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  • php + MySQL editing table data.

    - by Jacksta
    This question is relating to 2 php scripts. The first script is called pick_modcontact.php where I choose a contact (from a contact book like phone book), then posts to the script show_modcontact.php When I click the submit button on the form on pick.modcontact.php. As a result of submitting the form I am then taken to show_modcontact.php. As the variables are not present the user is directed back to pick_modcontact.php I can not work out how to correct the code so that it will show the results of the script show_modcontact.php This script shows all contacts in a database which is an "address book" this part works fine. please see below. Name:pick_modcontact.php if ($_SESSION['valid'] != "yes") { header( "Location: contact_menu.php"); exit; } $db_name = "testDB"; $table_name = "my_contacts"; $connection = @mysql_connect("localhost", "admin", "user") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT id, f_name, l_name FROM $table_name ORDER BY f_name"; $result = @mysql_query($sql, $connection) or die(mysql_error()); $num = @mysql_num_rows($result); if ($num < 1) { $display_block = "<p><em>Sorry No Results!</em></p>"; } else { while ($row = mysql_fetch_array($result)) { $id = $row['id']; $f_name = $row['f_name']; $l_name = $row['l_name']; $option_block .= "<option value\"$id\">$f_name, $l_name</option>"; } $display_block = "<form method=\"POST\" action=\"show_modcontact.php\"> <p><strong>Contact:</strong> <select name=\"id\">$option_block</select> <input type=\"submit\" name=\"submit\" value=\"Select This Contact\"></p> </form>"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Modify A Contact</title> </head> <body> <h1>My Contact Management System</h1> <h2><em>Modify a Contact</em></h2> <p>Select a contact from the list below, to modify the contact's record.</p> <? echo "$display_block"; ?> <br> <p><a href="contact_menu.php">Return to Main Menu</a></p> </body> </html> This script is for modifying the contact: named show_modcontact.php <?php if (!$_POST['id']) { header( "Location: pick_modcontact.php"); exit; } else { session_start(); } if ($_SESSION['valid'] != "yes") { header( "Location: pick_modcontact.php"); exit; } $db_name = "testDB"; $table_name = "my_contacts"; $connection = @mysql_connect("localhost", "admin", "pass") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT f_name, l_name, address1, address2, address3, postcode, prim_tel, sec_tel, email, birthday FROM $table_name WHERE id = '" . $_POST['id'] . "'"; $result = @mysql_query($sql, $connection) or die(mysql_error()); while ($row = mysql_fetch_array($result)) { $f_name = $row['f_name']; $l_name = $row['l_name']; $address1 = $row['address1']; $address2 = $row['address2']; $address3 = $row['address3']; $country = $row['country']; $prim_tel = $row['prim_tel']; $sec_tel = $row['sec_tel']; $email = $row['email']; $birthday = $row['birthday']; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Modify A Contact</title> </head> <body> <form action="do_modcontact.php" method="post"> <input type="text" name="id" value="<? echo $_POST['id']; ?>" /> <table cellpadding="5" cellspacing="3"> <tr> <th>Name & Address Information</th> <th> Other Contact / Personal Information</th> </tr> <tr> <td align="top"> <p><strong>First Name:</strong><br /> <input type="text" name="f_name" value="<? echo "$f_name"; ?>" size="35" maxlength="75" /></p> <p><strong>Last Name:</strong><br /> <input type="text" name="l_name" value="<? echo "$l_name"; ?>" size="35" maxlength="75" /></p> <p><strong>Address1:</strong><br /> <input type="text" name="f_name" value="<? echo "$address1"; ?>" size="35" maxlength="75" /></p> <p><strong>Address2:</strong><br /> <input type="text" name="f_name" value="<? echo "$address2"; ?>" size="35" maxlength="75" /></p> <p><strong>Address3:</strong><br /> <input type="text" name="f_name" value="<? echo "$address3"; ?>" size="35" maxlength="75" /> </p> <p><strong>Postcode:</strong><br /> <input type="text" name="f_name" value="<? echo "$postcode"; ?>" size="35" maxlength="75" /></p> <p><strong>Country:</strong><br /> <input type="text" name="f_name" value="<? echo "$country"; ?>" size="35" maxlength="75" /> </p> <p><strong>First Name:</strong><br /> <input type="text" name="f_name" value="<? echo "$f_name"; ?>" size="35" maxlength="75" /></p> </td> <td align="top"> <p><strong>Prim Tel:</strong><br /> <input type="text" name="f_name" value="<? echo "$prim_tel"; ?>" size="35" maxlength="75" /></p> <p><strong>Sec Tel:</strong><br /> <input type="text" name="f_name" value="<? echo "$sec_tel"; ?>" size="35" maxlength="75" /></p> <p><strong>Email:</strong><br /> <input type="text" name="f_name" value="<? echo "$email;" ?>" size="35" maxlength="75" /> </p> <p><strong>Birthday:</strong><br /> <input type="text" name="f_name" value="<? echo "$birthday"; ?>" size="35" maxlength="75" /> </p> </td> </tr> <tr> <td align="center"> <p><input type="submit" name="submit" value="Update Contact" /></p> <br /> <p><a href="contact_menu.php">Retuen To Menu</a></p> </td> </tr> </table> </form> </body> </html> note for site admin, I am re posting this question with the hope of someone else reading over it. older questions seem to go dead after a while.

