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• Decrementing/Incrementing loop variable inside for loop. Is this code smell?

  - by FairDune

  I have to read lines from a text file in sequential order. The file is a custom text format that contains sections. If some sections are out of order, I would like to look for the starting of the next valid section and continue processing. Currently, I have some code that looks like this: for (int currentLineIndex=0; currentLineIndex < lines.Count; currentLineIndex++ ) { //Process section here if( out_of_order_condition ) { currentLineIndex--;//Stay on the same line in the next iteration because this line may be the start of a valid section. continue; } } Is this code smell?

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• Stuck in logon loop

  - by MJeffryes

  Here's the deal. I set up a computer with Ubuntu 10.04 for my grandmother. Everything worked fine. I connected it to the internet at her house today. After rebooting the computer I found that the computer would kick you back to the logon screen if you attempted to logon to her account. It worked fine logging on to my admin account, and also in Gnome's safe mode. I thought it had resolved itself, but turns out it hadn't, and now I don't have physical access to the computer, plus the remote connection I'd hoped to use only works intermittently. I need some suggestions for troubleshooting for when I'm at her house at some point next week. Ask for any more details, but I'm afraid I won't be able to provide many more until I've checked it out in person, since she is basically unable to use a computer beyond web browsing. Thanks in advance!

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• How do I stop infinite loop? [closed]

  - by SystemNetworks

  As you see, I have a stack overflow error. I wanted to use a class (goldArmor.java) which has all its own stuffs and uses some booleans, int, double from my main class(play.java). Now I want to call my other class(goldArmor.java) to my main class(play.java). When I press run, it says stackoverflow. How do I fix it? For My goldArmor.java: Play playI = new Play(); This is what I tried: Created another class(connect) to connect from my sub-class to my play.class: goldArmor goldArm = new goldArmor(); THen in my play.java: connect con = new connect();

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• Arrays and For Loop [closed]

  - by java

  I wrote this code: import java.io.*; public class dictionary { public static void main(String args[]) { String[] MyArrayE=new String[5]; String[] MyArrayS=new String[5]; String[] MyArrayA=new String[5]; MyArrayE[0]="Language"; MyArrayE[1]="Computer"; MyArrayE[2]="Engineer"; MyArrayE[3]="Home"; MyArrayE[4]="Table"; MyArrayS[0]="Lingua"; MyArrayS[1]="Computador"; MyArrayS[2]="Ing."; MyArrayS[3]="Casa"; MyArrayS[4]="Mesa"; System.out.println("Please enter a word"); BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); String word= null; try { word= br.readLine(); } catch (IOException e) { System.out.println("Error!"); System.exit(1); } System.out.println("Your word is " + word); for(int i=0; i<MyArrayA.length; i++) { if(word.equals(MyArrayS[i])) { System.out.println(MyArrayE[i]); } } } } My Question: What about if the user inputs a word not in MyArrayS, I want to check that and print a statement like "Word does not exist". I think that it might look like: if(word!=MyArrayS) { System.out.println("Word does not exist"); }

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• Recursive Perl detail need help

  - by Catarrunas

  Hi everybody, i think this is a simple problem, but i'm stuck with it for some time now! I need a fresh pair of eyes on this. The thing is i have this code in perl: #!c:/Perl/bin/perl use CGI qw/param/; use URI::Escape; print "Content-type: text/html\n\n"; my $directory = param ('directory'); $directory = uri_unescape ($directory); my @contents; readDir($directory); foreach (@contents) { print "$_\n"; } #------------------------------------------------------------------------ sub readDir(){ my $dir = shift; opendir(DIR, $dir) or die $!; while (my $file = readdir(DIR)) { next if ($file =~ m/^\./); if(-d $dir.$file) { #print $dir.$file. " ----- DIR\n"; readDir($dir.$file); } push @contents, ($dir . $file); } closedir(DIR); } I've tried to make it recursive. I need to have all the files of all of the directories and subdirectories, with the full path, so that i can open the files in the future. But my output only returns the files in the current directory and the files in the first directory that it finds. If i have 3 folders inside the directory it only shows the first one. Ex. of cmd call: "perl readDir.pl directory=C:/PerlTest/" Thanks

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• Whats faster in Javascript a bunch of small setInterval loops, or one big one?

