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  • Getting unhandled error and connection get lost when a client tries to communicate with chat server in twisted

    - by user2433888
    from twisted.internet.protocol import Protocol,Factory from twisted.internet import reactor class ChatServer(Protocol): def connectionMade(self): print "A Client Has Connected" self.factory.clients.append(self) print"clients are ",self.factory.clients self.transport.write('Hello,Welcome to the telnet chat to sign in type aim:YOUR NAME HERE to send a messsage type msg:YOURMESSAGE '+'\n') def connectionLost(self,reason): self.factory.clients.remove(self) self.transport.write('Somebody was disconnected from the server') def dataReceived(self,data): #print "data is",data a = data.split(':') if len(a) > 1: command = a[0] content = a[1] msg="" if command =="iam": self.name + "has joined" elif command == "msg": ma=sg = self.name + ":" +content print msg for c in self.factory.clients: c.message(msg) def message(self,message): self.transport.write(message + '\n') factory = Factory() factory.protocol = ChatServer factory.clients = [] reactor.listenTCP(80,factory) print "Iphone Chat server started" reactor.run() The above code is running succesfully...but when i connect the client (by typing telnet localhost 80) to this chatserver and try to write message ,connection gets lost and following errors occurs : Iphone Chat server started A Client Has Connected clients are [<__main__.ChatServer instance at 0x024AC0A8>] Unhandled Error Traceback (most recent call last): File "C:\Python27\lib\site-packages\twisted\python\log.py", line 84, in callWithLogger return callWithContext({"system": lp}, func, *args, **kw) File "C:\Python27\lib\site-packages\twisted\python\log.py", line 69, in callWithContext return context.call({ILogContext: newCtx}, func, *args, **kw) File "C:\Python27\lib\site-packages\twisted\python\context.py", line 118, in callWithContext return self.currentContext().callWithContext(ctx, func, *args, **kw) File "C:\Python27\lib\site-packages\twisted\python\context.py", line 81, in callWithContext return func(*args,**kw) --- --- File "C:\Python27\lib\site-packages\twisted\internet\selectreactor.py", line 150, in _doReadOrWrite why = getattr(selectable, method)() File "C:\Python27\lib\site-packages\twisted\internet\tcp.py", line 199, in doRead rval = self.protocol.dataReceived(data) File "D:\chatserverultimate.py", line 21, in dataReceived content = a[1] exceptions.IndexError: list index out of range Where am I going wrong?

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  • Display all feeds using simplexml_load() using PHP

    - by Jean
    Hello, I want to loop to get all feeds, but displaying only one $url = "http://localhost/feeds/feeds.rss"; $xml = simplexml_load_file($url); foreach($xml->item as $result){ echo $result->description."<br>"; } RSS Feed is - <channel> <title>/</title> <link>/</link> <atom:link type="application/rss+xml" href="/" rel="self"/> <description>/</description> <language>/</language> <ttl>/</ttl> <item> <title>/</title> <description>/</description> <pubDate>/</pubDate> <guid>/</guid> <link>/</link> </item> <item> <title>/</title> <description>/</description> <pubDate>/</pubDate> <guid>/</guid> <link>/</link> </item> </channel> Thanks Jean

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  • How i can convert client for Axis 1.4 to Axis2 in JAVA ?

    - by dahevos
    Hello, First of all i success to programming a client for an Axis 1.2 web service, but for Axis2 i don't know how i can do, and the tutorial in Apache don't really help me. Here my code : import java.net.URL; import javax.xml.namespace.QName; import org.apache.axis.client.Call; import org.apache.axis.client.Service; public class EmployeClient { public static void main(String [] args) throws Exception { Service service = new Service(); Call call = (Call)service.createCall(); String endpoint = "http://localhost:8080/axis/services/EmployeService"; call.setTargetEndpointAddress(new URL(endpoint)); call.setOperationName(new QName("getCurrentID")); String dept = "marketing"; String name = "sacha"; String position = (String)call.invoke(new Object [] {new String(dept), new String(name)}); System.out.println("Résultat de la recherche : " + position ); } } So how can i do for convert this code in Axis2 ? Thanks you very much. ps : i'm french, sorry for my bad english !

