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  • Mysql Left Join Null Result

    - by Ozzy
    I have this query SELECT articles.*, users.username AS `user` FROM `articles` LEFT JOIN `users` ON articles.user_id = users.id ORDER BY articles.timestamp Basically it returns the list of articles and the username that the article is associated to. Now if there is no entry in the users table for a particular user id, the users var is NULL. Is there anyway to make it that if its null it returns something like "User Not Found"? or would i have to do this using php?

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  • Can you set SQL Server permissions from a Silverlight app?

    - by Paul Smith
    Is it possible to create a Silverlight application which can be used to provide a nice user interface for managing SQL Server permissions? We want to create a simple admin app to allow certain users to create new users, disable old users, and manage specific permissions for those users, but we feel that SQL Server Management Studio is perhaps too complex.

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  • untar from run command in fibre

    - by Shah.Bhavesh
    i want to untar file from source to destination with below statement `def untar(source,destination): run("tar -xf {0} {1}".format(source,destination)) ` i am getting Error C:\Users\test\Desktop\fabric>fab -H user@host-p pass untar:source =/shared/sample.tar,destination=/home/ Traceback (most recent call last): File "C:\Users\shasmukh\AppData\Roaming\Python\Python27\site-packages\fabric\m ain.py", line 630, in main docstring, callables, default = load_fabfile(fabfile) File "C:\Users\shasmukh\AppData\Roaming\Python\Python27\site-packages\fabric\m ain.py", line 163, in load_fabfile imported = importer(os.path.splitext(fabfile)[0]) File "C:\Users\shasmukh\Desktop\fabric\fabfile.py", line 11 def copy(source,destination) ^ SyntaxError: invalid syntax

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  • Error building C program

    - by John
    Here are my 2 source files: main.c: #include <stdio.h> #include "part2.c" extern int var1; extern int array1[]; int main() { var1 = 4; array1[0] = 2; array1[1] = 4; array1[2] = 5; array1[3] = 7; display(); printf("---------------"); printf("Var1: %d", var1); printf("array elements:"); int x; for(x = 0;x < 4;++x) printf("%d: %d", x, array1[x]); return 0; } part2.c #include <stdio.h> int var1; int array1[4]; void display(void); void display(void) { printf("Var1: %d", var1); printf("array elements:"); int x; for(x = 0;x < 4;++x) printf("%d: %d", x, array1[x]); } When i try to compile the program this is what i get: Ld /Users/John/Library/Developer/Xcode/DerivedData/Test-blxrdmnozbbrbwhcekmouessaprf/Build/Products/Debug/Test normal x86_64 cd /Users/John/Xcode/Test setenv MACOSX_DEPLOYMENT_TARGET 10.7 /Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/clang -arch x86_64 -isysroot /Applications/Xcode.app/Contents/Developer/Platforms/MacOSX.platform/Developer/SDKs/MacOSX10.7.sdk -L/Users/John/Library/Developer/Xcode/DerivedData/Test-blxrdmnozbbrbwhcekmouessaprf/Build/Products/Debug -F/Users/John/Library/Developer/Xcode/DerivedData/Test-blxrdmnozbbrbwhcekmouessaprf/Build/Products/Debug -filelist /Users/John/Library/Developer/Xcode/DerivedData/Test-blxrdmnozbbrbwhcekmouessaprf/Build/Intermediates/Test.build/Debug/Test.build/Objects-normal/x86_64/Test.LinkFileList -mmacosx-version-min=10.7 -o /Users/John/Library/Developer/Xcode/DerivedData/Test-blxrdmnozbbrbwhcekmouessaprf/Build/Products/Debug/Test ld: duplicate symbol _display in /Users/John/Library/Developer/Xcode/DerivedData/Test-blxrdmnozbbrbwhcekmouessaprf/Build/Intermediates/Test.build/Debug/Test.build/Objects-normal/x86_64/part2.o and /Users/John/Library/Developer/Xcode/DerivedData/Test-blxrdmnozbbrbwhcekmouessaprf/Build/Intermediates/Test.build/Debug/Test.build/Objects-normal/x86_64/main.o for architecture x86_64 clang: error: linker command failed with exit code 1 (use -v to see invocation) I am using Xcode and both files are inside of a C project called Test What is causing the error and how do i fix it?

