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  • Ensure exclusive access to webservice

    - by Henrik P. Hessel
    Just to be on the safe side, what's the best practice to ensure that only my application has access to my webservice, which is hosted on a public server? Should I implement I shared key or something? My webservice is hosted on Googles App Engine and my Application runs on iPhones and iPads. If you need further information, just ask. Thanks, Henrik

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  • Access Google Chrome's cache

    - by jldupont
    Is it possible to access Google Chrome's cache from within an extension? I'd like to write an extension that loads a cached version of a page when the online one can't be accessed (e.g. Internet connectivity issue). Updated: I know I could write an NPAPI plugin accessible through an extension to accomplish this but I'd rather not suffer writing one... I am after a solution without resorting to NPAPI, please.

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  • How can UNIX access control create compromise problems ?

    - by Berkay
    My system administrators advice me to be careful when setting access control to files and directories. He gave me an example and i got confused, here it is: a file with protection mode 644 (octal) contained in a directory with protection mode 730. so it means: File:101 100 100 (owner, group,other: r-x r-- r--) Directory:111 011 000 (owner, group,other: rwx -wx ---) How can file be compromised in this case ?

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  • How to access a String that is in JSON array format

    - by Sayem Ahmed
    I have an asp.net page which is returning a list of object as a json string to an ajax request. The string is as follows : [ {"Name":"Don","Age":23,"Description":"Tall man with no glasses"} ,{"Name":"Charlie","Age":24,"Description":"Short man with glasses"} ] I want to access each field individually, like the name of the person, his age, his description etc. How can I do that? I am using JQuery for client-side scripting.

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  • access django session from a decorator

    - by ed1t
    I have a decorator that I use for my views @valid_session from django.http import Http404 def valid_session(the_func): """ function to check if the user has a valid session """ def _decorated(*args, **kwargs): if ## check if username is in the request.session: raise Http404('not logged in.') else: return the_func(*args, **kwargs) return _decorated I would like to access my session in my decoartor. When user is logged in, I put the username in my session.

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  • How to access a named element in a control that inherits from a templated control

    - by Mrt
    Hello this is similar to http://stackoverflow.com/questions/2620165/how-to-access-a-named-element-of-a-derived-user-control-in-silverlight with the difference is inheriting from a templated control, not a user control. I have a templated control called MyBaseControl <Style TargetType="Problemo:MyBaseControl"> <Setter Property="Template"> <Setter.Value> <ControlTemplate TargetType="Problemo:MyBaseControl"> <Grid x:Name="LayoutRoot" Background="White"> <Border Name="HeaderControl" Background="Red" /> </Grid> </ControlTemplate> </Setter.Value> </Setter> </Style> public class MyBaseControl : Control { public UIElement Header { get; set; } public MyBaseControl() { DefaultStyleKey = typeof(MyBaseControl); } public override void OnApplyTemplate() { base.OnApplyTemplate(); var headerControl = GetTemplateChild("HeaderControl") as ContentPresenter; if (headerControl != null) headerControl.Content = Header; } } I have another control called myControl which inherits from MyBaseControl Control <me:MyBaseControl x:Class="Problemo.MyControl" xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation" xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml" xmlns:d="http://schemas.microsoft.com/expression/blend/2008" xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006" mc:Ignorable="d" xmlns:me="clr-namespace:Problemo" d:DesignHeight="300" d:DesignWidth="400"> <me:MyBaseControl.Header> <TextBlock Name="xxx" /> </me:MyBaseControl.Header> </me:MyBaseControl> public partial class MyControl : MyBaseControl { public string Text { get; set; } public MyControl(string text) { InitializeComponent(); Text = text; Loaded += MyControl_Loaded; } void MyControl_Loaded(object sender, RoutedEventArgs e) { base.ApplyTemplate(); xxx.Text = Text; } } The issue is xxx is null. How do I access the xxx control in the code behind ?

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  • Using memorystream and DotNetZip in MVC gives "Cannot access a closed Stream"

    - by Frode Lillerud
    I'm trying to create a zipfile in a MVC method using the DotNetZip components. Here is my code: public FileResult DownloadImagefilesAsZip() { using (var memoryStream = new MemoryStream()) { using (var zip = new ZipFile()) { zip.AddDirectory(Server.MapPath("/Images/")); zip.Save(memoryStream); return File(memoryStream, "gzip", "images.zip"); } } } When I run it I get a "Cannot access a closed Stream" error, and I'm not sure why.

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  • Access denied when start sqlexpress from command line

    - by Brettski
    Windows 7 computer running SQL Server Express 2008. When I try to run net start mssql$sqlexpress I receive an error: System error 5 has occurred. Access is denied The SQL service is running under "Network Service" account The service starts fine if I use the services window. Could somebody help me figure out why I am receiving this error?

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  • ASP MVC Access ViewData Array?

    - by Jacob Huggart
    I have some viewdata that is generated by going through my repository to the database to grab some scheduling info. When the information is stored in the Viewdata, I noticed that the viewdata is enumerated. How could I access the enumerated items and generate a table/list based on the viewdata? Most of the information just needs to be spit out into a table, but one item will have a link generated for it. Thanks!

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  • how to access owl file inside jsp

    - by Udayanga
    hi, I'm trying to access owl file using jsp.I've successfully load the owl file and quering that using SPARQL.But still I couldn't success with JSP. I'm always getting error "java.lang.ClassNotFoundException: com.hp.hpl.jena.util.FileManager " help me! Thank in advance!

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  • Allow access to my server.

    - by Zachary Brown
    I have a server, did some of the programming myself. It ison my home network, but I need to be able to access it from anywhere over the internet. I have done the port forwarding like I am supposed to, but I still cant get to it from an outside computer. It just displays Internet Explorer cannot display the webpage. I don't know what else to do. I am on a Linksys WRT54G v8 router running ddWRT v24 micro firmware.

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  • How to access a simple file or folder from Tomcat webapps folder

    - by Ankur
    I want to be able to access a folder from my tomcat webapps folder so that I can give someone a URL like: http://localhost:8080/myFolder/myFile.f And in a web browser if they point to this they should start downloading the file. But in reality I get a 404 error when I try to point to this location. How can I solve this or get around it.

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  • PHP API for access social networks

    - by War Coder
    hello guys, am looking for a php api that can access social networks like facebook,myspace,orkut and so many more of them. I should be able to upload pictures and videos to my favourite social networking account and also post get updates from thems, change status, make comments and similar stuffs. Just wondering if there is anything like that or similar to it. Thanks.

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