defining it in a new user control?

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  • realizing number ...how?? [closed]

    - by gcc
    i hold input like that A is char pointer A[0]=n A[1]=j A[2]=n A[3]=d . there is one number in A[] and every A[i] is important for me because what will i do in next step is determined by input in A[i] or A[n] A[j]=$ . A[i]=14(any number) . . int func(int temp) { if(temp=='n') ..do something then return 10; if(temp=='j') .. return 11; if(temp=='d') .. return 12; if(........) when temp find/realize number ,i wanna return 13; in if statement, what code should i write } how i can do }

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  • How to get top/left x/y of image map with javascript / jquery?

    - by jpea
    Using jQuery's position() or offset(), I can't seeme to get the top/left coordinates of an image map area. It works in FF, but nothing else - no webkit, IE, Opera. $('area').bind("click",function(){ alert($(this).position().left); }); <area shape="rect" coords="14,25,205,150" href="#"> Anyone know of a different way to access these? Normally I would just take the coords and split(",") but there are a bunch of multi-faceted area's on these pages.

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  • Updating multiple rows with an array

    - by Copephobia
    I have a table that holds user information. One of the columns holds the position of the user in the game they are in. When a game is being created, I need to update the positions of the users of each team. Here is an example: Game id : 7 Team 1 users : 1,2 Team 2 users : 3,4 team1_position : array(1,2) team2_position : array(13,14) What I want to do is update the user table using the array of positions in the SET area. My goal is to be able to update the users without the need for their id (I have different size game boards, so I have multiple position arrays for each board size) How can I do something like this: UPDATE user SET position='(team1_position)' WHERE game = '7' AND team = '1' I feel like it would be a waste of resources to select all the id's of each team and update them separately.

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  • MySql. What is there error on this insert+select+where statement?

    - by acidzombie24
    What is wrong with this statement? It works with sqlite and MS sql (last time i tested it) I would like a select unless there is no match with name or email. Then i would like to insert. I have it as one statement because its easy to keep concurrent safe as one statement. (its not that complex of a statement). INSERT INTO `user_data` ( `last_login_ip`, `login_date`, ...) SELECT @0, @14 WHERE not exists (SELECT * FROM `user_data` WHERE `name` = @15 OR `unconfirmed_email` = @16 LIMIT 1); Exception You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE not exists (SELECT * FROM `user_data` WHERE `name` = 'thename' OR `unconfirmed' at line 31

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  • Why is this writing part of the text to a new line? (Python)

    - by whatsherface
    I'm adding some new bits to one of the lines in a text file and then writing it along with the rest of the lines in the file to a new file. Referring to the if statement, I that to be all on the same line: x = 13.55553e9 y = 14.55553e9 z = 15.55553e9 infname = 'afilename' outfname = 'anotherone' oldfile = open(infname) lnum=1 for line in oldfile: if (lnum==18): line = "{0:.2e}".format(x)+' '+line+' '+"{0:.2e}".format(y)+' '+ {0:.2e}".format(z) newfile = open(outfname,'w') newfile.write(line) lnum=lnum+1 oldfile.close() newfile.close() but y and z are being written on the line below the rest of it. What am I missing here?

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  • Humanizing time

    - by keruilin
    I have a number of products that are perishable. Therefore, each product has an attribute called hours_expiration that tells how many hours the product can be used before it goes bad. For ex, apple expires in 168 hours; nut expires in 4320 hours. Given, the product's hours-to-expiration and the current time (Time.now or Date.now), how can I humanize the time-to-expiration in some of the following sample ways? Your item is set to expire in about: 6 months and 14 days 1 month and 13 days 1 month and 1 day 27 days 1 day 23 hours 1 hour 50 minutes 1 minute Looking for something robust and simple!

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  • Problem installing Ubuntu 10.04 64 bit side by side with Vista by using a bootable USB drive. What n