  - by RobertWHurst

  Just wondering if its worth it to make a monolithic loop function or just add loops were they're needed. The big loop option would just be a loop of callbacks that are added dynamically with an add function. adding a function would look like this setLoop(function(){ alert('hahaha! I\'m a really annoying loop that bugs you every tenth of a second'); }); setLoop would add the function to the monolithic loop. so is the is worth anything in performance or should I just stick to lots of little loops using setInterval?

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• Active Directory Membership Provider - how to expand on this?

  - by Jaxidian

  I'm working on getting an MVC app up and running via AD Membership Provider and I'm having some issues figuring this out. I have a base configuration setup and working when I login as [email protected] + password. <connectionStrings> <add name="MyConnString" connectionString="LDAP://domaincontroller/OU=Product Users,DC=my,DC=domain,DC=com" /> </connectionStrings> <membership defaultProvider="MyProvider"> <providers> <clear /> <add name="MyProvider" connectionStringName="MyConnString" connectionUsername="my.domain.com\service_account" connectionPassword="biguglypassword" type="System.Web.Security.ActiveDirectoryMembershipProvider, System.Web, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a" /> </providers> </membership> However, I'd LIKE to do some other things and I'm not sure how to go about them. Login without typing the domain (i.e. the "@my.domain.com"). I realize that this could only work if I limit myself to just one domain - that's fine. Organize users in up to N different OUs within a single OU. As you can tell from my current connection string, I'm authenticating users in my Product Users OU. I would LIKE to create OUs for various companies within this OU and put the users into those OUs. How can I authenticate across all of these different OUs? I'm trying to figure out how the Active Directory Membership Provider ties in with the Profile and Role providers. Are there AD versions of those too or am I stuck with SQL, home-grown, or finding something somebody else has coded up? Many thanks!!

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• Breaking the loop in Jython

  - by kdev

  Hi everyone, I have a written a loop for and if condition and I want to break to loop completely when if gets satisfied. while count: i=0 for line in read_record: #print str.strip(count[1:28]) #print str.strip(read_record[i]) if string.find(str.strip(read_record[i]),str.strip(count[1:28]))>0: code=str.strip(read_record[i+19])+str.strip(read_record[i+20]) print code[25:] break i=i+1 So here if the if string.find condition gets satisfied I want to go to while loop flow. Please tell me what will be the best place to break and how should I modify the program so that once the if condition is satisfied I am out of the loop and start again with while loop.

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• Shell script exiting the loop after calling another script

  - by Johnyy

  Hi Guys, I have a shell script s1 calling another script s2 in a loop. However s1 can not seem to continue the loop after s2 returns. Commenting out the line that calls s2 will enable the loop to continue. s2 does copy of one file, s1 checks conditions and copy several files using s2. Can anyone give a pointer what is going on here? ... while read line s2 param1 param2 param3 echo "copy done" done < $tempfile echo "out of loop" ... "copy done" is printed, so is "out of loop"

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• How can I remove all users in an Active Directory group?

  - by Beavis

  I'm trying to remove all users from an AD group with the following code: private void RemoveStudents() { foreach (DirectoryEntry childDir in rootRefreshDir.Children) { DirectoryEntry groupDE = new DirectoryEntry(childDir.Path); for (int counter = 0; counter < groupDE.Properties["member"].Count; counter++) { groupDE.Properties["member"].Remove(groupDE.Properties["member"][counter]); groupDE.CommitChanges(); groupDE.Close(); } } } The rootRefreshDir is the directory that contains all the AD groups (childDir). What I'm finding here is that this code does not behave correctly. It removes users, but it doesn't do it after the first run. It does "some". Then I run it again, and again, and again - depending on how many users need to be deleted in a group. I'm not sure why it's functioning this way. Can someone help fix this code or provide an alternative method to delete all users in a group? Thanks!

