Search Results

Search found 61449 results on 2458 pages for 'base class library'.

Page 555/2458 | < Previous Page | 551 552 553 554 555 556 557 558 559 560 561 562  | Next Page >

  • HtmlHelper Getting the route name

    - by Simon G
    Hi, I've created a html helper that adds a css class property to a li item if the user is on the current page. The helper looks like this: public static string MenuItem( this HtmlHelper helper, string linkText, string actionName, string controllerName, object routeValues, object htmlAttributes ) { string currentControllerName = ( string )helper.ViewContext.RouteData.Values["controller"]; string currentActionName = ( string )helper.ViewContext.RouteData.Values["action"]; var builder = new TagBuilder( "li" ); // Add selected class if ( currentControllerName.Equals( controllerName, StringComparison.CurrentCultureIgnoreCase ) && currentActionName.Equals( actionName, StringComparison.CurrentCultureIgnoreCase ) ) builder.AddCssClass( "active" ); // Add link builder.InnerHtml = helper.ActionLink( linkText, actionName, controllerName, routeValues, htmlAttributes ); // Render Tag Builder return builder.ToString( TagRenderMode.Normal ); } I want to expand this class so I can pass a route name to the helper and if the user is on that route then it adds the css class to the li item. However I'm having difficulty finding the route the user is on. Is this possible? The code I have so far is: public static string MenuItem( this HtmlHelper helper, string linkText, string routeName, object routeValues, object htmlAttributes ) { string currentControllerName = ( string )helper.ViewContext.RouteData.Values["controller"]; string currentActionName = ( string )helper.ViewContext.RouteData.Values["action"]; var builder = new TagBuilder( "li" ); // Add selected class // Some code for here // if ( routeName == currentRoute ) AddCssClass; // Add link builder.InnerHtml = helper.RouteLink( linkText, routeName, routeValues, htmlAttributes ); // Render Tag Builder return builder.ToString( TagRenderMode.Normal ); } BTW I'm using MVC 1.0. Thanks

    Read the article

  • Create 2 connection pools using c3p0 in Jetty

    - by Mike
    Hello, I'm trying to set up a maven web project that runs Jetty. In this project, I need 2 JNDIs... my plan is to configure 2 connection pools using c3p0 in Jetty. So, I created WEB-INF/jetty-env.xml, and I have the following:- <Configure class="org.mortbay.jetty.webapp.WebAppContext"> <New id="ds1" class="org.mortbay.jetty.plus.naming.Resource"> <Arg>jdbc/ds1</Arg> <Arg> <New class="com.mchange.v2.c3p0.ComboPooledDataSource"> // ... JTDS to SQL Server - omitted for brevity </New> </Arg> </New> <New id="ds2" class="org.mortbay.jetty.plus.naming.Resource"> <Arg>jdbc/ds2</Arg> <Arg> <New class="com.mchange.v2.c3p0.ComboPooledDataSource"> // ... JTDS to Sybase - omitted for brevity </New> </Arg> </New> </Configure> When I run jetty, I get this exception:- May 14, 2010 1:16:56 PM com.mchange.v2.c3p0.impl.AbstractPoolBackedDataSource getPoolManager INFO: Initializing c3p0 pool... com.mchange.v2.c3p0.ComboPooledDataSource [ acquireIncrement -> ... ... ... Exception in thread "com.mchange.v2.async.ThreadPoolAsynchronousRunner$PoolThread-#0" java.lang.LinkageError: net.sourceforge.jtds.jdbc.DefaultProperties at java.lang.ClassLoader.defineClassImpl(Native Method) at java.lang.ClassLoader.defineClass(ClassLoader.java:258) It seems to me that I can't create 2 connection pools using c3p0. If I remove either one of the connection pool, it worked. What am I doing wrong? How do I create 2 connection pools in Jetty? Thanks much.

    Read the article

  • Why isn't django-nose running the doctests in my models?

    - by Conley Owens
    I'm trying to use doctests with django-nose. All my doctests are running, except not any doctests within a model (unless it is abstract). class TestModel1(models.Model): """ >>> print 'pass' pass """ pass class TestModel2(models.Model): """ >>> print 'pass' pass """ class Meta: abstract = True pass The first doctest does not run and the second does. Why is this?