    - by Adam Siddhi
    What happened I decided to install Ubuntu 10.04 64 bit side by side with Vista Home Premium (I guess on another partition) with a USB stick. I found instructions on how to do this here: https://help.ubuntu.com/community/Installation/FromUSBStick To create the bootable USB drive I had to download a program called Unetbootin. That process was simple enough. All I had to do was just choose the disk image option, select the ubuntu-10.04-desktop-amd64.iso image, make sure it recognizes my USB drive and then press OK. It takes only like a few minutes to create a working bootable USB drive. Then I have to restart my computer, enter the BIOS, select my USB drive as the first boot drive, save options and continue with booting up. After this Ubuntu actually loads up. I think this is known as the Live version of Ubuntu so you can try it out before fully installing it. Any ways, on the Ubuntu 10.04 desktop I saw an installer. I click it and begin the installation process. Just so you know, I tried installing it 2 times. I will explain what happened each time: The first time I tried installing Ubuntu 10.04 I got stuck at step 4 of 7. I remember selecting the last option in the window which was Specify Partitions Manually (Advanced) I made my partition for Ubuntu like 52 gigs. I clicked forward and a little pop up window appeared saying Please Wait. So the installation process stalled on this window so I closed out of it and quit the installation process. So at this point I was worried because I had already selected the partition size and assumed it started making it. Since it stalled I had to quit out though. Anyways, once again I reached step 4 of 7 a decided to select the first option which is Install them side by side choosing between them each startup. I figured this was the safe way to go. I did that and the pop up window saying Please Wait popped up again but lasted only like 10 seconds. Then I got to I guess step 6 where it asks you to enter your desired name and password. Did that and clicked forward. The Ubuntu 10.04 installation load screen appeared and the loading bar at the bottom started filling up. So I got to 83% and stalled during the Importing other profile information (I think it was called this. I had the option to do this during I think step 6) process. So at this point I decided to get stop the installation process. I was getting very nervous. I tried to restart the computer but all that happened was that Ubuntu restarted. I finally got the computer to restart. I was pretty sure I had screwed something up big time by this point. As my computer was restarting I entered BIOS again and switched back to it booting from my main hard drive containing Vista. Saved it and continued the boot process. My worst fears were confirmed as Vista would not boot up. I mean I saw the little Microsoft Windows choppy animated green loading bar at the bottom of the screen and then boom! It decided to restart. When it restarted I had the option to run a memory test check to see if there was anything that needed to be repaired. That took like 20 minutes and at the end I saw that I did indeed have to repair something. I had to go through 2 repair processes. After each I had to restart the computer. The 2nd time it went through the repair process it said that it could not fully repair the damage. I was scared and restarted but Vista did load up. I got to my desktop and saw a message saying something like Repairs have been made, Please restart for changes to take effect I noticed that some Notification icons were missing and I could not hear volume in a video. Things were a bit funky. So I did restart and here I am. Now what?! So since I got back into Vista and thankfully have a working Internet connection I am trying to find answers to my problem (that is why I am writing this post). I am scared that I have partioned my hard drive 2 times after researching Installing Ubuntu 10.04 and seeing this post http://techie-buzz.com/foss/ubuntu-10-04-lts-installation-guide.html The author shows screen shots of installing Ubuntu 10.04. He shows the image of step 4 of 7 with a caption at the bottom. I will recreate it below: Select a partitioning option. Unless you want to format all the hard drive and install Ubuntu afresh, select the last option and proceed. Questions If I have indeed partitioned my HD 2 times (which I am sure it is), how do I get to a point where I can see all my bad, unfinished Ubuntu partitions and get rid of them? How do I clean this big mess up? & How can I ensure that this mess will not happen next time I try installing Ubuntu 10.04? Thank you Adam

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  • SharePoint 2010 – SQL Server has an unsupported version 10.0.2531.0