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  • php oop and mysql

    - by gloris
    I need to get data, to check and send to db. Programming with PHP OOP. Could you tell me if my class structure is good and how dislpay all data?. Thanks <?php class Database{ private $DBhost = 'localhost'; private $DBuser = 'root'; private $DBpass = 'root'; private $DBname = 'blog'; public function connect(){ //Connect to mysql db } public function select($rows){ //select data from db } public function insert($rows){ //Insert data to db } public function delete($rows){ //Delete data from db } } class CheckData{ public $number1; public $number2; public function __construct(){ $this->number1 = $_POST['number1']; $this->number2 = $_POST['number2']; } function ISempty(){ if(!empty($this->$number1)){ echo "Not Empty"; $data = new Database(); $data->insert($this->$number1); } else{ echo "Empty1"; } if(!empty($this->$number2)){ echo "Not Empty"; $data = new Database(); $data->insert($this->$number2); } else{ echo "Empty2"; } } } class DisplayData{ //How print all data? function DisplayNumber(){ $data = new Database(); $data->select(); } } $check = new CheckData(); $check->ISempty(); $display = new DisplayData() $display->DisplayNumber(); ?>

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  • Any suggestions to improve my PDO connection class?

    - by Scarface
    Hey guys I am pretty new to pdo so I basically just put together a simple connection class using information out of the introductory book I was reading but is this connection efficient? If anyone has any informative suggestions, I would really appreciate it. class PDOConnectionFactory{ public $con = null; // swich database? public $dbType = "mysql"; // connection parameters public $host = "localhost"; public $user = "user"; public $senha = "password"; public $db = "database"; public $persistent = false; // new PDOConnectionFactory( true ) <--- persistent connection // new PDOConnectionFactory() <--- no persistent connection public function PDOConnectionFactory( $persistent=false ){ // it verifies the persistence of the connection if( $persistent != false){ $this->persistent = true; } } public function getConnection(){ try{ $this->con = new PDO($this->dbType.":host=".$this->host.";dbname=".$this->db, $this->user, $this->senha, array( PDO::ATTR_PERSISTENT => $this->persistent ) ); // carried through successfully, it returns connected return $this->con; // in case that an error occurs, it returns the error; }catch ( PDOException $ex ){ echo "We are currently experiencing technical difficulties. We have a bunch of monkies working really hard to fix the problem. Check back soon: ".$ex->getMessage(); } } // close connection public function Close(){ if( $this->con != null ) $this->con = null; } }

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  • Can not issue data manipulation statements with executeQuery in java

    - by user225269
    I'm trying to insert records in mysql database using java, What do I place in this code so that I could insert records: String id; String name; String school; String gender; String lang; Scanner inputs = new Scanner(System.in); System.out.println("Input id:"); id=inputs.next(); System.out.println("Input name:"); name=inputs.next(); System.out.println("Input school:"); school= inputs.next(); System.out.println("Input gender:"); gender= inputs.next(); System.out.println("Input lang:"); lang=inputs.next(); Class.forName("com.mysql.jdbc.Driver"); Connection con = DriverManager.getConnection("jdbc:mysql://localhost:3306/employee_record", "root", "MyPassword"); PreparedStatement statement = con.prepareStatement("insert into employee values('id', 'name', 'school', 'gender', 'lang');"); statement.executeUpdate();

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  • Mail function wont send eMail. ERROR

    - by Peter
    I think i tried to fix this issue fr 3 days now and cant seem to find the problem. I use XAMPP and use this code: <?php $to = "[email protected]"; $subject = "Test mail"; $message = "Hello! This is a simple email message."; $from = "[email protected]"; $headers = "From: $from"; $res= mail($to,$subject,$message,$headers); echo " $res Mail Sent."; ?> when i enter that page i get an error that says: Warning: mail() [function.mail]: Failed to connect to mailserver at "localhost" port 25, verify your "SMTP" and "smtp_port" setting in php.ini or use ini_set( My php.init file in xampp are as follow: [mail function] ; For Win32 only. ; http://php.net/smtp SMTP = smpt.gmail.com ; http://php.net/smtp-port smtp_port = 25 That is all my codes.