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  • Rails. Putting update logic in your migrations

    - by Daniel Abrahamsson
    A couple of times I've been in the situation where I've wanted to refactor the design of some model and have ended up putting update logic in migrations. However, as far as I've understood, this is not good practice (especially since you are encouraged to use your schema file for deployment, and not your migrations). How do you deal with these kind of problems? To clearify what I mean, say I have a User model. Since I thought there would only be two kinds of users, namely a "normal" user and an administrator, I chose to use a simple boolean field telling whether the user was an adminstrator or not. However, after I while I figured I needed some third kind of user, perhaps a moderator or something similar. In this case I add a UserType model (and the corresponding migration), and a second migration for removing the "admin" flag from the user table. And here comes the problem. In the "add_user_type_to_users" migration I have to map the admin flag value to a user type. Additionally, in order to do this, the user types have to exist, meaning I can not use the seeds file, but rather create the user types in the migration (also considered bad practice). Here comes some fictional code representing the situation: class CreateUserTypes < ActiveRecord::Migration def self.up create_table :user_types do |t| t.string :name, :nil => false, :unique => true end #Create basic types (can not put in seed, because of future migration dependency) UserType.create!(:name => "BASIC") UserType.create!(:name => "MODERATOR") UserType.create!(:name => "ADMINISTRATOR") end def self.down drop_table :user_types end end class AddTypeIdToUsers < ActiveRecord::Migration def self.up add_column :users, :type_id, :integer #Determine type via the admin flag basic = UserType.find_by_name("BASIC") admin = UserType.find_by_name("ADMINISTRATOR") User.all.each {|u| u.update_attribute(:type_id, (u.admin?) ? admin.id : basic.id)} #Remove the admin flag remove_column :users, :admin #Add foreign key execute "alter table users add constraint fk_user_type_id foreign key (type_id) references user_types (id)" end def self.down #Re-add the admin flag add_column :users, :admin, :boolean, :default => false #Reset the admin flag (this is the problematic update code) admin = UserType.find_by_name("ADMINISTRATOR") execute "update users set admin=true where type_id=#{admin.id}" #Remove foreign key constraint execute "alter table users drop foreign key fk_user_type_id" #Drop the type_id column remove_column :users, :type_id end end As you can see there are two problematic parts. First the row creation part in the first model, which is necessary if I would like to run all migrations in a row, then the "update" part in the second migration that maps the "admin" column to the "type_id" column. Any advice?

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  • Content management system similar to StackOverflow.com

    - by Ghostrider
    I'm looking content management system that would have features similar to stackoverflow: Users can ask and answer/questions. Users can search within existing threads. The system is access controlled - so some users can be readers/writers/admins. Some or all QnA threads should be visible only to logged in users. So far I've mostly used wordpress as CMS but it looks like it's not well-suited for this particular task

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  • Kohanav3 ORM: calling where->find_all twice

    - by David Lawson
    When I do something like the following: $site = ORM::factory('site')->where('name', '=', 'Test Site')->find(); $users = $site->users; $deletedusers = $users->where('deleted', '=', '1')->find_all(); $nondeletedusers = $users->where('deleted', '=', '0')->find_all(); The contents of $deletedusers is correct, but the $nondeletedusers contains every non-deleted user, not just the ones in the loaded $site. What am I doing wrong?

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  • Retrieving all objects in code upfront for performance reasons

    - by ming yeow
    How do you folks retrieve all objects in code upfront? I figure you can increase performance if you bundle all the model calls together? This makes for a bigger deal, especially if your DB cannot keep everything in memory def hitDBSeperately { get X users ...code get Y users... code get Z users... code } Versus: def hitDBInSingleCall { get X+Y+Z users code for X code for Y... }

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  • javascript and jsp

    - by akellakarthik1254
    i am new bee to java. i have a html which has a show users button, on clicking this button i should redirect the user to a users.jsp page. How do i achieve that. will a function like this will help function msg() { alert("List of Users");<br/> jsp:forward page="Users.jsp"<br/> }

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  • unknown action with will_paginate

    - by merlin
    In my users controller I have this in a method: @users = User.paginate :page => params[:page], :per_page => 10, The results are rendered on users/search. The 2nd page link points to users/search?page=2, but it leads to an unknown action error.

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  • Changing the id parameter in Rails routing

    - by japancheese
    Hello, Using rails3 new routing system, is it possible to change the default :id parameter resources :users, :key => :username come out with the following routes /users/new /users/:username /users/:username/edit ...etc I'm asking because although the above example is simple, it would be really helpful to do in a current project I'm working on. Is it possible to change this parameter, and if not, is there a particular reason as to why not?