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• Hiding elements based on last closed element jquery script

  - by Jared

  Hi my question is, how can I make this jquery script close all previously opened children when entering a new parent? At the moment it traverses thru all the tree structure fine, but switching from one parent to another does not close the previous children, but rather only the each individual parents elements as a user browses. Here is the jquery I'm using: <script type="text/javascript"> $(document).ready($(function(){ $('#nav>li>ul').hide(); $('.children').hide(); $('#nav>li').mousedown(function(){ // check that the menu is not currently animated if ($('#nav ul:animated').size() == 0) { // create a reference to the active element (this) // so we don't have to keep creating a jQuery object $heading = $(this); // create a reference to visible sibling elements // so we don't have to keep creating a jQuery object $expandedSiblings = $heading.siblings().find('ul:visible'); if ($expandedSiblings.size() > 0) { $expandedSiblings.slideUp(0, function(){ $heading.find('ul').slideDown(0); }); } else { $heading.find('ul').slideDown(0); } } }); $('#nav>li>ul>li').mousedown(function(){ // check that the menu is not currently animated if ($('#nav ul:animated').size() == 0) { // create a reference to the active element (this) // so we don't have to keep creating a jQuery object $heading2 = $(this); // create a reference to visible sibling elements // so we don't have to keep creating a jQuery object $expandedSiblings2 = $heading2.siblings().find('.children:visible'); if ($expandedSiblings2.size() > 0) { $expandedSiblings2.slideUp(0, function(){ $heading2.find('.children').slideDown(0); }); } else { $heading2.find('.children').slideDown(0); } } }); })); </script> and here is my html output <ul id="nav"> <li><a href="#">folder 4</a> <ul><li><a href="#">2001</a> <ul><li class="children"><a href="./directory//folder 4/2001/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder 4/2001/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder 4/2001/doc3.txt">doc3.txt</a></li> </ul> </li> <li><a href="#">2002</a> <ul><li class="children"><a href="./directory//folder 4/2002/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder 4/2002/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder 4/2002/doc3.txt">doc3.txt</a></li> <li class="children"><a href="./directory//folder 4/2002/doc4.txt">doc4.txt</a></li> </ul> </li> <li><a href="#">2003</a> <ul><li class="children"><a href="./directory//folder 4/2003/Copy of doc1.txt">Copy of doc1.txt</a></li> <li class="children"><a href="./directory//folder 4/2003/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder 4/2003/doc2.txt">doc2.txt</a></li> </ul> </li> <li><a href="#">2004</a> <ul><li class="children"><a href="./directory//folder 4/2004/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder 4/2004/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder 4/2004/doc3.txt">doc3.txt</a></li> <li class="children"><a href="./directory//folder 4/2004/doc4.txt">doc4.txt</a></li> </ul> </li> </ul> </li> <li><a href="#">folder1</a> <ul><li><a href="#">2001</a> <ul><li class="children"><a href="./directory//folder1/2001/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder1/2001/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder1/2001/doc3.txt">doc3.txt</a></li> </ul> </li> <li><a href="#">2002</a> <ul><li class="children"><a href="./directory//folder1/2002/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder1/2002/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder1/2002/doc3.txt">doc3.txt</a></li> <li class="children"><a href="./directory//folder1/2002/doc4.txt">doc4.txt</a></li> </ul> </li> <li><a href="#">2003</a> <ul><li class="children"><a href="./directory//folder1/2003/Copy of doc1.txt">Copy of doc1.txt</a></li> <li class="children"><a href="./directory//folder1/2003/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder1/2003/doc2.txt">doc2.txt</a></li> </ul> </li> <li><a href="#">2004</a> <ul><li class="children"><a href="./directory//folder1/2004/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder1/2004/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder1/2004/doc3.txt">doc3.txt</a></li> <li class="children"><a href="./directory//folder1/2004/doc4.txt">doc4.txt</a></li> </ul> </li> </ul> </li> <li><a href="#">folder2</a> <ul><li><a href="#">2001</a> <ul><li class="children"><a href="./directory//folder2/2001/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder2/2001/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder2/2001/doc3.txt">doc3.txt</a></li> </ul> </li> <li><a href="#">2002</a> <ul><li class="children"><a href="./directory//folder2/2002/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder2/2002/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder2/2002/doc3.txt">doc3.txt</a></li> <li class="children"><a href="./directory//folder2/2002/doc4.txt">doc4.txt</a></li> </ul> </li> <li><a href="#">2003</a> <ul><li class="children"><a href="./directory//folder2/2003/Copy of doc1.txt">Copy of doc1.txt</a></li> <li class="children"><a href="./directory//folder2/2003/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder2/2003/doc2.txt">doc2.txt</a></li> </ul> </li> <li><a href="#">2004</a> <ul><li class="children"><a href="./directory//folder2/2004/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder2/2004/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder2/2004/doc3.txt">doc3.txt</a></li> <li class="children"><a href="./directory//folder2/2004/doc4.txt">doc4.txt</a></li> </ul> </li> </ul> </li> <li><a href="#">folder3</a> <ul><li><a href="#">2001</a> <ul><li class="children"><a href="./directory//folder3/2001/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder3/2001/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder3/2001/doc3.txt">doc3.txt</a></li> </ul> </li> <li><a href="#">2002</a> <ul><li class="children"><a href="./directory//folder3/2002/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder3/2002/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder3/2002/doc3.txt">doc3.txt</a></li> <li class="children"><a href="./directory//folder3/2002/doc4.txt">doc4.txt</a></li> </ul> </li> <li><a href="#">2003</a> <ul><li class="children"><a href="./directory//folder3/2003/Copy of doc1.txt">Copy of doc1.txt</a></li> <li class="children"><a href="./directory//folder3/2003/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder3/2003/doc2.txt">doc2.txt</a></li> </ul> </li> <li><a href="#">2004</a> <ul><li class="children"><a href="./directory//folder3/2004/doc1.txt">doc1.txt</a></li> <li class="children"><a href="./directory//folder3/2004/doc2.txt">doc2.txt</a></li> <li class="children"><a href="./directory//folder3/2004/doc3.txt">doc3.txt</a></li> <li class="children"><a href="./directory//folder3/2004/doc4.txt">doc4.txt</a></li> </ul> </li> </ul> </li> </ul> I assume my problem is, jquery isn't closing the children between each new parent so I need to make a call, but I'm a bit lost on how to do that. I know the code is pretty messy, this project was done in a huge rush and a very tight timeframe. Appreciate your answers and any other constructive comments, cheers :)