    Read the article

  • Accessing django choice field

    - by Hulk
    there is a module as header , from test.models import SEL_VALUES class rubrics_header(models.Model): sel_values = models.IntegerField(choices=SEL_VALUES) So when SEL_VALUES is imported from test.modules.What is the code that has to go in views to get the choices in sel_values . And the test.modules has the following, class SEL_VALUES: vaue = 0 value2 = 1 class Entries(forms.Form) : models.IntegerField(choices=SEL_VALUES) SEL_VALUES = ((ACCESS.value,'NAME'),(ACCESS.value2,'DESIGNATION'))

    Read the article

  • How to change UIBarButtonItem Title - iPhone SDK

    - by user340226
    Hi Can anyone help me in that I am only trying to change the Title on an UIBarButtonItem from a different class. My code is: -(IBAction)spanishPush { SafetyTalks *bbiTitle= [[SafetyTalks alloc]init]; bbiTitle.bbiOpenPopOver.title = @"Spanish"; } SafetyTalks = the class I am trying to reference bbiOpenPopOver = the UIBarButtonItem. I can change the Title when in the SafetyTalks class by simple: bbiOpenPopOver.title = @"Talk Topics"; but cannot do it when I am out of that class. Please help. Andy

    Read the article

  • No EJB receiver available for handling [appName:,modulename:HelloWorldSessionBean,distinctname:]

    - by zoit
    I'm trying to develop my first EJB with an Example I found, I have the next mistake: Exception in thread "main" java.lang.IllegalStateException: No EJB receiver available for handling [appName:,modulename:HelloWorldSessionBean,distinctname:] combination for invocation context org.jboss.ejb.client.EJBClientInvocationContext@41408b80 at org.jboss.ejb.client.EJBClientContext.requireEJBReceiver(EJBClientContext.java:584) at org.jboss.ejb.client.ReceiverInterceptor.handleInvocation(ReceiverInterceptor.java:119) at org.jboss.ejb.client.EJBClientInvocationContext.sendRequest(EJBClientInvocationContext.java:181) at org.jboss.ejb.client.EJBInvocationHandler.doInvoke(EJBInvocationHandler.java:136) at org.jboss.ejb.client.EJBInvocationHandler.doInvoke(EJBInvocationHandler.java:121) at org.jboss.ejb.client.EJBInvocationHandler.invoke(EJBInvocationHandler.java:104) at $Proxy0.sayHello(Unknown Source) at com.ibytecode.client.EJBApplicationClient.main(EJBApplicationClient.java:16) I use JBOSS 7.1, and the code is this: HelloWorld.java package com.ibytecode.business; import javax.ejb.Remote; @Remote public interface HelloWorld { public String sayHello(); } HelloWorldBean.java package com.ibytecode.businesslogic; import com.ibytecode.business.HelloWorld; import javax.ejb.Stateless; /** * Session Bean implementation class HelloWorldBean */ @Stateless public class HelloWorldBean implements HelloWorld { /** * Default constructor. */ public HelloWorldBean() { } public String sayHello() { return "Hello World !!!"; } } EJBApplicationClient.java: package com.ibytecode.client; import javax.naming.Context; import javax.naming.NamingException; import com.ibytecode.business.HelloWorld; import com.ibytecode.businesslogic.HelloWorldBean; import com.ibytecode.clientutility.ClientUtility; public class EJBApplicationClient { public static void main(String[] args) { // TODO Auto-generated method stub HelloWorld bean = doLookup(); System.out.println(bean.sayHello()); // 4. Call business logic } private static HelloWorld doLookup() { Context context = null; HelloWorld bean = null; try { // 1. Obtaining Context context = ClientUtility.getInitialContext(); // 2. Generate JNDI Lookup name String lookupName = getLookupName(); // 3. Lookup and cast bean = (HelloWorld) context.lookup(lookupName); } catch (NamingException e) { e.printStackTrace(); } return bean; } private static String getLookupName() { /* The app name is the EAR name of the deployed EJB without .ear suffix. Since we haven't deployed the application as a .ear, the app name for us will be an empty string */ String appName = ""; /* The module name is the JAR name of the deployed EJB without the .jar suffix. */ String moduleName = "HelloWorldSessionBean"; /*AS7 allows each deployment to have an (optional) distinct name. This can be an empty string if distinct name is not specified. */ String distinctName = ""; // The EJB bean implementation class name String beanName = HelloWorldBean.class.getSimpleName(); // Fully qualified remote interface name final String interfaceName = HelloWorld.class.getName(); // Create a look up string name String name = "ejb:" + appName + "/" + moduleName + "/" + distinctName + "/" + beanName + "!" + interfaceName; return name; } } ClientUtility.java package com.ibytecode.clientutility; import java.util.Properties; import javax.naming.Context; import javax.naming.InitialContext; import javax.naming.NamingException; public class ClientUtility { private static Context initialContext; private static final String PKG_INTERFACES = "org.jboss.ejb.client.naming"; public static Context getInitialContext() throws NamingException { if (initialContext == null) { Properties properties = new Properties(); properties.put("jboss.naming.client.ejb.context", true); properties.put(Context.URL_PKG_PREFIXES, PKG_INTERFACES); initialContext = new InitialContext(properties); } return initialContext; } } properties.file: remote.connectionprovider.create.options.org.xnio.Options.SSL_ENABLED=false remote.connections=default remote.connection.default.host=localhost remote.connection.default.port = 4447 remote.connection.default.connect.options.org.xnio.Options.SASL_POLICY_NOANONYMOUS=false This is what I have. Why I have this?. Thanks so much. Regards