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  • How to use WebSockets refresh multi-window for play framework 1.2.7

    - by user2468652
    My code can work.But only refresh a page of one window. If I open window1 and window2 , both open websocket connect. I keyin word "test123" in window1, click sendbutton. Only refresh window1. How to refresh window1 and window2 ? Client <script> window.onload = function() { document.getElementById('sendbutton').addEventListener('click', sendMessage,false); document.getElementById('connectbutton').addEventListener('click', connect, false); } function writeStatus(message) { var html = document.createElement("div"); html.setAttribute('class', 'message'); html.innerHTML = message; document.getElementById("status").appendChild(html); } function connect() { ws = new WebSocket("ws://localhost:9000/ws?name=test"); ws.onopen = function(evt) { writeStatus("connected"); } ws.onmessage = function(evt) { writeStatus("response: " + evt.data); } } function sendMessage() { ws.send(document.getElementById('messagefield').value); } </script> </head> <body> <button id="connectbutton">Connect</button> <input type="text" id="messagefield"/> <button id="sendbutton">Send</button> <div id="status"></div> </body> Play Framework WebSocketController public class WebSocket extends WebSocketController { public static void test(String name) { while(inbound.isOpen()) { WebSocketEvent evt = await(inbound.nextEvent()); if(evt instanceof WebSocketFrame) { WebSocketFrame frame = (WebSocketFrame)evt; System.out.println("received: " + frame.getTextData()); if(!frame.isBinary()) { if(frame.getTextData().equals("quit")) { outbound.send("Bye!"); disconnect(); } else { outbound.send("Echo: %s", frame.getTextData()); } } } } } }

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  • log4j rootLogger seems to inherit log level of other logger. Why?

    - by AndrewR
    I've got a log4J setup in which the root logger is supposed to log ERROR level messages and above to the console and another logger logs everything to syslog. log4j.properties is: # Root logger option log4j.rootLogger=ERROR,R log4j.appender.R=org.apache.log4j.ConsoleAppender log4j.appender.R.layout=org.apache.log4j.PatternLayout log4j.appender.R.layout.ConversionPattern=%d %p %t %c - %m%n log4j.logger.SGSearch=DEBUG,SGSearch log4j.appender.SGSearch=org.apache.log4j.net.SyslogAppender log4j.appender.SGSearch.SyslogHost=localhost log4j.appender.SGSearch.Facility=LOCAL6 log4j.appender.SGSearch.layout=org.apache.log4j.PatternLayout log4j.appender.SGSearch.layout.ConversionPattern=[%-5p] %m%n In code I do private static final Logger logger = Logger.getLogger("SGSearch"); . . . logger.info("Commencing snapshot index [" + args[1] + " -> " + args[2] + "]"); What is happening is that I get the console logging for all logging levels. What seems to be happening is that the level for SGSearch overrides the level set for the root logger somehow. I can't figure it out. I have confirmed that Log4J is reading then properties file I think it is, and no other (via the -Dlog4j.debug option)