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• Enabling Session Directory under Terminal Server Configuration Tool and Server Settings

  - by LPE

  Yello, I'm trying to add up a Terminal Server Session Directory client to an already fully functional Session Directory cluster which today runs two clients as well as the server. I've been reading up on both Google, Microsoft KB's as well as old documentation from an earlier employee but to no avail. The step I'm stuck at is when I open up Terminal Server Configuration Tool (tscc.msc), chooses ServerSettings. I know there should be an option saying "Session Directory" on the right hand side along with Active Desktop, Licensing and whatnot, but it's not there. I've logged on to both the other already functional clients and checked the same list and there the Session Directory option sure is both visible as well as working good with the specified information. This picture is the same view that I'm looking at at the moment, but mine is missing the bottom option that says "Session Directory" http://www.inetnj.com/doc/images/TerminalServerConfiguration.jpg Any help would be greatly appriciated. Regards LPE

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• Directory Not Found Error

  - by noobguy

  I am trying to verify tails and when I get to the command prompt portion of the verification some difficulties seem to have arose. Below is the script: noob@noob-System-Product-Name:~$ cd [/media/noob/UUI] bash: cd: [/media/noob/UUI]: No such file or directory noob@noob-System-Product-Name:~$ gpg --keyid-format long --import tails-signing.key gpg: can't open `tails-signing.key': No such file or directory gpg: Total number processed: 0 Same thing happens when I try from download directory; noob@noob-System-Product-Name:~$ cd [/home/noob/Downloads] bash: cd: [/home/noob/Downloads]: No such file or directory noob@noob-System-Product-Name:~$ gpg --keyid-format long --import tails-signing.key gpg: can't open `tails-signing.key': No such file or directory gpg: Total number processed: 0 Any suggestions would be greatly appreciated.