    Read the article

  • How to select form input based on label inner HTML?

    - by Shane
    I have multiple forms that are dynamically created with different input names and id's. The only thing unique they will have is the inner HTML of the label. Is it possible to select the input via the label inner HTML with jQuery? Here is an example of one of my patient date of birth blocks, there are many and all unique except for innerHTML. <div class="iphorm-element-spacer iphorm-element-spacer-text iphorm_1_8-element-spacer"> <label for="iphorm_081a9e2e6b9c83d70496906bb4671904150cf4b43c0cb1_8">events=Object { mouseover=[1], mouseout=[1]}handle=function()data=Object { InFieldLabels={...}} Patient DOB <span class="iphorm-required">*</span> </label> <div class="iphorm-input-wrap iphorm-input-wrap-text iphorm_1_8-input-wrap"> <input id="iphorm_081a9e2e6b9c83d70496906bb4671904150cf4b43c0cb1_8" class="iphorm-element-text iphorm_1_8" type="text" value="" name="iphorm_1_8">events=Object { focus=[1], blur=[1], keydown=[1], more...}handle=function() </div> <div class="iphorm-errors-wrap iphorm-hidden"> </div> This is in a Wordpress Plugin and because we are building to allow employees to edit their sites (this is actually a Wordpress Network), we do not want to alter the plugin if possible. Note that the label "for" and the input "id" share the same dynamic key, so this might be a way to maybe get the id, but wanted to see if there is a shorter way of doing this.

    Read the article

  • How to implement login page using Spring Security so that it works with Spring web flow?

    - by simon
    I have a web application using Spring 2.5.6 and Spring Security 2.0.4. I have implemented a working login page, which authenticates the user against a web service. The authentication is done by defining a custom authentincation manager, like this: <beans:bean id="customizedFormLoginFilter" class="org.springframework.security.ui.webapp.AuthenticationProcessingFilter"> <custom-filter position="AUTHENTICATION_PROCESSING_FILTER" /> <beans:property name="defaultTargetUrl" value="/index.do" /> <beans:property name="authenticationFailureUrl" value="/login.do?error=true" /> <beans:property name="authenticationManager" ref="customAuthenticationManager" /> <beans:property name="allowSessionCreation" value="true" /> </beans:bean> <beans:bean id="customAuthenticationManager" class="com.sevenp.mobile.samplemgmt.web.security.CustomAuthenticationManager"> <beans:property name="authenticateUrlWs" value="${WS_ENDPOINT_ADDRESS}" /> </beans:bean> The authentication manager class: public class CustomAuthenticationManager implements AuthenticationManager, ApplicationContextAware { @Transactional @Override public Authentication authenticate(Authentication authentication) throws AuthenticationException { //authentication logic return new UsernamePasswordAuthenticationToken(principal, authentication.getCredentials(), grantedAuthorityArray); } The essential part of the login jsp looks like this: <c:url value="/j_spring_security_check" var="formUrlSecurityCheck"/> <form method="post" action="${formUrlSecurityCheck}"> <div id="errorArea" class="errorBox"> <c:if test="${not empty param.error}"> ${sessionScope["SPRING_SECURITY_LAST_EXCEPTION"].message} </c:if> </div> <label for="loginName"> Username: <input style="width:125px;" tabindex="1" id="login" name="j_username" /> </label> <label for="password"> Password: <input style="width:125px;" tabindex="2" id="password" name="j_password" type="password" /> </label> <input type="submit" tabindex="3" name="login" class="formButton" value="Login" /> </form> Now the problem is that the application should use Spring Web Flow. After the application was configured to use Spring Web Flow, the login does not work anymore - the form action to "/j_spring_security_check" results in a blank page without error message. What is the best way to adapt the existing login process so that it works with Spring Web Flow?