    - by Jeff Widmer
    I am trying to perform a database attach upgrade to SharePoint Foundation 2010. At this point I am trying to attach the content database to a Web application by using Windows Powershell: Mount-SPContentDatabase -Name <DatabaseName> -DatabaseServer <ServerName> -WebApplication <URL> [-Updateuserexperience] I am following the directions from this TechNet article: Attach databases and upgrade to SharePoint Foundation 2010.  When I go to mount the content database I am receiving this error: Mount-SPContentDatabase : Could not connect to [DATABASE_SERVER] using integrated security: SQL server at [DATABASE_SERVER] has an unsupported version 10.0.2531.0. Please refer to “http://go.microsoft.com/fwlink/?LinkId=165761” for information on the minimum required SQL Server versions and how to download them. At first this did not make sense because the default SharePoint Foundation 2010 website was running just fine.  But then I realized that the default SharePoint Foundation site runs off of SQL Server Express and that I had just installed SQL Server Web Edition (since the database is greater than 4GB) and restored the database to this version of SQL Server. Checking the documentation link above I see that SharePoint Server 2010 requires a 64-bit edition of SQL Server with the minimum required SQL Server versions as follows: SQL Server 2008 Express Edition Service Pack 1, version number 10.0.2531 SQL Server 2005 Service Pack 3 cumulative update package 3, version number 9.00.4220.00 SQL Server 2008 Service Pack 1 cumulative update package 2, version number 10.00.2714.00 The version of SQL Server 2008 Web Edition with Service Pack 1 (the version I installed on this machine) is 10.0.2531.0. SELECT @@VERSION: Microsoft SQL Server 2008 (SP1) - 10.0.2531.0 (X64)   Mar 29 2009 10:11:52   Copyright (c) 1988-2008 Microsoft Corporation  Web Edition (64-bit) on Windows NT 6.1 <X64> (Build 7600: ) (VM) But I had to read the article several times since the minimum version number for SQL Server Express is 10.0.2531.0.  At first I thought I was good with the version of SQL Server 2008 Web that I had installed, also 10.0.2531.0.  But then I read further to see that there is a cumulative update (hotfix) for SQL Server 2008 SP1 (NOT the Express edition) that is required for SharePoint 2010 and will bump the version number to 10.0.2714.00. So the solution was to install the Cumulative update package 2 for SQL Server 2008 Service Pack 1 on my SQL Server 2008 Web Edition to allow SharePoint 2010 to work with SQL Server 2008 (other than the SQL Server 2008 Express version). SELECT @@VERSION (After installing Cumulative update package 2): Microsoft SQL Server 2008 (SP1) - 10.0.2714.0 (X64)   May 14 2009 16:08:52   Copyright (c) 1988-2008 Microsoft Corporation  Web Edition (64-bit) on Windows NT 6.1 <X64> (Build 7600: ) (VM)

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  • SQL SERVER – Summary of Month – Wait Type – Day 28 of 28

    - by pinaldave
    I am glad to announce that the month of Wait Types and Queues very successful. I am glad that it was very well received and there was great amount of participation from community. I am fortunate to have some of the excellent comments throughout the series. I want to dedicate this series to all the guest blogger – Jonathan, Jacob, Glenn, and Feodor for their kindness to take a participation in this series. Here is the complete list of the blog posts in this series. I enjoyed writing the series and I plan to continue writing similar series. Please offer your opinion. SQL SERVER – Introduction to Wait Stats and Wait Types – Wait Type – Day 1 of 28 SQL SERVER – Signal Wait Time Introduction with Simple Example – Wait Type – Day 2 of 28 SQL SERVER – DMV – sys.dm_os_wait_stats Explanation – Wait Type – Day 3 of 28 SQL SERVER – DMV – sys.dm_os_waiting_tasks and sys.dm_exec_requests – Wait Type – Day 4 of 28 SQL SERVER – Capturing Wait Types and Wait Stats Information at Interval – Wait Type – Day 5 of 28 SQL SERVER – CXPACKET – Parallelism – Usual Solution – Wait Type – Day 6 of 28 SQL SERVER – CXPACKET – Parallelism – Advanced Solution – Wait Type – Day 7 of 28 SQL SERVER – SOS_SCHEDULER_YIELD – Wait Type – Day 8 of 28 SQL SERVER – PAGEIOLATCH_DT, PAGEIOLATCH_EX, PAGEIOLATCH_KP, PAGEIOLATCH_SH, PAGEIOLATCH_UP – Wait Type – Day 9 of 28 SQL SERVER – IO_COMPLETION – Wait Type – Day 10 of 28 SQL SERVER – ASYNC_IO_COMPLETION – Wait Type – Day 11 of 28 SQL SERVER – PAGELATCH_DT, PAGELATCH_EX, PAGELATCH_KP, PAGELATCH_SH, PAGELATCH_UP – Wait Type – Day 12 of 28 SQL SERVER – FT_IFTS_SCHEDULER_IDLE_WAIT – Full Text – Wait Type – Day 13 of 28 SQL SERVER – BACKUPIO, BACKUPBUFFER – Wait Type – Day 14 of 28 SQL SERVER – LCK_M_XXX – Wait Type – Day 15 of 28 SQL SERVER – Guest Post – Jonathan Kehayias – Wait Type – Day 16 of 28 SQL SERVER – WRITELOG – Wait Type – Day 17 of 28 SQL SERVER – LOGBUFFER – Wait Type – Day 18 of 28 SQL SERVER – PREEMPTIVE and Non-PREEMPTIVE – Wait Type – Day 19 of 28 SQL SERVER – MSQL_XP – Wait Type – Day 20 of 28 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 SQL SERVER – Guest Post – Jacob Sebastian – Filestream – Wait Types – Wait Queues – Day 22 of 28 SQL SERVER – OLEDB – Link Server – Wait Type – Day 23 of 28 SQL SERVER – 2000 – DBCC SQLPERF(waitstats) – Wait Type – Day 24 of 28 SQL SERVER – 2011 – Wait Type – Day 25 of 28 SQL SERVER – Guest Post – Glenn Berry – Wait Type – Day 26 of 28 SQL SERVER – Best Reference – Wait Type – Day 27 of 28 SQL SERVER – Summary of Month – Wait Type – Day 28 of 28 Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL Optimization, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, SQL Wait Stats, SQL Wait Types, SQLServer, T SQL, Technology