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• Jail Linux user to directory for FTP login

  - by Greg

  I'm planning on using vsftpd to act as a secure ftp server, but I am having difficulty controlling the linux users that will be used as ftp logins. The users are required to be "jailed" into a specific directory (and subdirectories) and have full read/write access. Requirements: - User account "admin_ftp" should be jailed to /var/www directory. - Other accounts will be added as needed, for each site... e.g: - User account "picturegallery_ftp" should be jailed to /var/www/picturegallery.com directory. I have tried the following, but to no avail: # Group to store all ftp accounts in. groupadd ftp_accounts # Group for single user, with the same name as the username. groupadd admin_ftp useradd -g admin_ftp -G ftp_accounts admin_ftp chgrp -R ftp_accounts /var/www chmod -R g+w /var/www When I log into FTP using account admin_ftp, I am given the error message: 500 OOPS: cannot change directory:/home/admin_ftp But didn't I specify the home directory? Extra internets for a guide how to do this specifically for vsftpd :)

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• How to enable directory browsiing in IIS7?

  - by frankadelic

  How I enable directory browsing in IIS7? MS technet says this can be done in the IIS console: Open IIS Manager and navigate to the level you want to manage. In Features View, double-click Directory Browsing. In the Actions pane, click Enable if the Directory Browsing feature is disabled and you want to enable it. Or, click Disable if the Directory Browsing feature is enabled and you want to disable it. http://technet.microsoft.com/en-us/library/cc731109%28WS.10%29.aspx However, my IIS console doesn't have the Directory Browsing option mentioned in Step 2. How can this option be made available. Note, this is for a static HTML site, so I don't have any web.config or ASPX files.

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• wget recursively with -np option still ascends to parent directory

  - by vectra

  tl;dr: will `wget --no-parrent -r ' download from a directory above the given url's directory? when using wget to download, say images, recursively from example.com/a/b with the -r and -np options, will a picture that is under example.com/a/c/ be downloaded when example.com/a/b/ delivers a html-file containing a link to the picture? if so, how do i get all pictures, that are in a folder and it's subfolders and only those? the description of the option --no-parent says "Do not ever ascend to the parent directory when retrieving recursively". anyway directory browsing delivers a link to the parent directory, which wget will follow, despite mentioned option. now what did i miss? edit: using GNU Wget 1.12

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• Can't read directory owned by my group

  - by Jonathan

  I moved the postgres data directory to a separate partition and it works great. The directory is owned by postgres user and postgres group. d-wx------ 11 postgres postgres 4.0K 2010-06-11 08:28 data/ I added myself to the group > sudo addgroup me postgres > groups me me : me adm dialout cdrom plugdev lpadmin admin sambashare postgres And gave the group read and execute permissions to everything in the directory. sudo chmod -R g+rx ./data d-wxr-x--- 11 postgres postgres 4.0K 2010-06-11 08:28 data/ But I still can not CD or LS the directory. > ls data ls: cannot open directory data: Permission denied What beginner mistake am I making?

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• Can't get sync with SVN to work to put a directory under version control

  - by Chamster

  I've created a directory on our SVN server and added a local directory with my project to it. It got uploaded and seems to be OK. However, I didn't get those cute green (or any others) icons on the file. So I clicked on the project's directory and used "checkout". That wasn't good because I've downloaded ALL of the repository. How can I synchronize my uploaded directory with it's corresponding directory on the server? (It's probably something easy that I've forgot to check in or so...)

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• Jail user to home directory while still allowing permission to create and delete files/folders

  - by Sevenupcan

  I'm trying to give a client SFTP access to the root directory of their site on my server (Ubuntu 10.10) so they can manager their website themselves. While I have been successful in jailing a user to a directory and giving them SFTP access; they are only allowed to create and delete new files in sub directories (the directories they own). This means that I must give them access to the parent directory to the root of their site. How can I limit them to the root of their site (for example public_html) while still allowing them the ability create and delete files. All the tutorials I have read suggest that the root must be the owner of the user's home directory, which prevents them from write access inside that directory. I'm relatively new to managing my own server so any advice would be very grateful. Many thanks.