    Read the article

  • Is it possible to change the model name in the django admin site?

    - by luc
    Hello, I am translating a django app and I would like to translate also the homepage of the django admin site. On this page are listed the application names and the model class names. I would like to translate the model class name but I don't find how to give a user-friendly name for a model class. Does anybody know how to do that?

    Read the article

  • Django loading mysql data into template correctly

    - by user805981
    I'm new to django and I'm trying to get display a list of buildings and sort them alphabetically, then load it into an html document. Is there something that I am not doing correctly? below is models.py class Class(models.Model): building = models.CharField(max_length=20) class Meta: db_table = u'class' def __unicode__(self): return self.building below is views.py views.py def index(request): buildinglist = Class.objects.all().order_by('building') c = {'buildinglist': buildinglist} t = loader.get_template('index.html') return HttpResponse(t.render(c)) below is index.html index.html {% block content%} <h3>Buildings:</h3> <ul> {% for building in buildinglist %} <li> <a href='www.{% building %}.com'> # ex. www.searstower.com </li> {% endfor %} </ul> {% endblock %} Can you guys point me in the right direction? Thank you in advance guys! I appreciate your help very much.

    Read the article

  • MVC - Calling Controller Methods

    - by JT703
    Hello, My application is following the MVC design pattern. The problem I keep running into is needing to call methods inside a Controller class from outside that Controller class (ex. A View class wants to call a Controller method, or a Manager class wants to call a Controller method). Is calling Controller methods in this way allowed in MVC? If it's allowed, what's the proper way to do it? According to the version of MVC that I am following (there seems to be so many different versions out there), the View knows of the Model, and the Controller knows of the View. Doing it this way, I can't access the controller. Here's the best site I've found and the one describing the version of MVC I'm following: http://leepoint.net/notes-java/GUI/structure/40mvc.html. The Main Program code block really shows how this works. Thanks for any answers.

    Read the article

  • Is it possible to override List accessors in Grails domain classes?

    - by Ali G
    If I have a List in a Grails domain class, is there a way to override the addX() and removeX() accessors to it? In the following example, I'd expect MyObject.addThing(String) to be called twice. In fact, the output is: Adding thing: thing 2 class MyObject { static hasMany = [things: String] List things = [] void addThing(String newThing) { println "Adding thing: ${newThing}" things << newThing } } class BootStrap { def init = { servletContext -> MyObject o = new MyObject().save() o.things << 'thing 1' o.addThing('thing 2') } def destroy = { } }

    Read the article

  • Project setup for creating third party libraries for Android

    - by Jarle Hansen
    Hi all, I am creating a library for Android that others can include in their own project. So far I have been working on it as a normal Java project with JDK 1.6 setup as system library. This works just fine in Eclipse when I add the android.jar. The issue comes when I try to my build script. I am running Gradle and doing a normal compile and test build cycle. My thoughts were that it does not matter if I compile it with a normal JDK, since this is not a standalone application. The benefits by creating a normal Java project is that Gradle does support this much better. My project also does not contain any UI at all. However, the problem is that of course android.jar and the JDK contains lots of the same classes and I think that this is what messes up my build script. Everything crashes when running the tests (the tests are in the same project under src/test/java). My question is, how should I create this project that is meant to be included in Android projects as a third party library? Should I create it as an Android project in Eclipse even though I am only creating a library that does not use any of the UI features? Also, should the tests be in a separate project? Thanks for all responses!