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  • getting data from dynamic schema

    - by coure2011
    I am using mongoose/nodejs to get data as json from mongodb. For using mongoose I need to define schema first like this var mongoose = require('mongoose'); var Schema = mongoose.Schema; var GPSDataSchema = new Schema({ createdAt: { type: Date, default: Date.now } ,speed: {type: String, trim: true} ,battery: { type: String, trim: true } }); var GPSData = mongoose.model('GPSData', GPSDataSchema); mongoose.connect('mongodb://localhost/gpsdatabase'); var db = mongoose.connection; db.on('open', function() { console.log('DB Started'); }); then in code I can get data from db like GPSData.find({"createdAt" : { $gte : dateStr, $lte: nextDate }}, function(err, data) { res.writeHead(200, { "Content-Type": "application/json", "Access-Control-Allow-Origin": "*" }); var body = JSON.stringify(data); res.end(body); }); How to define scheme for a complex data like this, you can see that subSection can go to any deeper level. [ { 'title': 'Some Title', 'subSection': [{ 'title': 'Inner1', 'subSection': [ {'titile': 'test', 'url': 'ab/cd'} ] }] }, .. ]

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  • python on apache - getting 404

    - by Kirby
    I edited this question after i found a solution... i need to understand why the solution worked instead of my method? This is likely to be a silly question. I tried searching other questions that are related... but to no avail. i am running Apache/2.2.11 (Ubuntu) DAV/2 SVN/1.5.4 PHP/5.2.6-3ubuntu4.5 with Suhosin-Patch mod_python/3.3.1 Python/2.6.2 i have a script called test.py #! /usr/bin/python print "Content-Type: text/html" # HTML is following print # blank line, end of headers print "hello world" running it as an executable works... /var/www$ ./test.py Content-Type: text/html hello world when i run http://localhost/test.py i get a 404 error. What am i missing? i used this resource to enable python parsing on apache. http://ubuntuforums.org/showthread.php?t=91101 From that same thread... the following code worked.. why? #!/usr/bin/python import sys import time def index(req): # Following line causes error to be sent to browser # rather than to log file (great for debug!) sys.stderr = sys.stdout #print "Content-type: text/html\n" #print """ blah1 = """<html> <head><title>A page from Python</title></head> <body> <h4>This page is generated by a Python script!</h4> The current date and time is """ now = time.gmtime() displaytime = time.strftime("%A %d %B %Y, %X",now) #print displaytime, blah1 += displaytime #print """ blah1 += """ <hr> Well House Consultants demonstration </body> </html> """ return blah1

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  • Ajax request gets to server but page doesn't update - Rails, jQuery

    - by Jesse
    So I have a scenario where my jQuery ajax request is hitting the server, but the page won't update. I'm stumped... Here's the ajax request: $.ajax({ type: 'GET', url: '/jsrender', data: "id=" + $.fragment().nav.replace("_link", "") }); Watching the rails logs, I get the following: Processing ProductsController#jsrender (for 127.0.0.1 at 2010-03-17 23:07:35) [GET] Parameters: {"action"=>"jsrender", "id"=>"products", "controller"=>"products"} ... Rendering products/jsrender.rjs Completed in 651ms (View: 608, DB: 17) | 200 OK [http://localhost/jsrender?id=products] So, it seems apparent to me that the ajax request is getting to the server. The code in the jsrender method is being executed, but the code in the jsrender.rjs doesn't fire. Here's the method, jsrender: def jsrender @currentview = "shared/#{params[:id]}" respond_to do |format| format.js {render :template => 'products/jsrender.rjs'} end end For the sake of argument, the code in jsrender.rjs is: page<<"alert('this works!');" Why is this? I see in the params that there is no authenticity_token, but I have tried passing an authenticity_token as well with the same result. Thanks in advance.