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  • Customize Team Build 2010 – Part 11: Speed up opening my build process template

    In the series the following parts have been published Part 1: Introduction Part 2: Add arguments and variables Part 3: Use more complex arguments Part 4: Create your own activity Part 5: Increase AssemblyVersion Part 6: Use custom type for an argument Part 7: How is the custom assembly found Part 8: Send information to the build log Part 9: Impersonate activities (run under other credentials) Part 10: Include Version Number in the Build Number Part 11: Speed up opening my build process template Part 12: How to debug my custom activities Part 13: Get control over the Build Output Part 14: Execute a PowerShell script Part 15: Fail a build based on the exit code of a console application       When you open the build process template, it takes 15 – 30 seconds until it opens. When you are in the process of creating your custom build process template, this can be very frustrating. Thanks to Ed Blankenship how has found a little trick to speed up the opening of the template. It now only takes a few seconds. Create a file called empty.xaml and place the following text in it: <Activity http://www.edsquared.com/ct.ashx?id=1746c587-59ce-45eb-85af-8ea167862617&url=http%3a%2f%2fschemas.microsoft.com%2fnetfx%2f2009%2fxaml%2factivities"http://schemas.microsoft.com/netfx/2009/xaml/activities"> </Activity> Open this file in Visual Studio. In the toolbox panel, add a new tab called “Team Foundation Build Activities”.  Note that it is important to get the tab name correct because if it is not correct then the activities will be reloaded. Inside the new tab, right click and select “Choose Items” Click the Browse button Load the file C:\Windows\Microsoft.NET\assembly\GAC_MSIL\Microsoft.TeamFoundation.Build.Workflow\v4.0_10.0.0.0__b03f5f7f11d50a3a\Microsoft.TeamFoundation.Build.Workflow.dll Click OK to add the toolbox items to the tab. Create another new tab called “Team Foundation LabManagement Activities”. Inside the new tab, right click and select “Choose Items” Click the Browse button Load the file C:\Windows\Microsoft.NET\assembly\GAC_MSIL\Microsoft.TeamFoundation.Lab.Workflow.Activities\v4.0_10.0.0.0__b03f5f7f11d50a3a\Microsoft.TeamFoundation.Lab.Workflow.Activities.dll Click OK to add the toolbox items to the tab. You can download the full solution at BuildProcess.zip. It will include the sources of every part and will continue to evolve.

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  • Oracle Coherence, Split-Brain and Recovery Protocols In Detail