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  • Sencha : how to pass parameter to php using Ext.data.HttpProxy?

    - by Lauraire Jérémy
    I have successfully completed this great tutorial : http://www.sencha.com/learn/ext-js-grids-with-php-and-sql/ I just can't use the baseParams field specified with the proxy... Here is my code that follows tutorial description : __ My Store : Communes.js ____ Ext.define('app.store.Communes', { extend: 'Ext.data.Store', id: 'communesstore', requires: ['app.model.Commune'], config: { model: 'app.model.Commune', departement:'var', // the proxy with POST method proxy: new Ext.data.HttpProxy({ url: 'app/php/communes.php', // File to connect to method: 'POST' }), // the parameter passed to the proxy baseParams:{ departement: "VAR" }, // the JSON parser reader: new Ext.data.JsonReader({ // we tell the datastore where to get his data from rootProperty: 'results' }, [ { name: 'IdCommune', type: 'integer' }, { name: 'NomCommune', type: 'string' } ]), autoLoad: true, sortInfo:{ field: 'IdCommune', direction: "ASC" } } }); _____ The php file : communes.php _____ <?php /** * CREATE THE CONNECTION */ mysql_connect("localhost", "root", "pwd") or die("Could not connect: " . mysql_error()); mysql_select_db("databasename"); /** * INITIATE THE POST */ $departement = 'null'; if ( isset($_POST['departement'])){ $departement = $_POST['departement']; // Get this from Ext } getListCommunes($departement); /** * */ function getListCommunes($departement) { [CODE HERE WORK FINE : just a connection and query but $departement is NULL] } ?> There is no parameter passed as POST method... Any idea?

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  • django test client trouble

    - by Anton Koval'
    I've got a problem... we're writing project using django, and i'm trying to use django.test.client with nose test-framework for tests. Our code is like this: from simplejson import loads from urlparse import urljoin from django.test.client import Client TEST_URL = "http://smakly.localhost:9090/" def test_register(): cln = Client() ref_data = {"email": "[email protected]", "name": "???????", "website": "http://hot.bear.com", "xhr": "true"} print urljoin(TEST_URL, "/accounts/register/") response = loads(cln.post(urljoin(TEST_URL, "/accounts/register/"), ref_data)) print response["message"] and in nose output I catch: Traceback (most recent call last): File "/home/psih/work/svn/smakly/eggs/nose-0.11.1-py2.6.egg/nose/case.py", line 183, in runTest self.test(*self.arg) File "/home/psih/work/svn/smakly/src/smakly.tests/smakly/tests/frontend/test_profile.py", line 25, in test_register response = loads(cln.post(urljoin(TEST_URL, "/accounts/register/"), ref_data)) File "/home/psih/work/svn/smakly/parts/django/django/test/client.py", line 313, in post response = self.request(**r) File "/home/psih/work/svn/smakly/parts/django/django/test/client.py", line 225, in request response = self.handler(environ) File "/home/psih/work/svn/smakly/parts/django/django/test/client.py", line 69, in __call__ response = self.get_response(request) File "/home/psih/work/svn/smakly/parts/django/django/core/handlers/base.py", line 78, in get_response urlconf = getattr(request, "urlconf", settings.ROOT_URLCONF) File "/home/psih/work/svn/smakly/parts/django/django/utils/functional.py", line 273, in __getattr__ return getattr(self._wrapped, name) AttributeError: 'Settings' object has no attribute 'ROOT_URLCONF' My settings.py file does have this attribute. If I get the data from the server with standard urllib2.urllopen().read() it works in the proper way. Any ideas how I can solve this case?

