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  • Cannot Call WordPress Plugin Files Under wp-content

    - by Volomike
    I have a client who has many blog customers. Each of these WordPress blogs calls a plugin that provides a product link. The way that link is composed looks like this: {website}/wp-content/plugins/prodx/product?id=432320 This works fine on all blogs except two. On those two, when you try to call the URL, you get a 404. So, I disabled all plugins except prodx and reverted the theme to default (Kubrick), thinking perhaps a plugin intercept with add_action() API was doing this, such as intercepting URLs and redirecting them. However, this did not help. So, I upgraded the WordPress to the latest version. Again, didn't fix. So, I checked permissions, comparing with a blog that worked just fine. Again, didn't fix. So I replaced the .htaccess, using one from a working blog. Again, didn't fix. So I replaced all the files using some from a working blog that was identical to this one, and then restored the wp-config.php file back so that it talked to the right blog database. Again, didn't fix. Again I checked permissions meticulously, comparing to a perfectly working blog. Again, didn't fix. So, I created a test.php that looks like so: <?php print_r($_GET); echo "hello world"; I then copied it into another plugin folder and used my browser to get to it -- again, 404. So I copied it into the root of wp-content/plugins and tried to call it there -- again, 404. So I copied it into wp-content -- again, 404. Last, I copied it into the root of the WordPress blog website, and this time, it worked! Doesn't make sense. I started to think that perhaps something was going on with /etc/httpd/conf/httpd.conf for this customer, but the only thing I saw different in their for this customer was the IP address was different than the customer's blog that worked. Each customer gets their own IP in this environment my client has built. My client sysop is baffled too. What do you think is going on? Is there something wrong in the WP database for this customer? Is there something wrong in httpd.conf?

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  • Database design - How can I have a foreign key of the primary key in the same table?

    - by Sergio Tapia
    My database has to store all the available departments in my company. Some departments are sub-departments on another existing department. I've decided to solve this like this: Departments ID Description HeadOfDepartment ParentDepartment ParentDepartment can be null, indicating it is a root department. If it has a parent I'll act accordingly, my question is how can I code this in Microsoft SQL?

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  • iPhone: Grouped tables and navigation controller issues

    - by Jack Griffiths
    Hi there, I've set up a grouped table on my app, and pushing to new views works fine. Except for one thing. The titles and stack layout are all weird. Here's the breakdown: I have two sections in my table. When I tap on the first row in the first section, it takes me to the correct view, but the title of the new view is the name of the first row in the second section of the table. In turn, the second row in the first section's title is the second row in the second section's title. If I tap on the second row in the second section of the root view table, the navigation button goes to the second row in the first section of the table. So here's a diagram of my table: Table Section 1 Row 1 Row 2 Row 3 Table Section 2 Row A Row B Row C So if I tap on row 3, the title of the pushed view is Row C. The navigation button tells me to go back to Row 3, then eventually ending up at the root view. Here's my implementation file pushing the views: - (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath { //CSS if ([[arryClientSide objectAtIndex:indexPath.row] isEqual:@"CSS"]) { CSSViewController *css = [[CSSViewController alloc] initWithNibName:@"CSSViewController" bundle:nil]; [css setTitle:@"CSS"]; [self.navigationController pushViewController:css animated:YES]; } //HTML if ([[arryClientSide objectAtIndex:indexPath.row] isEqual:@"HTML"]) { HTMLViewController *html = [[HTMLViewController alloc] initWithNibName:@"HTMLViewController" bundle:nil]; [html setTitle:@"HTML"]; [self.navigationController pushViewController:html animated:YES]; } //JS if ([[arryClientSide objectAtIndex:indexPath.row] isEqual:@"JavaScript"]) { JSViewController *js = [[JSViewController alloc] initWithNibName:@"JSViewController" bundle:nil]; [js setTitle:@"JavaScript"]; [self.navigationController pushViewController:js animated:YES]; } //PHP if ([[arryServerSide objectAtIndex:indexPath.row] isEqual:@"PHP"]) { PHPViewController *php = [[PHPViewController alloc] initWithNibName:@"PHPViewController" bundle:nil]; [php setTitle:@"PHP"]; [self.navigationController pushViewController:php animated:YES]; } //SQL if ([[arryServerSide objectAtIndex:indexPath.row] isEqual:@"SQL"]) { SQLViewController *sql = [[SQLViewController alloc] initWithNibName:@"SQLViewController" bundle:nil]; [sql setTitle:@"SQL"]; [self.navigationController pushViewController:sql animated:YES]; } & the array feeding the table's data: - (void)viewDidLoad { [super viewDidLoad]; arryClientSide = [[NSArray alloc] initWithObjects:@"CSS",@"HTML",@"JavaScript",nil]; arryServerSide = [[NSArray alloc] initWithObjects:@"Objective-C", @"PHP",@"SQL",nil]; // arryResources = [[NSArray alloc] initWithObjects:@"HTML Colour Codes", @"Useful Resources", @"About",nil]; self.title = @"Select a Language"; [super viewDidLoad]; // Uncomment the following line to display an Edit button in the navigation bar for this view controller. // self.navigationItem.rightBarButtonItem = self.editButtonItem; } Any help would be greatly appreciated. Jack