    - by Ricardo Ferreira
    This article provides a high level conceptual overview of Split-Brain scenarios in distributed systems. It will focus on a specific example of cluster communication failure and recovery in Oracle Coherence. This includes a discussion on the witness protocol (used to remove failed cluster members) and the panic protocol (used to resolve Split-Brain scenarios). Note that the removal of cluster members does not necessarily indicate a Split-Brain condition. Oracle Coherence does not (and cannot) detect a Split-Brain as it occurs, the condition is only detected when cluster members that previously lost contact with each other regain contact. Cluster Topology and Configuration In order to create an good didactic for the article, let's assume a cluster topology and configuration. In this example we have a six member cluster, consisting of one JVM on each physical machine. The member IDs are as follows: Member ID  IP Address  1  10.149.155.76  2  10.149.155.77  3  10.149.155.236  4  10.149.155.75  5  10.149.155.79  6  10.149.155.78 Members 1, 2, and 3 are connected to a switch, and members 4, 5, and 6 are connected to a second switch. There is a link between the two switches, which provides network connectivity between all of the machines. Member 1 is the first member to join this cluster, thus making it the senior member. Member 6 is the last member to join this cluster. Here is a log snippet from Member 6 showing the complete member set: 2010-02-26 15:27:57.390/3.062 Oracle Coherence GE 3.5.3/465p2 <Info> (thread=main, member=6): Started DefaultCacheServer... SafeCluster: Name=cluster:0xDDEB Group{Address=224.3.5.3, Port=35465, TTL=4} MasterMemberSet ( ThisMember=Member(Id=6, Timestamp=2010-02-26 15:27:58.635, Address=10.149.155.78:8088, MachineId=1102, Location=process:228, Role=CoherenceServer) OldestMember=Member(Id=1, Timestamp=2010-02-26 15:27:06.931, Address=10.149.155.76:8088, MachineId=1100, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:511, Role=CoherenceServer) ActualMemberSet=MemberSet(Size=6, BitSetCount=2 Member(Id=1, Timestamp=2010-02-26 15:27:06.931, Address=10.149.155.76:8088, MachineId=1100, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:511, Role=CoherenceServer) Member(Id=2, Timestamp=2010-02-26 15:27:17.847, Address=10.149.155.77:8088, MachineId=1101, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:296, Role=CoherenceServer) Member(Id=3, Timestamp=2010-02-26 15:27:24.892, Address=10.149.155.236:8088, MachineId=1260, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:32459, Role=CoherenceServer) Member(Id=4, Timestamp=2010-02-26 15:27:39.574, Address=10.149.155.75:8088, MachineId=1099, Location=process:800, Role=CoherenceServer) Member(Id=5, Timestamp=2010-02-26 15:27:49.095, Address=10.149.155.79:8088, MachineId=1103, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:3229, Role=CoherenceServer) Member(Id=6, Timestamp=2010-02-26 15:27:58.635, Address=10.149.155.78:8088, MachineId=1102, Location=process:228, Role=CoherenceServer) ) RecycleMillis=120000 RecycleSet=MemberSet(Size=0, BitSetCount=0 ) ) At approximately 15:30, the connection between the two switches is severed: Thirty seconds later (the default packet timeout in development mode) the logs indicate communication failures across the cluster. In this example, the communication failure was caused by a network failure. In a production setting, this type of communication failure can have many root causes, including (but not limited to) network failures, excessive GC, high CPU utilization, swapping/virtual memory, and exceeding maximum network bandwidth. In addition, this type of failure is not necessarily indicative of a split brain. Any communication failure will be logged in this fashion. Member 2 logs a communication failure with Member 5: 2010-02-26 15:30:32.638/196.928 Oracle Coherence GE 3.5.3/465p2 <Warning> (thread=PacketPublisher, member=2): Timeout while delivering a packet; requesting the departure confirmation for Member(Id=5, Timestamp=2010-02-26 15:27:49.095, Address=10.149.155.79:8088, MachineId=1103, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:3229, Role=CoherenceServer) by MemberSet(Size=2, BitSetCount=2 Member(Id=1, Timestamp=2010-02-26 15:27:06.931, Address=10.149.155.76:8088, MachineId=1100, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:511, Role=CoherenceServer) Member(Id=4, Timestamp=2010-02-26 15:27:39.574, Address=10.149.155.75:8088, MachineId=1099, Location=process:800, Role=CoherenceServer) ) The Coherence clustering protocol (TCMP) is a reliable transport mechanism built on UDP. In order for the protocol to be reliable, it requires an acknowledgement (ACK) for each packet delivered. If a packet fails to be acknowledged within the configured timeout period, the Coherence cluster member will log a packet timeout (as seen in the log message above). When this occurs, the cluster member will consult with other members to determine who is at fault