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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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  • 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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  • tomcat resource missing, servlet not running

    - by user2837260
    import javax.servlet.*; import java.io.*; public class MyServlet implements Servlet { public void init(ServletConfig con) {} public void service(ServletRequest req, ServletResponse res) throws IOException,ServletException { res.setContentType("text/html"); PrintWriter out=res.getWriter(); String s="blah"; String s1="blah"; out.println("<html><body>"); if((s.equals(req.getParameter("firstname")))&&(s1.equals(req.getParameter("pwd")))) out.println("passwords match"); else out.println("password and name combo does not match"); out.println("</body></html>"); } public void destroy() {} public ServletConfig getServletConfig() { return null;} public String getServletInfo() { return null;} } this is my java file with the servlet class.its saved with the name MyServlet.java and so is the class file. and here is the xml file: <web-app> <servlet> <servlet-name>demoo</servlet-name> <servlet-class>MyServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>demoo</servlet-name> <url-pattern>/demo</url-pattern> </servlet-mapping> </web-app> i have made the folder as WEB-INF and then classes... WEB-INF also contains the .xml file but when i try to run the servlet , it says resource not found ps- i am already looking for the servlet with the name :demo localhost:8081/s1/demo* s1 is the war file * a html file in the war file seems to run fine on the server though. *

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  • how to make connection pool in spring application using BasicDataSource.

    - by vipin
    hi friend, I have created the application in which I need to configure the connection pool.In which I am configuring the connection pooling in the spring_Config file. using the Basicdatasource. but there is some problem to create the connection pool. Please tell me how to create the connection pooling in spring application using BasicDatasource. I tried this one code in spring config ;- bean id="datasource" class="org.apache.commons.dbcp.BasicDataSource" com.mysql.jdbc.Driver jdbc:mysql://192.168.1.12:3306/revup?noAccessToProcedureBodies=true jdbc:mysql://localhost:3306/revup?noAccessToProcedureBodies=true-- revuser root-- kjacob gme997FK-- <property name="poolPreparedStatements"> <value>true</value> </property> <property name="initialSize"> <value>2</value> </property> <property name="maxActive"> <value>15</value> </property> Is there any modification of code please tell me. thanks in advance.

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  • Error in connection in ruby.

    - by piemesons
    require 'rubygems' require 'mysql' db = Mysql.connect('localhost', 'root', '', 'mohit') //db.rb:4: undefined method `connect' for Mysql:Class (NoMethodError) //undefined method `real_connect' for Mysql:Class (NoMethodError) db.query("CREATE TABLE people ( id integer primary key, name varchar(50), age integer)") db.query("INSERT INTO people (name, age) VALUES('Chris', 25)") begin query = db.query('SELECT * FROM people') puts "There were #{query.num_rows} rows returned" query.each_hash do |h| puts h.inspect end rescue puts db.errno puts db.error end error i am geting is: undefined method `connect' for Mysql:Class (NoMethodError) OR undefined method `real_connect' for Mysql:Class (NoMethodError) EDIT return value of Mysql.methods ["private_class_method", "inspect", "name", "tap", "clone", "public_methods", "object_id", "__send__", "method_defined?", "instance_variable_defined?", "equal?", "freeze", "extend", "send", "const_defined?", "methods", "ancestors", "module_eval", "instance_method", "hash", "autoload?", "dup", "to_enum", "instance_methods", "public_method_defined?", "instance_variables", "class_variable_defined?", "eql?", "constants", "id", "instance_eval", "singleton_methods", "module_exec", "const_missing", "taint", "instance_variable_get", "frozen?", "enum_for", "private_method_defined?", "public_instance_methods", "display", "instance_of?", "superclass", "method", "to_a", "included_modules", "const_get", "instance_exec", "type", "<", "protected_methods", "<=>", "class_eval", "==", "class_variables", ">", "===", "instance_variable_set", "protected_instance_methods", "protected_method_defined?", "respond_to?", "kind_of?", ">=", "public_class_method", "to_s", "<=", "const_set", "allocate", "class", "new", "private_methods", "=~", "tainted?", "__id__", "class_exec", "autoload", "untaint", "nil?", "private_instance_methods", "include?", "is_a?"] return value of Mysql.methods(false) is []... blank array

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  • Why does grails use hsqldb when I ask for mysql?