    Read the article

  • PerformancePoint dashboard permissions problem in MOSS

    - by Nathan DeWitt
    I have a PerformancePoint dashboard running in MOSS 2007 portal. The dashboard consists of one SSRS 2005 report, running in SharePoint Integrated mode. NT Authority\Authenticated Users have read permissions to the report library containing the SSRS report, the dashboard, and the report library containing the dashboard. Users that attempt to access the dashboard receive the following error message: The permissions granted to user 'DOMAIN\firstname.lastname' are insufficient for performing this operation. (rsAccessDenied) Users that then click on the direct link to the report in MOSS will see the report with no problem. Subsequent visits to the dashboard show the report with no problem. The report is using a data source that is located one folder up from the report location. The report has been updated to point to the correct shared data source after deployment. Both the report and the data source have been published. The data source is using stored credentials, with a domain service account that has been set to Use as Windows credentials. This service account is serving other reports in other areas with no problem. Edit: Ok, I've gotten a lot more information on this problem. The request is never actually being made to the data source. The user comes in to the dashboard and requests a report for the first time using their kerberos token identifying themselves. The report looks in the Report Server database and finds that they are not listed in the users table and generates this rsAccessDenied error. Once they view the report directly their name is in this table and they never have the problem again. Unfortunately, removing the user from the Users table in the RS database doesn't actually cause this error to happen again. Everything I've read says that when you run a Report Server in MOSS integrated mode all your permissions are handled at the MOSS report library level, and all Auth users have permissions to the report library, as stated earlier. Any ideas?

    Read the article

  • jQuery fadein fadeout divs over set interval

    - by theDawckta
    I want to fadeOut the first div in a collection and then fadeIn the next div. The fade in out would trigger at a set time. The number of items in the collection is 1 to n. Here is an example of the html; <div class="contentPanel"> <div class="content"> <div style="border: solid 2px black; text-align: center"> This is first content </div> </div> <div class="content"> <div style="border: solid 2px black; text-align: center"> This is second content </div> </div> <div class="content"> <div style="border: solid 2px black; text-align: center"> This is third content </div> </div> </div> So on page load the first "content" class would be visible, after x amount of time, the current "content" would fadeout and the next "content" would fade in. When it got to the nth "content" it would start over, fadeout the nth "content" and fadein the first "content". This behavior would loop continuously.

    Read the article

  • More issues with IntelliJ 9.0.1 "Hello World" in Scala - Predef version 5.0 vs 4.1

    - by Alex R
    Any ideas what could cause this? Scala signature Predef has wrong version Expected 5.0 found: 4.1 in .... scala-library.jar I tried both versions 2.7.6 and 2.8 RC1 of scala-*.jar, the result was the same. JDK is 1.6.u20. UPDATE Today uninstalled IntelliJ 9.0.1, and installed 9.0.2 Early Availability, with the 4/14 stable version of the Scala plug-in. Then I setup a project from scratch through the wizards: new project from scratch JDK is 1.6.u20 accept the default (project) instead of global / module accept the download of Scala 2.8.0beta1 into project's lib folder Created a new class: object hello { def main(args: Array[String]) { println("hello: " + args); } } For my efforts, I now have a brand-new error :) Here it is: Scalac internal error: class java.lang.ClassNotFoundException [java.net.URLClassLoader$1.run(URLClassLoader.java:202), java.security.AccessController.doPrivileged(Native Method), java.net.URLClassLoader.findClass(URLClassLoader.java:190), java.lang.ClassLoader.loadClass(ClassLoader.java:307), sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301), java.lang.ClassLoader.loadClass(ClassLoader.java:248), java.lang.Class.forName0(Native Method), java.lang.Class.forName(Class.java:169), org.jetbrains.plugins.scala.compiler.rt.ScalacRunner.main(ScalacRunner.java:72)] Thanks

    Read the article

  • Limit foreign key choices in select in an inline form in admin

    - by mightyhal
    Edited :-) Hopefully a bit clearer now. The logic is of the model is: A Building has many Rooms A Room may be inside another Room (a closet, for instance--ForeignKey on 'self') A Room can only in inside of another Room in the same building (this is the tricky part) Here's the code I have: #spaces/models.py from django.db import models class Building(models.Model): name=models.CharField(max_length=32) def __unicode__(self): return self.name class Room(models.Model): number=models.CharField(max_length=8) building=models.ForeignKey(Building) inside_room=models.ForeignKey('self',blank=True,null=True) def __unicode__(self): return self.number and: #spaces/admin.py from ex.spaces.models import Building, Room from django.contrib import admin class RoomAdmin(admin.ModelAdmin): pass class RoomInline(admin.TabularInline): model = Room extra = 2 class BuildingAdmin(admin.ModelAdmin): inlines=[RoomInline] admin.site.register(Building, BuildingAdmin) admin.site.register(Room) The inline will display only rooms in the current building (which is what I want). The problem, though, is that for the inside_room drop down, it displays all of the rooms in the Rooms table (including those in other buildings). In the inline of rooms, I need to limit the inside_room choices to only rooms which are in the current building being displayed by the main form. I can't figure out a way to do it with either a limit_choices_to in the model, nor can I figure out how exactly to override the admin's inline formset properly (I feel like I should be somehow create a custom inline form, pass the building_id of the main form to the custom inline, then limit the queryset for the field's choices based on that--but I just can't wrap my head around how to do it). Maybe this is too complex for the admin site, but it seems like something that would be generally useful... Thanks again for your help!