for the communication failure. If the witness members agree that the suspect member is at fault, the suspect is removed from the cluster. If the witnesses unanimously disagree, the accuser is removed. This process is known as the witness protocol. Since Member 2 cannot communicate with Member 5, it selects two witnesses (Members 1 and 4) to determine if the communication issue is with Member 5 or with itself (Member 2). However, Member 4 is on the switch that is no longer accessible by Members 1, 2 and 3; thus a packet timeout for member 4 is recorded as well: 2010-02-26 15:30:35.648/199.938 Oracle Coherence GE 3.5.3/465p2 <Warning> (thread=PacketPublisher, member=2): Timeout while delivering a packet; requesting the departure confirmation for Member(Id=4, Timestamp=2010-02-26 15:27:39.574, Address=10.149.155.75:8088, MachineId=1099, Location=process:800, Role=CoherenceServer) by MemberSet(Size=2, BitSetCount=2 Member(Id=1, Timestamp=2010-02-26 15:27:06.931, Address=10.149.155.76:8088, MachineId=1100, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:511, Role=CoherenceServer) Member(Id=6, Timestamp=2010-02-26 15:27:58.635, Address=10.149.155.78:8088, MachineId=1102, Location=process:228, Role=CoherenceServer) ) Member 1 has the ability to confirm the departure of member 4, however Member 6 cannot as it is also inaccessible. At the same time, Member 3 sends a request to remove Member 6, which is followed by a report from Member 3 indicating that Member 6 has departed the cluster: 2010-02-26 15:30:35.706/199.996 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Cluster, member=2): MemberLeft request for Member 6 received from Member(Id=3, Timestamp=2010-02-26 15:27:24.892, Address=10.149.155.236:8088, MachineId=1260, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:32459, Role=CoherenceServer) 2010-02-26 15:30:35.709/199.999 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Cluster, member=2): MemberLeft notification for Member 6 received from Member(Id=3, Timestamp=2010-02-26 15:27:24.892, Address=10.149.155.236:8088, MachineId=1260, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:32459, Role=CoherenceServer) The log for Member 3 determines how Member 6 departed the cluster: 2010-02-26 15:30:35.161/191.694 Oracle Coherence GE 3.5.3/465p2 <Warning> (thread=PacketPublisher, member=3): Timeout while delivering a packet; requesting the departure confirmation for Member(Id=6, Timestamp=2010-02-26 15:27:58.635, Address=10.149.155.78:8088, MachineId=1102, Location=process:228, Role=CoherenceServer) by MemberSet(Size=2, BitSetCount=2 Member(Id=1, Timestamp=2010-02-26 15:27:06.931, Address=10.149.155.76:8088, MachineId=1100, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:511, Role=CoherenceServer) Member(Id=2, Timestamp=2010-02-26 15:27:17.847, Address=10.149.155.77:8088, MachineId=1101, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:296, Role=CoherenceServer) ) 2010-02-26 15:30:35.165/191.698 Oracle Coherence GE 3.5.3/465p2 <Info> (thread=Cluster, member=3): Member departure confirmed by MemberSet(Size=2, BitSetCount=2 Member(Id=1, Timestamp=2010-02-26 15:27:06.931, Address=10.149.155.76:8088, MachineId=1100, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:511, Role=CoherenceServer) Member(Id=2, Timestamp=2010-02-26 15:27:17.847, Address=10.149.155.77:8088, MachineId=1101, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:296, Role=CoherenceServer) ); removing Member(Id=6, Timestamp=2010-02-26 15:27:58.635, Address=10.149.155.78:8088, MachineId=1102, Location=process:228, Role=CoherenceServer) In this case, Member 3 happened to select two witnesses that it still had connectivity with (Members 1 and 2) thus resulting in a simple decision to remove Member 6. Given the departure of Member 6, Member 2 is left with a single witness to confirm the departure of Member 4: 2010-02-26 15:30:35.713/200.003 Oracle Coherence GE 3.5.3/465p2 <Info> (thread=Cluster, member=2): Member departure confirmed by MemberSet(Size=1, BitSetCount=2 Member(Id=1, Timestamp=2010-02-26 15:27:06.931, Address=10.149.155.76:8088, MachineId=1100, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:511, Role=CoherenceServer) ); removing Member(Id=4, Timestamp=2010-02-26 15:27:39.574, Address=10.149.155.75:8088, MachineId=1099, Location=process:800, Role=CoherenceServer) In the meantime, Member 4 logs a missing heartbeat from the senior member. This message is also logged on Members 5 and 6. 