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  • Adobe sort Project Parfait, un outil en ligne pour extraire le CSS de votre PSD

    Adobe Project Parfait : extraire le CSS de votre PSDAdobe signe un bel outil qui promet de rendre de bons services aux webmaster en permettant d'extraire le code CSS d'un fichier PSD dans le navigateur. Dans le navigateur? C'est à dire que la version actuelle (beta) est gratuite et ne nécessite pas de posséder Photoshop pour l'utiliser.On peut imaginer que le webmaster a reçu la maquette d'une page en PSD qu'il lui suffit d'uploader sur Project Parfait pour en extraire le CSS.L'interface intuitive...

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  • Windows 7 Enterprise : Microsoft sort une version d'évaluation gratuite de l'édition "réservée aux professionnels de l'informatique"

    Microsoft propose une version d'évaluation gratuite de Windows 7 Enterprise Que pensez-vous de cette édition spécialement dédiée aux professionnels de l'informatique Le 8 avril 2014, Windows XP sera de l'histoire ancienne pour Microsoft. Plus aucune mise à jour et plus aucun support ne seront proposés pour le système d'exploitation n°1 dans le monde. Vista a suivi mais a déboussolé tant d'utilisateurs que pour beaucoup de professionnels IT, le vrai successeur de Windows XP est en fait Windows 7. Et plus exactement Windows 7 Enterprise. Au moment où Microsoft vient de lancer une offre spéciale de 90 jours d'essai gratuit pour cette édition spéciale (justement en rapport avec...

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  • Installing packages into local directory?

    - by Gili
    I'd like to install software packages, similar to apt-get install <foo> but: Without sudo, and Into a local directory The purpose of this exercise is to isolate independent builds in my continuous integration server. I don't mind compiling from source, if that's what it takes, but obviously I'd prefer the simplest approach possible. I tried apt-get source --compile <foo> as mentioned here but I can't get it working for packages like autoconf. I get the following error: dpkg-checkbuilddeps: Unmet build dependencies: help2man I've got help2man compiled in a local directory, but I don't know how to inform apt-get of that. Any ideas? UPDATE: I found an answer that almost works at http://askubuntu.com/a/350/23678. The problem with chroot is that it requires sudo. The problem with apt-get source is that I don't know how to resolve dependencies. I must say, chroot looks very appealing. Is there an equivalent command that doesn't require sudo?

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  • How to sort time column by value instead of alphabetically

    - by Turch
    I'm creating a pivot table by connecting to an SSAS tabular model (Data - From Other Sources - From Analysis Services) . The model has a "time" column that I want to sort by. The default (database) sorting is earliest to latest: When I click the triangle next to 'Row Labels' and select "Sort A to Z", I get alphabetically sorted times: How can I get the times to sort by time? Changing the number format from "General" to "Time" does nothing. The times aren't stored as text either - the data type of the column in the SSAS model is Auto (Date)

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  • Sort rectangles in a grid based on a comparison of the center point of each

    - by Mrwolfy
    If I have a grid of rectangles and I move one of the rectangles, say above and to the left of another rectangle, how would I resort the rectangles? Note the rectangles are in an array, so each rectangle has an index and a matching tag. All I really need to do is set the proper index based on the rectangles new center point position within the rectangle, as compared with the center point position of the other rectangles in the grid. Here is what I am doing now in pseudo code (works somewhat, but not accurate): -(void)sortViews:myView { int newIndex; // myView is the view that was moved. [viewsArray removeObject:myView]; [viewsArray enumerate:obj*view]{ if (myView.center.x > view.center.x) { if (myView.center.y > view.center.y) { newIndex = view.tag -1; *stop = YES; } else { newIndex = view.tag +1; *stop = YES; } } else if (myView.center.x < view.center.x) { if (myView.center.y > view.center.y) { newIndex = view.tag -1; *stop = YES; } else { newIndex = view.tag +1; *stop = YES; } } }]; if (newIndex < 0) { newIndex = 0; } else if (newIndex > 5) { newIndex = 5; } [viewsArray insertObject:myView atIndex:newIndex]; [self arrangeGrid]; }