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  • Advise guidance on how to form this jQuery script for show/hide fade element

    - by Rick
    Hey guys.. I basically have several links on the left side of the screen and on the right is a preview window. Below the preview window is another box for the affiliate link code. So what I am trying to do is create an affiliate page where you choose the banner size on the left by clicking on the link and on the right you see it dynamically change to the banner size and the code changes accordingly as well. So far I have the following code and it works but it seems very very cumbersome and bloated. Can you see if I can trim this down? jQuery(".banner-style li").click(function() { jQuery(".banner-style li").removeClass("selected"); jQuery(this).addClass("selected"); var $banner = jQuery(this).attr("class"); $banner = $banner.replace(" selected",""); jQuery(".preview img").fadeOut('fast',function() { jQuery(".preview img").attr("src", "http://localhost/site/banners/"+$banner+".jpg") .fadeIn('slow'); }); jQuery(".code p").removeClass('hide').hide(); jQuery(".code p."+$banner).show(); }); Also to note the funny thing is in FF, when you click for the first to on any link, the original image on the right fades out and in real quick and then it loads the "clicked" image. This does not happen in other browsers...

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  • Passing list of values to django view via jQuery ajax call

    - by finspin
    I'm trying to pass a list of numeric values (ids) from one web page to another with jQuery ajax call. I can't figure out how to pass and read all the values in the list. I can successfully post and read 1 value but not multiple values. Here is what I have so far: jQuery: var postUrl = "http://localhost:8000/ingredients/"; $('li').click(function(){ values = [1, 2]; $.ajax({ url: postUrl, type: 'POST', data: {'terid': values}, traditional: true, dataType: 'html', success: function(result){ $('#ingredients').append(result); } }); }); /ingredients/ view: def ingredients(request): if request.is_ajax(): ourid = request.POST.get('terid', False) ingredients = Ingredience.objects.filter(food__id__in=ourid) t = get_template('ingredients.html') html = t.render(Context({'ingredients': ingredients,})) return HttpResponse(html) else: html = '<p>This is not ajax</p>' return HttpResponse(html) With Firebug I can see that POST contains both ids but probably in the wrong format (terid=1&terid=2). So my ingredients view picks up only terid=2. What am I doing wrong? EDIT: To clarify, I need the ourid variable pass values [1, 2] to the filter in the ingredients view.

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  • Can no longer debug ActiveX controls in Visual Studio 2008

    - by Phenglei Kai
    For a long time I would throw up a DebugBreak() or ASSERT(false) in the startup code of my ActiveX control, load up IE, go to a localhost page hosting my control, wait for the dialog to show up, then debug my application. I could also launch it under the debugger by setting IE as the container. I tried again for the first time in 2 months and now this no longer works. If I use the ASSERT(false) method, when I get the Visual C++ Debug dialog and click "retry", IE simply closes without any debugger activity. When I try launching from VS2008 and hoping the DebugBreak() will kick in after I load the page, VS2008 does break, but it says either the "RPC Server is Unavailable" or the "RPC Client Call failed." I am never allowed to have my application in the debugger and it doesn't show up in the modules list of VS. The stack trace in VS2008 only contains Microsoft DLLs and modules and not a hint of my code. I assume it's something I've picked up through Windows Update that broke this. Has anyone else ever seen this issue and know how to make it go away? As it stands, I'm now completely unable to debug my ActiveX control.

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  • I can't insert data into my database

    - by Ken
    I don't know why, but my data doesn't go into my database 'users' with the table 'data'. <html> <body> <?php date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "g00dfor@boy"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db(`users`, $con) or die(mysql_error()); mysql_query(INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password')) or die(mysql_error()); mysql_close($con) echo("Thank you for registering!"); ?> </body> </html> All i get is a blank page.