2010-02-26 15:30:07.906/150.453 Oracle Coherence GE 3.5.3/465p2 <Info> (thread=PacketListenerN, member=4): Scheduled senior member heartbeat is overdue; rejoining multicast group. Next, Member 4 logs a TcpRing failure with Member 2, thus resulting in the termination of Member 2: 2010-02-26 15:30:21.421/163.968 Oracle Coherence GE 3.5.3/465p2 <D4> (thread=Cluster, member=4): TcpRing: Number of socket exceptions exceeded maximum; last was "java.net.SocketTimeoutException: connect timed out"; removing the member: 2 For quick process termination detection, Oracle Coherence utilizes a feature called TcpRing which is a sparse collection of TCP/IP-based connections between different members in the cluster. Each member in the cluster is connected to at least one other member, which (if at all possible) is running on a different physical box. This connection is not used for any data transfer, only heartbeat communications are sent once a second per each link. If a certain number of exceptions are thrown while trying to re-establish a connection, the member throwing the exceptions is removed from the cluster. Member 5 logs a packet timeout with Member 3 and cites witnesses Members 4 and 6: 2010-02-26 15:30:29.791/165.037 Oracle Coherence GE 3.5.3/465p2 <Warning> (thread=PacketPublisher, member=5): Timeout while delivering a packet; requesting the departure confirmation for Member(Id=3, Timestamp=2010-02-26 15:27:24.892, Address=10.149.155.236:8088, MachineId=1260, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:32459, Role=CoherenceServer) by MemberSet(Size=2, BitSetCount=2 Member(Id=4, Timestamp=2010-02-26 15:27:39.574, Address=10.149.155.75:8088, MachineId=1099, Location=process:800, Role=CoherenceServer) Member(Id=6, Timestamp=2010-02-26 15:27:58.635, Address=10.149.155.78:8088, MachineId=1102, Location=process:228, Role=CoherenceServer) ) 2010-02-26 15:30:29.798/165.044 Oracle Coherence GE 3.5.3/465p2 <Info> (thread=Cluster, member=5): Member departure confirmed by MemberSet(Size=2, BitSetCount=2 Member(Id=4, Timestamp=2010-02-26 15:27:39.574, Address=10.149.155.75:8088, MachineId=1099, Location=process:800, Role=CoherenceServer) Member(Id=6, Timestamp=2010-02-26 15:27:58.635, Address=10.149.155.78:8088, MachineId=1102, Location=process:228, Role=CoherenceServer) ); removing Member(Id=3, Timestamp=2010-02-26 15:27:24.892, Address=10.149.155.236:8088, MachineId=1260, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:32459, Role=CoherenceServer) Eventually we are left with two distinct clusters consisting of Members 1, 2, 3 and Members 4, 5, 6, respectively. In the latter cluster, Member 4 is promoted to senior member. The connection between the two switches is restored at 15:33. Upon the restoration of the connection, the cluster members immediately receive cluster heartbeats from the two senior members. In the case of Members 1, 2, and 3, the following is logged: 2010-02-26 15:33:14.970/369.066 Oracle Coherence GE 3.5.3/465p2 <Warning> (thread=Cluster, member=1): The member formerly known as Member(Id=4, Timestamp=2010-02-26 15:30:35.341, Address=10.149.155.75:8088, MachineId=1099, Location=process:800, Role=CoherenceServer) has been forcefully evicted from the cluster, but continues to emit a cluster heartbeat; henceforth, the member will be shunned and its messages will be ignored. Likewise for Members 4, 5, and 6: 2010-02-26 15:33:14.343/336.890 Oracle Coherence GE 3.5.3/465p2 <Warning> (thread=Cluster, member=4): The member formerly known as Member(Id=1, Timestamp=2010-02-26 15:30:31.64, Address=10.149.155.76:8088, MachineId=1100, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:511, Role=CoherenceServer) has been forcefully evicted from the cluster, but continues to emit a cluster heartbeat; henceforth, the member will be shunned and its messages will be ignored. This message indicates that a senior heartbeat is being received from members that were previously removed from the cluster, in other words, something that should not be possible. For this reason, the recipients of these messages will initially ignore them. After several iterations of these messages, the existence of multiple clusters is acknowledged, thus triggering the panic protocol to reconcile this situation. When the presence of more than one cluster (i.e. Split-Brain) is detected by a Coherence member, the panic protocol is invoked in order to resolve the conflicting clusters and consolidate into a single cluster. The protocol consists of the removal of smaller clusters until there is one cluster remaining. In the case of equal size clusters, the one with the older Senior Member will survive. Member 1, being the oldest member, initiates the protocol: 2010-02-26 15:33:45.970/400.066 Oracle Coherence GE 3.5.3/465p2 <Warning> (thread=Cluster, member=1): An existence of a cluster island with senior Member(Id=4, Timestamp=2010-02-26 15:27:39.574, Address=10.149.155.75:8088, MachineId=1099, Location=process:800, Role=CoherenceServer) containing 3 nodes have been detected. Since this Member(Id=1, Timestamp=2010-02-26 15:27:06.931, Address=10.149.155.76:8088, MachineId=1100, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:511, Role=CoherenceServer) is the senior of an older cluster island, the panic protocol is being activated to stop the other island's senior and all junior nodes that belong to it. Member 3 receives the panic: 2010-02-26 15:33:45.803/382.336 Oracle Coherence GE 3.5.3/465p2 <Error> (thread=Cluster, member=3): Received panic from