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  • MySQL subqueries

    - by swamprunner7
    Can we do this query without subqueries? SELECT login, post_n, (SELECT SUM(vote) FROM votes WHERE votes.post_n=posts.post_n)AS votes, (SELECT COUNT(comments.post_n) FROM comments WHERE comments.post_n=posts.post_n)AS comments_count FROM users, posts WHERE posts.id=users.id AND (visibility=2 OR visibility=3) ORDER BY date DESC LIMIT 0, 15 tables: Users: id, login Posts: post_n, id, visibility Votes: post_n, vote id — it`s user id, Users the main table.

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  • Basic Database Design

    - by Thurein
    Hi, This question would be a bit childish, I have three tables, users, usergroups and contacts. In my system, the end user can create a contact and subsequently he\she may define the visibility of the contact by setting only for that user, or for a set of users or a set of usergroups. So I am wondering, how my database design would be, it should be many to many between users and contacts or many to many between usergroups and contacts. Definitely there is a one to many relationship between users and usergroups. Thanks Thurein

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  • Spring MVC, REST, and HATEOAS

    - by SingleShot
    I'm struggling with the correct way to implement Spring MVC 3.x RESTful services with HATEOAS. Consider the following constraints: I don't want my domain entities polluted with web/rest constructs. I don't want my controllers polluted with view constructs. I want to support multiple views. Currently I have a nicely put together MVC app without HATEOAS. Domain entities are pure POJOs without any view or web/rest concepts embedded. For example: class User { public String getName() {...} public String setName(String name) {...} ... } My controllers are also simple. They provide routing and status, and delegate to Spring's view resolution framework. Note my application supports JSON, XML, and HTML, yet no domain entities or controllers have embedded view information: @Controller @RequestMapping("/users") class UserController { @RequestMapping public ModelAndView getAllUsers() { List<User> users = userRepository.findAll(); return new ModelAndView("users/index", "users", users); } @RequestMapping("/{id}") public ModelAndView getUser(@PathVariable Long id) { User user = userRepository.findById(id); return new ModelAndView("users/show", "user", user); } } So, now my issue - I'm not sure of a clean way to support HATEOAS. Here's an example. Let's say when the client asks for a User in JSON format, it comes out like this: { firstName: "John", lastName: "Smith" } Let's also say that when I support HATEOAS, I want the JSON to contain a simple "self" link that the client can then use to refresh the object, delete it, or something else. It might also have a "friends" link indicating how to get the user's list of friends: { firstName: "John", lastName: "Smith", links: [ { rel: "self", ref: "http://myserver/users/1" }, { rel: "friends", ref: "http://myserver/users/1/friends" } ] } Somehow I want to attach links to my object. I feel the right place to do this is in the controller layer as the controllers all know the correct URLs. Additionally, since I support multiple views, I feel like the right thing to do is somehow decorate my domain entities in the controller before they are converted to JSON/XML/whatever in Spring's view resolution framework. One way to do this might be to wrap the POJO in question with a generic Resource class that contains a list of links. Some view tweaking would be required to crunch it into the format I want, but its doable. Unfortunately nested resources could not be wrapped in this way. Other things that come to mind include adding links to the ModelAndView, and then customizing each of Spring's out-of-the-box view resolvers to stuff links into the generated JSON/XML/etc. What I don't want is to be constantly hand-crafting JSON/XML/etc. to accommodate various links as they come and go during the course of development. Thoughts?

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  • PHP, MySQL: Display only required parts of my website in sister website

    - by Devner
    Hi all, Now I have my website built on PHP & Mysql. Consider this like a forum. Now when a user posts a reply in my website 1 (ex. www.website1.com), I want to be able to show the starting thread and it's related replies in a sister website of mine. I want to do this in a way that it does not show the rest of the page & other page contents (like logo etc.). I don't think iframe would be a solution because an iframe would embed the whole page and the users visiting my sister website (totally different domain i.e. www.website2.com) would be able to see all the page contents, like logo etc. I want to avoid that. I want to make them see only limited information from website 1 and only the info. that I intend. I hope that makes sense. In a way, you could say that I am trying to replicate my 1 website, and show only a limited part of it. Users browsing 2nd website can post a reply in the 2nd website and it should automatically be posted & visible to the visitors of the website 1. Users of website 1 should not know that a user of website 2 has posted it. They would feel that some user from website 1 has posted it. Do I have to use 2 separate mysql DB or just 1? I think it would be problematic if I am trying to use different DB. I also feel I might have to face DB connectivity issues as I can connect to only 1 DB at a time. It's basically like users of website1.com should feel that they are replying to users of website1.com & users of website2.com should feel that they are replying to users of website2.com. (I need it this way to bridge the gap between them). At the same time I want to make the front end of the websites different so that they don't feel that they are replying to some other users outside the domain. These websites would be under my control and I will have access to the source code at any time. If I need to change the source code, these changes are welcome. Is this really possible? Thank you in advance.