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  • PHP and MySQL echoing out a Table

    - by user1631702
    Okay, so I've done this before, and it worked. I am trying to echo out specific rows on my database in a table. Here is my code: <?php $connect = mysql_connect("localhost", "xxx", "xxx") or die ("Hey loser, check your server connection."); mysql_select_db("xxx"); $quey1="select * from `Ad Requests`"; $result=mysql_query($quey1) or die(mysql_error()); ?> <table border=1 style="background-color:#F0F8FF;" > <caption><EM>Student Record</EM></caption> <tr> <th>Student ID</th> <th>Student Name</th> <th>Class</th> </tr> <?php while($row=mysql_fetch_array($result)){ echo "</td><td>"; echo $row['id']; echo "</td><td>"; echo $row['twitter']; echo "</td><td>"; echo $row['why']; echo "</td></tr>"; } echo "</table>"; ?> It gives me no errors, but It just shows a blank table with none of these rows. My Question: How come this wont show any rows in the table, what am I doing wrong?

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  • Error with MySQL Query

    - by Ken
    Okay, I must be an idiot, because this is my 3rd question for today. Here's my code: date_default_timezone_set("America/Los_Angeles"); include("mainmenu.php"); $con = mysql_connect("localhost", "root", "********"); if(!$con){ die(mysql_error()); } $usrname = $_POST['usrname']; $fname = $_POST['fname']; $lname = $_POST['lname']; $password = $_POST['password']; $email = $_POST['email']; mysql_select_db("`users`, $con) or die(mysql_error()"); $query = ("INSERT INTO `users`.`data` (`id`, `usrname`, `fname`, `lname`, `email`, `password`) VALUES (NULL, '$usrname', '$fname', '$lname', '$email', 'password'))"); mysql_query('$query') or die(mysql_error()); mysql_close($con); echo("Thank you for registering!"); I always get the error returned as: "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$query' at line 1. Help a newbie. I'm about to stab my monitor.

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  • Cleaning up PHP Code

    - by Michael
    Hi, I've noticed I am a very sloppy coder and do things out of the ordinary. Can you take a look at my code and give me some tips on how to code more efficiently? What can I do to improve? session_start(); /check if the token is correct/ if ($_SESSION['token'] == $_GET['custom1']){ /*connect to db*/ mysql_connect('localhost','x','x') or die(mysql_error()); mysql_select_db('x'); /*get data*/ $orderid = mysql_real_escape_string($_GET['order_id']); $amount = mysql_real_escape_string($_GET['amount']); $product = mysql_real_escape_string($_GET['product1Name']); $cc = mysql_real_escape_string($_GET['Credit_Card_Number']); $length = strlen($cc); $last = 4; $start = $length - $last; $last4 = substr($cc, $start, $last); $ipaddress = mysql_real_escape_string($_GET['ipAddress']); $accountid = $_SESSION['user_id']; $credits = mysql_real_escape_string($_GET['custom3']); /*insert history into db*/ mysql_query("INSERT into billinghistory (orderid, price, description, credits, last4, orderip, accountid) VALUES ('$orderid', '$amount', '$product', '$credits', '$last4', '$ipaddress', '$accountid')"); /*add the credits to the users account*/ mysql_query("UPDATE accounts SET credits = credits + $credits WHERE user_id = '$accountid'"); /*redirect is successful*/ header("location: index.php?x=1"); }else{ /*something messed up*/ header("location: error.php"); }

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  • Can anyone get app-engine plugin working with Grails on mac os x?