senior Member(Id=1, Timestamp=2010-02-26 15:27:06.931, Address=10.149.155.76:8088, MachineId=1100, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:511, Role=CoherenceServer) caused by Member(Id=4, Timestamp=2010-02-26 15:27:39.574, Address=10.149.155.75:8088, MachineId=1099, Location=process:800, Role=CoherenceServer) Member 4, the senior member of the younger cluster, receives the kill message from Member 3: 2010-02-26 15:33:44.921/367.468 Oracle Coherence GE 3.5.3/465p2 <Error> (thread=Cluster, member=4): Received a Kill message from a valid Member(Id=3, Timestamp=2010-02-26 15:27:24.892, Address=10.149.155.236:8088, MachineId=1260, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:32459, Role=CoherenceServer); stopping cluster service. In turn, Member 4 requests the departure of its junior members 5 and 6: 2010-02-26 15:33:44.921/367.468 Oracle Coherence GE 3.5.3/465p2 <Error> (thread=Cluster, member=4): Received a Kill message from a valid Member(Id=3, Timestamp=2010-02-26 15:27:24.892, Address=10.149.155.236:8088, MachineId=1260, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:32459, Role=CoherenceServer); stopping cluster service. 2010-02-26 15:33:43.343/349.015 Oracle Coherence GE 3.5.3/465p2 <Error> (thread=Cluster, member=6): Received a Kill message from a valid Member(Id=4, Timestamp=2010-02-26 15:27:39.574, Address=10.149.155.75:8088, MachineId=1099, Location=process:800, Role=CoherenceServer); stopping cluster service. Once Members 4, 5, and 6 restart, they rejoin the original cluster with senior member 1. The log below is from Member 4. Note that it receives a different member id when it rejoins the cluster. 2010-02-26 15:33:44.921/367.468 Oracle Coherence GE 3.5.3/465p2 <Error> (thread=Cluster, member=4): Received a Kill message from a valid Member(Id=3, Timestamp=2010-02-26 15:27:24.892, Address=10.149.155.236:8088, MachineId=1260, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:32459, Role=CoherenceServer); stopping cluster service. 2010-02-26 15:33:46.921/369.468 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Cluster, member=4): Service Cluster left the cluster 2010-02-26 15:33:47.046/369.593 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Invocation:InvocationService, member=4): Service InvocationService left the cluster 2010-02-26 15:33:47.046/369.593 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=OptimisticCache, member=4): Service OptimisticCache left the cluster 2010-02-26 15:33:47.046/369.593 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=ReplicatedCache, member=4): Service ReplicatedCache left the cluster 2010-02-26 15:33:47.046/369.593 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=DistributedCache, member=4): Service DistributedCache left the cluster 2010-02-26 15:33:47.046/369.593 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Invocation:Management, member=4): Service Management left the cluster 2010-02-26 15:33:47.046/369.593 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Cluster, member=4): Member 6 left service Management with senior member 5 2010-02-26 15:33:47.046/369.593 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Cluster, member=4): Member 6 left service DistributedCache with senior member 5 2010-02-26 15:33:47.046/369.593 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Cluster, member=4): Member 6 left service ReplicatedCache with senior member 5 2010-02-26 15:33:47.046/369.593 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Cluster, member=4): Member 6 left service OptimisticCache with senior member 5 2010-02-26 15:33:47.046/369.593 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Cluster, member=4): Member 6 left service InvocationService with senior member 5 2010-02-26 15:33:47.046/369.593 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Cluster, member=4): Member(Id=6, Timestamp=2010-02-26 15:33:47.046, Address=10.149.155.78:8088, MachineId=1102, Location=process:228, Role=CoherenceServer) left Cluster with senior member 4 2010-02-26 15:33:49.218/371.765 Oracle Coherence GE 3.5.3/465p2 <Info> (thread=main, member=n/a): Restarting cluster 2010-02-26 15:33:49.421/371.968 Oracle Coherence GE 3.5.3/465p2 <D5> (thread=Cluster, member=n/a): Service Cluster joined the cluster with senior service member n/a 2010-02-26 15:33:49.625/372.172 Oracle Coherence GE 3.5.3/465p2 <Info> (thread=Cluster, member=n/a): This Member(Id=5, Timestamp=2010-02-26 15:33:50.499, Address=10.149.155.75:8088, MachineId=1099, Location=process:800, Role=CoherenceServer, Edition=Grid Edition, Mode=Development, CpuCount=2, SocketCount=1) joined cluster "cluster:0xDDEB" with senior Member(Id=1, Timestamp=2010-02-26 15:27:06.931, Address=10.149.155.76:8088, MachineId=1100, Location=site:usdhcp.oraclecorp.com,machine:dhcp-burlington6-4fl-east-10-149,process:511, Role=CoherenceServer, Edition=Grid Edition, Mode=Development, CpuCount=2, SocketCount=2) Cool isn't it?

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