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  • Translate SQL statement into named_scope?

    - by keruilin
    How can I translate this SQL into a named_scope? Also, I want the total comments param to be passed through a lambda. "select users., count() as total_comments from users, comments where (users.id = comments.user_id) and (comments.public_comment = 1) and (comments.aasm_state = 'posted') and (comments.forum_user_id is null) group by users.id having total_comments 25"

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  • Data from two tables without repeating data from the first?

    - by Aran
    I have two tables. Users table and Users Meta Table I am looking for a way to get all the information out of both tables with one query. But without repeating the information from Users table. This is all information relating to the users id number as well. So for example user_id = 1. Is there a way to query the database and collect all the information I from both tables without repeating the information from the first?

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  • Information Builders lance WebFocus Mobile BI, sa solution BI pour Smarphones et tablettes

    Information Builders lance WebFocus Mobile BI Sa solution BI pour Smarphones et tablettes Information Builders annonce le lancement de WebFOCUS Mobile BI, sa nouvelle solution de Business Intelligence qui permet de bénéficier des fonctionnalités analytiques sur mobiles via n'importe quel terminal, application mobile ou navigateur. La solution utilise les rapports WebFOCUS Active Technology en les optimisant pour les rendre compatibles avec tous les types de smartphone tels que l'iPhone, les téléphones sous Android, les Blackberry et les tablettes. Les rapports Active Technology permettent aux utilisateurs de manipuler et d'analyser les informations métiers sur leurs terminaux mobi...

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  • Using JQuery tabs in an HTML 5 page

    - by nikolaosk
    In this post I will show you how to create a simple tabbed interface using JQuery,HTML 5 and CSS.Make sure you have downloaded the latest version of JQuery (minified version) from http://jquery.com/download.Please find here all my posts regarding JQuery.Also have a look at my posts regarding HTML 5.In order to be absolutely clear this is not (and could not be) a detailed tutorial on HTML 5. There are other great resources for that.Navigate to the excellent interactive tutorials of W3School.Another excellent resource is HTML 5 Doctor.Two very nice sites that show you what features and specifications are implemented by various browsers and their versions are http://caniuse.com/ and http://html5test.com/. At this times Chrome seems to support most of HTML 5 specifications.Another excellent way to find out if the browser supports HTML 5 and CSS 3 features is to use the Javascript lightweight library Modernizr.In this hands-on example I will be using Expression Web 4.0.This application is not a free application. You can use any HTML editor you like.You can use Visual Studio 2012 Express edition. You can download it here. Let me move on to the actual example.This is the sample HTML 5 page<!DOCTYPE html><html lang="en">  <head>    <title>Liverpool Legends</title>    <meta http-equiv="Content-Type" content="text/html;charset=utf-8" >    <link rel="stylesheet" type="text/css" href="style.css">    <script type="text/javascript" src="jquery-1.8.2.min.js"> </script>     <script type="text/javascript" src="tabs.js"></script>       </head>  <body>    <header>        <h1>Liverpool Legends</h1>    </header>     <section id="tabs">        <ul>            <li><a href="http://weblogs.asp.net/controlpanel/blogs/posteditor.aspx?SelectedNavItem=Posts§ionid=1153&postid=9143136#first-tab">Defenders</a></li>            <li><a href="http://weblogs.asp.net/controlpanel/blogs/posteditor.aspx?SelectedNavItem=Posts§ionid=1153&postid=9143136#second-tab">Midfielders</a></li>            <li><a href="http://weblogs.asp.net/controlpanel/blogs/posteditor.aspx?SelectedNavItem=Posts§ionid=1153&postid=9143136#third-tab">Strikers</a></li>        </ul>   <div id="first-tab">     <h3>Liverpool Defenders</h3>     <p> The best defenders that played for Liverpool are Jamie Carragher, Sami Hyypia , Ron Yeats and Alan Hansen.</p>   </div>   <div id="second-tab">     <h3>Liverpool Midfielders</h3>     <p> The best midfielders that played for Liverpool are Kenny Dalglish, John Barnes,Ian Callaghan,Steven Gerrard and Jan Molby.        </p>   </div>   <div id="third-tab">     <h3>Liverpool Strikers</h3>     <p>The best strikers that played for Liverpool are Ian Rush,Roger Hunt,Robbie Fowler and Fernando Torres.<br/>      </p>   </div> </div></section>            <footer>        <p>All Rights Reserved</p>      </footer>     </body>  </html>  This is very simple HTML markup. I have styled this markup using CSS.The contents of the style.css file follow* {    margin: 0;    padding: 0;}header{font-family:Tahoma;font-size:1.3em;color:#505050;text-align:center;}#tabs {    font-size: 0.9em;    margin: 20px 0;}#tabs ul {    float: left;    background: #777;    width: 260px;    padding-top: 24px;}#tabs li {    margin-left: 8px;    list-style: none;}* html #tabs li {    display: inline;}#tabs li, #tabs li a {    float: left;}#tabs ul li.active {    border-top:2px red solid;    background: #15ADFF;}#tabs ul li.active a {    color: #333333;}#tabs div {    background: #15ADFF;    clear: both;    padding: 15px;    min-height: 200px;}#tabs div h3 {    margin-bottom: 12px;}#tabs div p {    line-height: 26px;}#tabs ul li a {    text-decoration: none;    padding: 8px;    color:#0b2f20;    font-weight: bold;}footer{background-color:#999;width:100%;text-align:center;font-size:1.1em;color:#002233;}There are some CSS rules that style the various elements in the HTML 5 file. These are straight-forward rules. The JQuery code lives inside the tabs.js file $(document).ready(function(){$('#tabs div').hide();$('#tabs div:first').show();$('#tabs ul li:first').addClass('active'); $('#tabs ul li a').click(function(){$('#tabs ul li').removeClass('active');$(this).parent().addClass('active');var currentTab = $(this).attr('href');$('#tabs div').hide();$(currentTab).show();return false;});}); I am using some of the most commonly used JQuery functions like hide , show, addclass , removeClass I hide and show the tabs when the tab becomes the active tab. When I view my page I get the following result Hope it helps!!!!!

