Search Results

Search found 20187 results on 808 pages for 'directory tree'.

Page 50/808 | < Previous Page | 46 47 48 49 50 51 52 53 54 55 56 57  | Next Page >

  • C# expression tree for ordinary code

    - by rwallace
    It's possible to create an expression tree, if you declare it as such. But is it possible to get an expression tree for an ordinary chunk of code such as a method or property getter? What I'm trying to do is, let's say for an order processing system, I have a class for order items: class Item : Entity { [Cascade] public Document document { get; set; } public int line { get; set; } public Product product { get; set; } public string description { get; set; } public decimal qty { get; set; } public decimal price { get; set; } public decimal net { get { return qty * price; } } public VatCode vat_code { get; set; } } where the net value equals qty * price, so I'd like to declare it as such, either with a property or method, and then also have the framework introspect that expression so it can generate appropriate SQL for defining a corresponding calculated column in a corresponding database view. The most obvious way to do this would be to get the expression tree for a property getter or a method, but I can't find any indication how to do this, or that it is possible. (I have found a way to get a method body as a byte stream, but that's not what's desired here.) If that isn't possible, I suppose the recommended solution would be to declare something like a static field that is an expression tree, and compile/run it at run time for internal use, and also introspect as normal for SQL generation?

    Read the article

  • How to render all records from a nested set into a real html tree

    - by Christoph Schiessl
    I'm using the awesome_nested_set plugin in my Rails project. I have two models that look like this (simplified): class Customer < ActiveRecord::Base has_many :categories end class Category < ActiveRecord::Base belongs_to :customer # Columns in the categories table: lft, rgt and parent_id acts_as_nested_set :scope => :customer_id validates_presence_of :name # Further validations... end The tree in the database is constructed as expected. All the values of parent_id, lft and rgt are correct. The tree has multiple root nodes (which is of course allowed in awesome_nested_set). Now, I want to render all categories of a given customer in a correctly sorted tree like structure: for example nested <ul> tags. This wouldn't be too difficult but I need it to be efficient (the less sql queries the better). Update: Figured out that it is possible to calculate the number of children for any given Node in the tree without further SQL queries: number_of_children = (node.rgt - node.lft - 1)/2. This doesn't solve the problem but it may prove to be helpful.

    Read the article

  • How to improve Minecraft-esque voxel world performance?

    - by SomeXnaChump
    After playing Minecraft I marveled a bit at its large worlds but at the same time I found them extremely slow to navigate, even with a quad core and meaty graphics card. Now I assume Minecraft is fairly slow because: A) It's written in Java, and as most of the spatial partitioning and memory management activities happen in there, it would naturally be slower than a native C++ version. B) It doesn't partition its world very well. I could be wrong on both assumptions; however it got me thinking about the best way to manage large voxel worlds. As it is a true 3D world, where a block can exist in any part of the world, it is basically a big 3D array [x][y][z], where each block in the world has a type (i.e BlockType.Empty = 0, BlockType.Dirt = 1 etc.) Now, I am assuming to make this sort of world perform well you would need to: A) Use a tree of some variety (oct/kd/bsp) to split all the cubes out; it seems like an oct/kd would be the better option as you can just partition on a per cube level not a per triangle level. B) Use some algorithm to work out which blocks can currently be seen, as blocks closer to the user could obfuscate the blocks behind, making it pointless to render them. C) Keep the block object themselves lightweight, so it is quick to add and remove them from the trees. I guess there is no right answer to this, but I would be interested to see peoples' opinions on the subject. How would you improve performance in a large voxel-based world?

    Read the article

  • What is the best way to manage large 3d worlds (i.e minecraft style)?

    - by SomeXnaChump
    After playing minecraft I was marvelling a bit at their large worlds but at the same time finding it extremely slow to navigate, even with a quad core and meaty graphics card. Now I assume its fairly slow because: A) Its written in Java, and as most of the actual spatial partitioning and other memory management activities happen in there it would be slower than a native C++ version. B) They are not partitioning their world very well I could be wrong on both assumptions, however it got me thinking about the best way to manage large worlds. As it is more of a true 3d world, where a block can exist in any part of the world, it is basically a big 3d array [x][y][z], where each block in the world has a type (i.e BlockType.Empty = 0, BlockType.Dirt = 1 etc). Now I am assuming to make this sort of world performant you would need to: a) Use a tree of some variety (oct/kd/bsp) to split all the cubes out, it seems like an oct/kd would be the better option as you can just partition on a per cube level not a per triangle level. b) Use some algorithm to work out if the blocks within the scene can currently be seen, as blocks closer to the user could obfuscate the blocks behind, making it pointless to render them. c) Keep the block object themselves lightweight, so it is quick to add and remove them from the trees I guess there is no right answer to this, but I would be interested to see peoples opinions on the subject.

    Read the article

  • 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

    Read the article

  • Tree structured troubleshooting survey which gives solution based on answers

    - by japie1001
    Is there some online tool which let me design a tree structured questionaire/survey which suggests a solution to a problem. Example troubleshooting survey for the problem of black monitor screen: Q1: Is the powercable connected to the monitor? 0 Yes 0 NO If answer is no: "The solution to your problem is to connect the power cable" If answer is yes: Contiue with Q2 Q2: Is the computer turned on: 0 yes 0 no etc Is there something like Google Forms where I can achieve this with online?

    Read the article

  • 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!!

    Read the article

  • How Spanning Tree Protocol detects Loops

    - by AMIT
    For last few days I've been reading about Spanning Tree Protocol ,L2 protocol and understood how it prevents loop in network ,various steps in STP but one thing i wanted to know how STP actually detects the loops in network so that it can prevent it.Somewhere I read STP uses BPDU as probe and detects loops I mean how it happen is when switch send a BPDU with Destination Address as multicast and receive same BPDU again mean there is loop in network . But is it how STP detects loops in network?

    Read the article

  • 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!

    Read the article

  • 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 :)

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • 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 :)

    Read the article

  • 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.

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

  • 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...)

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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.

    Read the article

  • 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:

    Read the article

  • 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.

    Read the article

< Previous Page | 46 47 48 49 50 51 52 53 54 55 56 57  | Next Page >