    - by tim
    I have been trying for 4 days to get app-engine and grails working together on my mac to no avail. I am using latest groovy/grails and appengine sdk versions. Im following the app-engine plugin step by step on the grails site.. http://grails.org/plugin/app-engine Groovy Version: 1.7.1 JVM: 1.5.0_22 Grails 1.3.0.RC1 echo $APPENGINE_HOME reveals /Users/markstim/appengine-java-sdk-1.3.2 I perform the following steps 1. grails create-app myapp 2. cd myapp; grails list-plugins reveals hibernate 1.3.0.RC1 -- Hibernate for Grails tomcat 1.3.0.RC1 -- Apache Tomcat plugin for Grails add the following line to Config.groovy google.appengine.application="myapp" install the plugin for app-engine grails install-plugin app-engine and answer 'jpa' when asked (no errors yet) installed plugins list now looks like app-engine 0.8.9 -- Grails AppEngine plugin gorm-jpa 0.7.1 -- GORM-JPA Plugin then grails run-app and get this error as the server is coming up... [java] WARNING: Nested in org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'pluginManager' defined in ServletContext resource [/WEB-INF/applicationContext.xml]: Invocation of init method failed; nested exception is org.codehaus.groovy.grails.exceptions.NewInstanceCreationException: Could not create a new instance of class [GormJpaGrailsPlugin]!: [java] java.lang.NoClassDefFoundError: org.grails.jpa.JpaPluginSupport then if i navigate to localhost:8080 I get HTTP ERROR: 503 Problem accessing /myapp. Reason: SERVICE_UNAVAILABLE Powered by Jetty://

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  • MySQL Database is Indexed at Apache Solr, How to access it via URL

    - by Wasim
    data-config.xml <dataConfig> <dataSource encoding="UTF-8" type="JdbcDataSource" driver="com.mysql.jdbc.Driver" url="jdbc:mysql://localhost:3306/somevisits" user="root" password=""/> <document name="somevisits"> <entity name="login" query="select * from login"> <field column="sv_id" name="sv_id" /> <field column="sv_username" name="sv_username" /> </entity> </document> </dataConfig> schema.xml <?xml version="1.0" encoding="UTF-8" ?> <schema name="example" version="1.5"> <fields> <field name="sv_id" type="string" indexed="true" stored="true" required="true" multiValued="false" /> <field name="username" type="string" indexed="true" stored="true" required="true"/> <field name="_version_" type="long" indexed="true" stored="true" multiValued="false"/> <field name="text" type="string" indexed="true" stored="false" multiValued="true"/> </fields> <uniqueKey>sv_id</uniqueKey> <types> <fieldType name="string" class="solr.StrField" sortMissingLast="true" /> <fieldType name="long" class="solr.TrieLongField" precisionStep="0" positionIncrementGap="0"/> </types> </schema> Solr successfully imported mysql database using full http://[localSolr]:8983/solr/#/collection1/dataimport?command=full-import My question is, how to access that mysql imported database now?

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  • Problem with checkboxes, sql select statements & php

    - by smokey20
    I am trying to display some rows from a database table based on choices submitted by the user. Here is my form code <form action="choice.php" method="POST" > <input type="checkbox" name="variable[]" value="Apple">Apple <input type="checkbox" name="variable[]" value="Banana">Banana <input type="checkbox" name="variable[]" value="Orange">Orange <input type="checkbox" name="variable[]" value="Melon">Melon <input type="checkbox" name="variable[]" value="Blackberry">Blackberry From what I understand I am placing the values of these into an array called variable. Two of my columns are called receipe name and ingredients(each field under ingredients can store a number of fruits). What I would like to do is, if a number of checkboxes are selected then the receipe name/s is displayed. Here is my php code. <?php // Make a MySQL Connection mysql_connect("localhost", "*****", "*****") or die(mysql_error()); mysql_select_db("****") or die(mysql_error()); $variable=$_POST['variable']; foreach ($variable as $variablename) { echo "$variablename is checked"; } $query = "SELECT receipename FROM fruit WHERE $variable like ingredients"; $row = mysql_fetch_assoc($result); foreach ($_POST['variabble'] as $ingredients) echo $row[$ingredients] . '<br/>'; ?> I am very new to php and just wish to display the data, I do not need to perform any actions on it. I have tried many select statements but I cannot get any results to display. My db connection is fine and it does print out what variables are checked. Many thanks in advance.

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  • What does this model error says in asp.net mvc?