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  • Recommended way to develop using Jetty and Eclipse

    - by Ula Krukar
    Hi, I am currently developing a J2EE application and I would like to use Jetty. I would like to have iot integrated with Eclipse, so I could debug the appliaction. I've tried out couple of plugins (including WTP) but nothing works well enough. Run Jetty Run plugin is the best, but I cannot specify context-root in it, which makes it unusable for me. What would you recommend?

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  • Uppercase and lowercase urls in PHP

    - by Arjun
    I have created folders in my root example: http://www.zipholidays.co.uk/Cuba or http://www.zipholidays.co.uk/Florida When I type http://www.zipholidays.co.uk/cuba (Cube in lowercase), it shows page not found. I'm using Apache server. People are linking to pages with lowercase, uppercase, mixed case - whatever. What do I do to make the pages case insensitive?

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  • mysql_connect "bool $new_link = true" is very slow

    - by Mikk
    Hi, I'm using latest version of Xampp on 64bit Win7. The problem is that, when I use mysql_connect with "bool $new_link" set to true like so: mysql_connect('localhost', 'root', 'my_password', TRUE); script execution time increases dramatically (about 0,5 seconds per connection, and when I have 4 diffirent objects using different connections, it takes ~2 seconds). Is setting "bool $new_link" to true, generally a bad idea or could it just be some problem with my software configuration. Thank you.

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  • svn (with git frontend) branch merging with different directory structure

    - by Fu86
    I have a subversion repository with a directory structure: frontend backend + a + b In a other branch, someone had put the sub-folders a and b in the root directory and delete the other stuff (frontend, backend). a b Now i have to merge this branch back into the trunk (backend-folder). How can I do that to dont lose the history from the branches? I use git to access and work with the subversion repository.

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  • Simple startup on boot of misc application (Java based) on Ubuntu Linux 8+ as a daemon

    - by Berlin Brown
    What is the easiest way to have an application launch at startup on Ubuntu server as daemon? This is a java application (java com.run.run.Run) etc. How would I have it launch as a user and possibly have access to write to some log file where the user has permissions to write? And if I don't end up doing that, how would I launch the application as the root user at startup. Edited: It is a headless server, I don't have access to the desktop applications.

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  • Using single spring application context for web app

    - by Ramo
    Hi, I'm using org.springframework.web.servlet.DispatcherServlet and org.springframework.ws.transport.http.MessageDispatcherServlet In the same app but each is loading own application contexts, I need to load all beans in a single application context. The application consists of typical layers webappdao etc What I have tried is to use one single spring-root-context.xml by setting it in the contextConfigLocation. But didn't help, this has been an issue for me for a long time an I would appreciate any help with this. Any online references would be a great help. Regards Ramo

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  • Google search ajax api is to restrictive. Any alternatives?

    - by boomhauer
    The google search ajax api is terrific, and the .net wrapper available on codeplex makes using it from a .net project very simple. However, the api itself is crippled so that it only returns 64 results per query. Not very useful for many applications. Ignoring the likely TOS problems, are there known .net libraries that can query the root google website and scrape the results into a resultset? I'm assuming this could result in much larger result counts than the ajax version enabled.

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  • mysqldump command not working?

    - by Harish Kurup
    I am using mysqldump to take backup of my database, but the command is not working.. the command i am using is mysqldump -u root dbname> 'c:\backupdatafolder\backup.sql' i am running this command in MySQL cli but not running,..is there any thing wrong in the command?