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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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  • Lucene neo4j sort with boolean fields

    - by Daniele
    I have indexed some documents (nodes of neo4j) with a boolean property which not always is present. Eg. Node1 label : "label A" Node2: label : "label A" (note, same label of node1) special : true The goal is to get Node2 higher than node 1 for query "label A". Here the code: Index<Node> fulltextLucene = graphDb.index().forNodes( "my-index" ); Sort sort = new Sort(new SortField[] {SortField.FIELD_SCORE, new SortField("special", SortField.????, true) }); IndexHits<Node> results = fulltextLucene.query( "label", new QueryContext( "label A").sort(sort)); How can I accomplish that? Thanks

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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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  • Maintaining sort order of database table rows

    - by Lox
    Say I have at database table containing information about a news article in each row. The table has an integer "sort" column to dictate the order in which the articles are to be presented on a web site. How do I best implement and maintain this sort order. The problem I want to avoid is having the the articles numbered 1,2,3,4,..,100 and when article number 50 suddenly becomes interesting it gets its sort number set to 1 and then all articles between them must have their sort number increased by one. Sure, setting initial sort numbers to 100,200,300,400 etc. leaves some space for moving around but at some point it will break. Is there a correct way to do this, maybe a completely different approach? Added-1: All article titles are shown in a list linking to the contents, so yes all sorted items are show at once. Added-2: An item is not necessarily moved to the top of the list; any item can be placed anywhere in the ordered list.

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  • Prolog: using the sort/2 predicate

    - by Øyvind Hauge
    So I'm trying to get rid of the wrapper clause by using the sort library predicate directly inside split. What split does is just generating a list of numbers from a list that looks like this: [1:2,3:2,4:6] ---split-- [1,2,3,2,4,6]. But the generated list contains duplicates, and I don't want that, so I'm using the wrapper to combine split and sort, which then generates the desired result: [1,2,3,4,6]. I'd really like to get rid of the wrapper and just use sort within split, however I keep getting "ERROR: sort/2: Arguments are not sufficiently instantiated." Any ideas? Thanks :) split([],[]). split([H1:H2|T],[H1,H2|NT]) :- split(T,NT). wrapper(L,Processed) :- split(L,L2), sort(L2,Processed).

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  • Virtual methods as Comp function to sort

    - by wilsongoncalo.mp
    Hello everyone! I'm new to C++ and i'm trying to use std::sort function to sort a vector of Solutions. The code is something like this (solution list is a *vector): void SolutionSet::sort(Comparator &comparator) { std::sort(solutionsList_->begin(), solutionsList_->end(), &comparator::compare); } The comparator param is a Comparator´s child class instance , and the compare method is virtual at Comparator class and implemented by all Comparator's child classes. And i want to use that function as a comparator function at std:sort(). Is this possible? If it is, can someone tell me how? Because with the previous code, it doesn't work. If i've not made myself clear, please just ask! Thank you guys!

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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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  • Shell Sort problem

    - by user191603
    Show the result of running Shell Sort on the input 9,8,7,6,5,4,3,2,1 using increments { 1,3,7 }. I have done this part. the result is: 9 8 7 6 5 4 3 2 1 (original) 2 1 7 6 5 4 3 9 8 ( 7-sort ) 2 1 4 3 5 7 6 9 8 ( 3-sort ) 1 2 3 4 5 6 7 8 9 ( 1-sort ) Then the question requires me to determine the running time of Shell Sort using Shell's increments of N/2, N/4, ..., 1 for sorted input. I am not quite sure how to answer the second question as I don't understand the requirement of this question. So, would anyone give some hints to let me finish this question? Thank you for your help first!

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  • sort list(of string()) using a variable index into string() as key - vb.net

    - by tullynyguy
    I have a List(of String()). I have written a custom comparer (implements IComparer(of string)) to do an alphanumeric sort. Is there a way to sort the List using a given index to determine which position in the String() to sort by? In other words one time I might sort by Index = 0 and another time by Index = 3. The length of all String() in the list is the same. For reference this question is similar to Sort List<String[]> except I am using VB.net and that question is hardwired to Index=0.

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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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