    - by Pandiya Chendur
    My Html action link takes me to a view where i can see the details.. For Ex:http://localhost:1985/Materials/Details/4 But it shows error, The model item passed into the dictionary is of type 'System.Data.Linq.DataQuery`1[CrMVC.Models.ConstructionRepository+Materials]' but this dictionary requires a model item of type 'CrMVC.Models.ConstructionRepository+Materials'. And my model is... public IQueryable<Materials> GetMaterial(int id) { return from m in db.Materials join Mt in db.MeasurementTypes on m.MeasurementTypeId equals Mt.Id where m.Mat_id == id select new Materials() { Id = Convert.ToInt64(m.Mat_id), Mat_Name = m.Mat_Name, Mes_Name = Mt.Name, }; } public class Materials { private Int64 id; public string mat_Name; public string mes_Name; public Int64 Id { get { return id; } set { id = value; } } public string Mat_Name { get { return mat_Name; } set { mat_Name = value; } } public string Mes_Name { get { return mes_Name; } set { mes_Name = value; } } } } and my controller method... public ActionResult Details(int id) { var material = consRepository.GetMaterial(id).AsQueryable(); return View("Details", material); } Any suggestion what am i missing here?

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  • ASP.NET MVC 2 - Custom route doesn't find controller action

    - by mcfroob
    For some reason my application isn't routing to my controller method correctly. I have a routelink like this in my webpage - <%= Html.RouteLink("View", "Blog", new { id=(item.BlogId), slug=(item.Slug) }) %> In global.asax.cs I have the following routes - routes.IgnoreRoute("{resource}.axd/{*pathInfo}"); routes.MapRoute( "MoreBlogs", "Blog/Page/{page}", new { controller = "Blog", action = "Index" } ); routes.MapRoute( "Blog", "Blog/View/{id}/{slug}", new { controller = "Blog", action = "View"} ); routes.MapRoute( "Default", // Route name "{controller}/{action}/{id}", // URL with parameters new { controller = "Blog", action = "Index", id = UrlParameter.Optional } // Parameter defaults ); And then I have a class BlogController that has a method - public ActionResult View(int id, string slug) { ... etc. } I put a breakpoint in the first line of the View method but it's not getting hit at all. I checked with a route debugger for the format localhost/Blog/View/1/test and it matched my custom route. All I'm getting is a 404 while running this, I can't work out why the route won't post to the view method in my controller - any ideas?

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  • Rails 3, Devise and custom controller action

    - by Johnny Klassy
    routes.rb match 'agencies/stub' => 'agencies#stub', :via => :get resources :agencies Here's the rake routes dump agencies_stub GET /agencies/stub(.:format) {:controller=>"agencies", :action=>"stub"} agencies GET /agencies(.:format) {:action=>"index", :controller=>"agencies"} POST /agencies(.:format) {:action=>"create", :controller=>"agencies"} new_agency GET /agencies/new(.:format) {:action=>"new", :controller=>"agencies"} edit_agency GET /agencies/:id/edit(.:format) {:action=>"edit", :controller=>"agencies"} agency GET /agencies/:id(.:format) {:action=>"show", :controller=>"agencies"} PUT /agencies/:id(.:format) {:action=>"update", :controller=>"agencies"} DELETE /agencies/:id(.:format) {:action=>"destroy", :controller=>"agencies"} Devise is setup to have all agenciesroutes only accessible as admin. The call I'm testing with is http://xyz:12345@localhost:3000/agencies/stub but it doesn't authenticate properly, ie, it doesn't recognize it as admin and throws me back to the Devise login page. The creds are a valid admin account. I'm baffled and have no idea why this is happening. Any insights will be much appreciated.