    Read the article

  • Why would someone use ob_start in this manner and what is the point?

    - by meder
    Something is failing in the class I copied over. It's not my class, but the relevant bit that fails is: class foo { function process() { ob_start( array( &$this, 'parseTemplate' ) ); } function parseTemplate(){} } Does anyone know what the ob_start expression is supposed to do? Call the parse_template method in the context of a copy of &$this? PHP Version is 5.3.2-1. I suspect that the class was coded for 5.0-5.2 and it breaks in 5.3? or could it be something else?

    Read the article

  • How do I use a different image for each JQuery UI Slider handle

    - by Tom
    I'm using a JQuery UI slider which has two handles (a.k.a range slider). I know how to style the first handle: .ui-slider-horizontal .ui-slider-handle {background: white url(http://stackoverflow.com/content/img/so/vote-arrow-down.png) no-repeat scroll 50% 50%;} But how do I style the second handle differently? Using Firebug I can see Jquery does not uniquely identify each handle: <div id="hourlyRateSlider" class="ui-slider ui-slider-horizontal ui-widget ui-widget-content ui-corner-all"> <div class="ui-slider-range ui-widget-header" style="left: 26%; width: 46%;"/> <a class="ui-slider-handle ui-state-default ui-corner-all" href="#" style="left: 26%;"/> <a class="ui-slider-handle ui-state-default ui-corner-all" href="#" style="left: 72%;"/> </div> So I imagine I have to use either a CSS child selector which could be cross-browser problematic. Or I could use some JQuery trickery to add a CSS class to the second handle? Anyone done this in a neat way before?

    Read the article

  • servlet-mapping for Wordpress on Tomcat using Quercus

    - by Jeremy
    I have a web app running in Tomcat and I'm trying to add a Wordpress blog to it using Quercus. It works if I hit a .php file in my blog, but links to my articles are structured like http://myapp.com/blog/2011/01/my-first-post/ which don't work. Below is my web.xml: <welcome-file-list> <welcome-file>index.do</welcome-file> <welcome-file>index.php</welcome-file> </welcome-file-list> <context-param> <param-name>contextConfigLocation</param-name> <param-value>/WEB-INF/myapp-service.xml</param-value> </context-param> <listener> <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class> </listener> <servlet> <servlet-name>myapp</servlet-name> <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>myapp</servlet-name> <url-pattern>*.do</url-pattern> </servlet-mapping> <servlet> <servlet-name>Quercus Servlet</servlet-name> <servlet-class>com.caucho.quercus.servlet.QuercusServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>Quercus Servlet</servlet-name> <url-pattern>*.php</url-pattern> </servlet-mapping> I've tried many combos of url-pattern such as /blog, /blog/*, etc. I can't get anything to work. Any help is appreciated. Thanks!

    Read the article

  • Is this a valid XPath expression?

    - by sid_com
    Is this xpath a valid XPath expression? (It does what it should ). #!/usr/bin/env perl use strict; use warnings; use 5.012; use XML::LibXML; my $string =<<EOS; <result> <cd> <artists> <artist class="1">Pumkinsingers</artist> <artist class="2">Max and Moritz</artist> </artists> <title>Hello, Hello</title> </cd> <cd> <artists> <artist class="3">Green Trees</artist> <artist class="4">The Leons</artist> </artists> <title>The Shield</title> </cd> </result> EOS #/ my $parser = XML::LibXML->new(); my $doc = $parser->load_xml( string => $string ); my $root = $doc->documentElement; my $xpath = '/result/cd[artists[artist[@class="2"]]]/title'; my @nodes = $root->findnodes( $xpath ); for my $node ( @nodes ) { say $node->textContent; }