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• How do you redirect pages from a subdirectory up one level to the root directory

  - by kezzman11

  I have recently moved all the content on my website from being in the www.mysite.com/shop directory to being in the root directory. This means that now I need to redirect any request to visit a page with the /shop directory back to the same page in the root directory eg. www.mysite.com/shop/categories/washroom to www.mysite.com/categories/washroom This needs to happen with all pages in my site that were previously using the /shop directory. The closest thing to a solution that I have found so far is the following code RedirectMatch ^/shop/.* http://www.mysite.com/ however this redirects all pages back to the homepage instead of to the relevant matching page without the /shop. Can someone please point me in the right direction, or if this has already been answered in here can you please post the link to the answer.

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• cd ~ dumps me in a seemingly empty directory

  - by Davidos

  This is on a Linux mint box. I'm told everywhere to use the command cd ~ To switch to the root directory before doing some command line magic. For some reason though, it dumps me in a directory named ~ where ls gives nothing and I can't get back to my home directory; I have to restart the terminal session to get out of the empty root directory. I'm positive that everything is just hidden to me, but even as a super-user I can't get the folders to show themselves. I usually just fall back to using a graphical file browser to roam those forbidden files, but I've recently just been shut out of my root directory, and the machine refuses to allow me to change the permissions on the stupid thing even when I type the root password in. It may just be some over-rigorous end-user shielding on the part of the mint team, but it's getting to be really frustrating now.

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• Windows 7 CD Command only echoes directory

  - by Zobbl

  The path for every new instance of the shell starts in my user directory (C:\Users\user). Within this directory or rather drive (in this case C:) I can't use the cd command as I'm used to - it only echoes the specified directory. As soon as I change the directory to a parent-directory I can execute "cd D:" and it changes to the drive. But this behavious doesn't appear consistently in all instances of the shell. Sometimes I have to go to C: to change it. I'm quite sure I'm not using the command in the wrong way, since it's what I'm used to do to start grails.

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• Nautilus tags/labels/marks/columns for folders/files

  - by madox2

  Is there any way how to mark folders or files with tags(or labels, new columns or whatever) in Nautilus? It would be nice to sort marked folders or files through this tags. Especially my first idea was to mark folders in my Movie directory with tags seen, not seen, must see, and so on. Then I realized it would be useful in any other workspaces with any custom tags... Is there any nautilus extension for this? Or any other file manager which can do this? It might look like this:

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• Which kind of public sitemap should I build for a search based navigation site

  - by Noam

  I have a search based navigation web-site. Each query has filters as well as sort-by. The search results point to end-pages inside the site. Each of those pages has many outlinks to other end-pages. Currently I have a XML sitemap which directs crawlers to all the end pages. I'm trying to add a silo sitemap directory to improve SEO. Assuming this is the right direction I have a couple of options: end pages sorted alphabetically. Pages by major search filters, and then divide alphabetically. Pages for every filter and cross option between them and the sort-by. Which would you recommend and why? NOTE: I'm not referring to a XML sitemap.

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• ADSI, SQL, Exchange Server 2010

  - by WernerCD

  Maybe I'm barking up the wrong tree... We have a Domain Controller "DC1". We have Exchange Server 2010 "Postman". Say I have an Address Book: And I add a few contacts to it: How then do I get the data from that contact via ADSI? Say I want the Job Title or CustomerID field that I filled out in the Contacts list? SELECT * FROM OPENQUERY(ADSI, 'SELECT EXTENSIONATTRIBUTE15 ,DISPLAYNAME ,GIVENNAME ,NAME ,SN ,SAMACCOUNTNAME FROM ''LDAP://DC=ATLANTICGENERAL,DC=ORG'' WHERE USERACCOUNTCONTROL=512 AND SAMACCOUNTTYPE=805306368 AND OBJECTCLASS=''PERSON'' AND OBJECTCLASS=''USER'' ORDER BY SAMACCOUNTTYPE ') How can I tie the Contact Card to an Active Directory User, so that I can edit either the AD account information OR the Exchange information and have them synced up?

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