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  • Update function

    - by kosmaks
    Hello everyone! I try to learn java for android devices.. i have to create the update function. But still have one question: How???? in class root public void update(){ maindebug("update"); // This is my debug function } public void run(){ while(isRunning){ // isRunning is a boolean variable SystemClock.sleep(100); update(); } } and inside onCreate run(); but it doesnt work :(

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  • Using XAML + designer to edit Plain Old CLR Objects?

    - by Joe White
    I want to write a POCO in XAML, and use a DataTemplate to display that object in the GUI at runtime. So far, so good; I know how to do all that. Since I'll already have a DataTemplate that can transform my POCO into a WPF visual tree, is there any way to get the Visual Studio designer to play along, and have the Design View show me the POCO+DataTemplate's resulting GUI, as I edit the POCO's XAML? (Obviously the designer wouldn't know how to edit the "design view"; I wouldn't expect the Toolbox or click-and-drag to work on the design surface. That's fine -- I just want to see a preview as I edit.) If you're curious, the POCOs in question would be level maps for a game. (At this point, I'm not planning to ship an end-user map editor, so I'll be doing all the editing myself in Visual Studio.) So the XAML isn't WPF GUI objects like Window and UserControl, but it's still not something where I would want to blindly bang out some XAML and hope for the best. I want to see what I'm doing (the GUI map) as I'm doing it. If I try to make a XAML file whose root is my map object, the designer shows "Intentionally Left Blank - The document root element is not supported by the visual designer." It does this even if I've defined a DataTemplate in App.xaml's <Application.Resources>. But I know the designer can show my POCO, when it's inside a WPF object. One possible way of accomplishing what I want would be to have a ScratchUserControl that just contains a ContentPresenter, and write my POCO XAML inside that ContentPresenter's Content property, e.g.: <UserControl ...> <ContentPresenter> <ContentPresenter.Content> <Maps:Map .../> </ContentPresenter.Content> </ContentPresenter> </UserControl> But then I would have to be sure to copy the content back out into its own file when I was done editing, which seems tedious and error-prone, and I don't like tedious and error-prone. And since I can preview my XAML this way, isn't there some way to do it without the UserControl?

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  • Connection for control user as defined in your configuration failed. xampp

    - by Zann
    when i tried to uninstall xampp and reinstall xampp.I received below error message when i go phpmyadmin Need help and guide to solve it .thanks Error MySQL said: Documentation 1045 - Access denied for user 'root'@'localhost' (using password: NO) Connection for controluser as defined in your configuration failed. phpMyAdmin tried to connect to the MySQL server, and the server rejected the connection. You should check the host, username and password in your configuration and make sure that they correspond to the information given by the administrator of the MySQL server.

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  • add a decorate function to a class

    - by wiso
    I have a decorated function (simplified version): class Memoize: def __init__(self, function): self.function = function self.memoized = {} def __call__(self, *args, **kwds): hash = args try: return self.memoized[hash] except KeyError: self.memoized[hash] = self.function(*args) return self.memoized[hash] @Memoize def _DrawPlot(self, options): do something... now I want to add this method to a pre-esisting class. ROOT.TChain.DrawPlot = _DrawPlot when I call this method: chain = TChain() chain.DrawPlot(opts) I got: self.memoized[hash] = self.function(*args) TypeError: _DrawPlot() takes exactly 2 arguments (1 given) why doesn't it propagate self?

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  • TouchCode XML parsing error

    - by itsaboutcode
    Hi, I have a xml document which has only one element in the document, which is <error>error string<error> But when i try to parse it, it says this document has no element at all. In other words when i try to access the rootElement it says "null" CXMLDocument *rssParser = [[[CXMLDocument alloc] initWithContentsOfURL:url options:0 error:nil] autorelease]; NSLog(@"Root: %@",[[rssParser rootElement] name]); Please tell me what is wroing with this. Thanks

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  • Changing Passwords Over Multiple Servers and Services

    - by tesmar
    Hi all, I am looking to automate the changing of passwords across multiple services and servers. For example: I want to change the root paassword to all of my web servers at once. I am thinking of writing a ruby script, but have you guys run across anything already written? If so, would that also give me the ability to change other system passwords like Database passwords and SVN passwords.

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  • If as assert fails, is there a bug?

    - by RichAmberale
    I've always followed the logic: if assert fails, then there is a bug. Root cause could either be: Assert itself is invalid (bug) There is a programming error (bug) (no other options) I.E. Are there any other conclusions one could come to? Are there cases where an assert would fail and there is no bug?

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