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  • Checking if an SSH tunnel is up and running

    - by Jarmund
    I have a perl script which, when destilled a bit, looks like this: my $randport = int(10000 + rand(1000)); # Random port as other scripts like this run at the same time my $localip = '192.168.100.' . ($port - 4000); # Don't ask... backwards compatibility system("ssh -NL $randport:$localip:23 root\@$ip -o ConnectTimeout=60 -i somekey &"); # create the tunnel in the background sleep 10; # Give the tunnel some time to come up # Create the telnet object my $telnet = new Net::Telnet( Timeout => 10, Host => 'localhost', Port => $randport, Telnetmode => 0, Errmode => \&fail, ); # SNIPPED... a bunch of parsing data from $telnet The thing is that the target $ip is on a link with very unpredictable bandwidth, so the tunnel might come up right away, it might take a while, it might not come up at all. So a sleep is necessary to give the tunnel some time to get up and running. So the question is: How can i test if the tunnel is up and running? 10 seconds is a really undesirable delay if the tunnel comes up straight away. Ideally, i would like to check if it's up and continue with creating the telnet object once it is, to a maximum of, say, 30 seconds.

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  • qTip pop ups come in from top left of screen (on first load)

    - by franko75
    Hi, not sure if i'm set things up incorrectly - I don't seem to see anyone else with this problem, but my qTip popups (all ajax loaded content) are loading quite erratically, in that they are often animating in from off screen before appearing in the correct position. Is there a simple solution to this which I may have missed? Thanks again for your help. HTML markup: <span class="formInfo"> <a href="http://localhost/httpdocs/index.php/help/kc_dob" class="jTip" name="" id="dob_help">?</a> </span> qTip initialisation.. //set up for qtip function initQtip() { $('a.jTip').each(function() { $(this).qtip( { content: { // Set the text to an image HTML string with the correct src URL to the loading image you want to use text: '<img src="/media/images/wait.gif" alt="Loading..." />', url: $(this).attr('href') // Use the rel attribute of each element for the url to load }, position: { adjust: { screen: true // Keep the tooltip on-screen at all times } }, show: { when: 'click', solo: true // Only show one tooltip at a time }, hide: 'unfocus', style: { tip: true, // Apply a speech bubble tip to the tooltip at the designated tooltip corner border: { width: 10, radius: 10 }, width: { min: 200, max: 500 }, name: 'light' // Use the default light style } }); //prevent default event on click }).bind('click', function(event){ event.preventDefault(); return false; }); }

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  • AngularJS - $routeParams Empty on $locationChangeSuccess

    - by Marc M.
    I configure my app in the following run block. Basically I want to preform an action that requires me to know the $routeParams every $locationChangeSuccess. However $routeParams is empty at this point! Are there any work rounds? What's going on? app.run(['$routeParams', function ($routeParams) { $rootScope.$on("$locationChangeSuccess", function () { console.log($routeParams); }); }]); UPDATE function configureApp(app, user) { app.config(['$routeProvider', function ($routeProvider) { $routeProvider. when('/rentroll', { templateUrl: 'rent-roll/rent-roll.html', controller: 'pwRentRollCtrl' }). when('/bill', { templateUrl: 'bill/bill/bill.html', controller: 'pwBillCtrl' }). when('/fileroom', { templateUrl: 'file-room/file-room/file-room.html', controller: 'pwFileRoomCtrl' }). when('/estate-creator', { templateUrl: 'estate/creator.html' }). when('/estate-manager', { templateUrl: 'estate/manager.html', controller: 'pwEstateManagerCtrl' }). when('/welcomepage', { templateURL: 'welcome-page/welcome-page.html', controller: 'welcomePageCtrl' }). otherwise({ redirectTo: '/welcomepage' }); }]); app.run(['$rootScope', '$routeParams', 'pwCurrentEstate','pwToolbar', function ($rootScope, $routeParams, pwCurrentEstate, pwToolbar) { $rootScope.user = user; $rootScope.$on("$locationChangeSuccess", function () { pwToolbar.reset(); console.log($routeParams); }); }]); } Accessing URL: http://localhost:8080/landlord/#/rentroll?landlord-account-id=ahlwcm9wZXJ0eS1tYW5hZ2VtZW50LXN1aXRlchwLEg9MYW5kbG9yZEFjY291bnQYgICAgICAgAoM&billing-month=2014-06

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