    Read the article

  • Blackberry read local properties file in project

    - by Dachmt
    Hi, I have a config.properties file at the root of my blackberry project (same place as Blackberry_App_Descriptor.xml file), and I try to access the file to read and write into it. See below my class: public class Configuration { private String file; private String fileName; public Configuration(String pathToFile) { this.fileName = pathToFile; try { // Try to load the file and read it System.out.println("---------- Start to read the file"); file = readFile(fileName); System.out.println("---------- Property file:"); System.out.println(file); } catch (Exception e) { System.out.println("---------- Error reading file"); System.out.println(e.getMessage()); } } /** * Read a file and return it in a String * @param fName * @return */ private String readFile(String fName) { String properties = null; try { System.out.println("---------- Opening the file"); //to actually retrieve the resource prefix the name of the file with a "/" InputStream is = this.getClass().getResourceAsStream(fName); //we now have an input stream. Create a reader and read out //each character in the stream. System.out.println("---------- Input stream"); InputStreamReader isr = new InputStreamReader(is); char c; System.out.println("---------- Append string now"); while ((c = (char)isr.read()) != -1) { properties += c; } } catch (Exception e) { } return properties; } } I call my class constructor like this: Configuration config = new Configuration("/config.properties"); So in my class, "file" should have all the content of the config.properties file, and the fileName should have this value "/config.properties". But the "name" is null because the file cannot be found... I know this is the path of the file which should be different, but I don't know what i can change... The class is in the package com.mycompany.blackberry.utils Thank you!

    Read the article

  • Eclipse doesn't build

    - by Christian
    A previously working Ecplise now gives me the error Java Virtual Machine Launcher Could not find main class: testing2. Program will exist. testing2 is my class and a source file exists but Ecplise doesn't seem to build the .class file. Maybe I hit the wrong hotkey and changed accidently some setting?

    Read the article

  • Event Handling for MFC Dialog

    - by Maksud
    This is my second question of the day, pardon me. I am writing a wrapper library to communicate with a scanner device. The source code was in C++ MFC. I am converting it to a plain Dll which will be invoked from C#. So, I am using DllImport in C# to call the wrapper library. Now I am provided with MFC code and the library is a ActiveX Object, at least I think so. class CDpocx : public CWnd { } So in my wrapper library I will have an instance of CDpocx and will call it via C# P/Invoke. But the problem is CDpocx also throws some events which I need to catch. In traditional app, I would just attach an function with it. But How would I attach the events on non MFC class. I have seen something like: BEGIN_EVENTSINK_MAP(CVC60Dlg, CDialog) //{{AFX_EVENTSINK_MAP(CVC60Dlg) ON_EVENT(CVC60Dlg, IDC_DPOCXCTRL1, 1 , OnReadyDpocxctrl1, VTS_NONE) //}}AFX_EVENTSINK_MAP END_EVENTSINK_MAP() OnReadyDpocxctrl1 is the function that handles 1 (Ready) event. How can I gain simmilar function in non MFC class. Regards, Maksud

    Read the article

  • extending django usermodel

    - by imran-glt
    Hi i am trying to create a signup form for my django app. for this i have extended the user model. This is my Forms.py from contact.models import register from django import forms from django.contrib import auth class registerForm(forms.ModelForm): class Meta: model=register fields = ('latitude', 'longitude', 'status') class Meta: model = auth.models.User # this gives me the User fields fields = ('username', 'first_name', 'last_name', 'email') and this is my model.py from django.db import models from django.contrib.auth.models import User STATUS_CHOICES = ( ('Online', 'Online.'), ('Busy', 'Busy.'), ('AppearOffline', 'AppearOffline.'),) class register(models.Model): user = models.ForeignKey('auth.User', unique = True) latitude = models.DecimalField(max_digits=8, decimal_places=6) longitude = models.DecimalField(max_digits=8, decimal_places=6) status = models.CharField(max_length=8,choices=STATUS_CHOICES, blank= True, null=True) i dont know where i am making a mistake. the users passwords are not accepted at the login and the latitude and logitude are not saved against the created user user. i am fiarly new to django and dont know what to do any body have any solution .?

    Read the article

< Previous Page | 551 552 553 554 555 556 557 558 559 560 561 562  | Next Page >