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  • nautilus selected item color

    - by shantanu
    See in the image, selected item "build" colour is black as background colour. How can i change the selected item colour gtk3 theme's nautilus.css script Which section colour need to modify: /* desktop mode */ .nautilus-desktop.nautilus-canvas-item { color: @bg_color; text-shadow: 1 1 alpha (#001B33, 0.8); } .nautilus-desktop.nautilus-canvas-item:active { background-image: none; background-color: alpha (@selected_bg_color, 0.84); border-radius: 4; color: @fg_color; } .nautilus-desktop.nautilus-canvas-item:selected { background-image: none; background-color: alpha (@bg_color, 0.84); border-radius: 4; color: @selected_fg_color; } .nautilus-desktop.nautilus-canvas-item:active, .nautilus-desktop.nautilus-canvas-item:prelight, .nautilus-desktop.nautilus-canvas-item:selected { text-shadow: none; } /* browser window */ NautilusTrashBar.info, NautilusXContentBar.info, NautilusSearchBar.info, NautilusQueryEditor.info { /* this background-color controls the symbolic icon in the entry */ background-color: mix (@fg_color, @base_color, 0.3); border-radius: 0; border-style: solid; border-width: 0 1 1 1; } NautilusSearchBar .entry { } .nautilus-cluebar-label { color: @fg_color; font: bold; } #nautilus-search-button *:active, #nautilus-search-button *:active:prelight { color: @dark_fg_color; } NautilusFloatingBar { background-color: @info_bg_color; border-radius: 3 3 0 0; border-style: solid; border-width: 1; border-color: darker (@info_bg_color); -unico-border-gradient: none; } NautilusFloatingBar .button { -GtkButton-image-spacing: 0; -GtkButton-inner-border: 0; } /* sidebar */ NautilusWindow .sidebar, NautilusWindow .sidebar .view { background-color: @bg_color; color: @fg_color; } NautilusWindow .sidebar .frame { } NautilusWindow > GtkTable > .pane-separator { background-color: @bg_color; border-color: @bg_color; border-width: 0 0 0 0; border-style: solid; }

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  • How do I change the background colour of Leafpad?

    - by Lao Tzu
    All the information that google returns says to change ~/.gtkrc2.0-mine. Here is my .gtkrc-2.0: # -- THEME AUTO-WRITTEN BY gtk-theme-switch2 DO NOT EDIT include "/usr/share/themes/Dust/gtk-2.0/gtkrc" include "/home/mars/.gtkrc-2.0.mine" # -- THEME AUTO-WRITTEN BY gtk-theme-switch2 DO NOT EDIT Here is my .gtkrc-2.0.mine: style "default" { GtkTextView::cursor_color = "#ffffff" base[NORMAL] = "#111111" base[ACTIVE] = "#111181" base[SELECTED] = "#808080" text[NORMAL] = "#c0c0c0" text[ACTIVE] = "#c0c0c0" text[SELECTED] = "#111111" } class "GtkTextView" style "default" Still appears with a white background!

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