    - by John
    I'm following the racetrack example from Jason Rudolph's book at InfoQ, using grails-1.2.1. I got up to the part where I was to switch from hsqldb to mysql. I think I've deleted every reference to hsqldb in the DataSource.groovy file, but I get an exception and the stack trace shows it's still using hsqldb. DataSource.groovy dataSource { boolean pooled = true String driverClassName = "com.mysql.jdbc.Driver" String url = "jdbc:mysql://localhost/dfpc2" String dbCreate = "create" String username = "dfpc2" String password = "dfpc2" dialect = org.hibernate.dialect.MySQL5InnoDBDialect } hibernate { cache.use_second_level_cache=true cache.use_query_cache=true cache.provider_class='net.sf.ehcache.hibernate.EhCacheProvider' } // environment specific settings environments { development { } test { } production { } } When I grails run-app it all starts up with no errors. I can navigate to the home page. But when I click on one of the links, I get a stack trace: java.sql.SQLException: Table not found in statement [select this_.id as id0_0_, this_.version as version0_0_, this_.name as name0_0_, this_.variant as variant0_0_ from domainObject this_ limit ?] at org.hsqldb.jdbc.Util.throwError(Unknown Source) at org.hsqldb.jdbc.jdbcPreparedStatement.<init>(Unknown Source) at org.hsqldb.jdbc.jdbcConnection.prepareStatement(Unknown Source) at dfpc2.domainObjectController$_closure2.doCall(script1269434425504953491149.groovy:13) at dfpc2.domainObjectController$_closure2.doCall(script1269434425504953491149.groovy) at java.lang.Thread.run(Thread.java:619) My mysql database shows no tables created. (I don't think groovy's connected to mysql yet.) Things I've checked: mysql-connector-java-5.1.6.jar is in lib directory. I've tried grails clean I tried putting the dataSource info in the development environment (I haven't graduated to test or prod yet), but it seemed to make no difference. The stdout shows I'm using development env. I've googled for solutions, but the only solution I've found is when people don't change the test or production environments.

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  • asp.net on linux/mono broken? weird load then 404/resource error

    - by acidzombie24
    I am testing out my asp.net on mono's VM Ware Image using opensuse From the home page it says Trying out your own code You can test your own applications by connecting with the file manager on this machine to the machine hosting your application, and copying over the directory containing application and its associated files. To test your ASP.NET applications, copy your code to a new directory in /srv/www/htdocs , then visit the following url: http://localhost/directoryname/page.aspx Where directoryname is the directory where you deployed your application, and page.aspx is the initial page for your software, typically Default.aspx. Mirror: http://tortillaflatscafe.com/ I saw my app had errors. After i restarted the VM i can no longer run my app. Instead of getting error messages or anything i get this error instead. Server Error in '/myfoldername' Application The resource cannot be found. Description: HTTP 404. The resource you are looking for (or one of its dependencies) could have been removed, had its name changed, or is temporarily unavailable. Please review the following URL and make sure that it is spelled correctly. Requested URL: /Default.aspx I tried /default.aspx, using sudo to set permissions to 777 recursively, restarting apache via terminal i could not get this error to go away. How do i fix this?

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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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  • User input in Perl with IO::Socket

    - by David
    I am trying to make a perl program which allows a user to input the host and the port number of a foreign host to connect to using IO::Socket. It allows me to run the program and input a host and a port but it never connects and says "Could not connect to [host] at c:\users\USER\Documents\code\perl\sql.pl line 18, line 2." What am i doing wrong with this code shown below? And also, how can i have input validation on my host, which can either be a host name or an ip address? Thanks a bunch! Code Below use IO::Socket print "Host to connect to: "; chomp ($host = <STDIN>); print "Port to connect with: "; chomp ($port = <STDIN>); while(($port > 65535) || ($port <= 0)){ print "Port to connect with [Port > 0 < 65535] : "; chomp ($port = <STDIN>); } print "\nConnecting to host $host on port $port\n"; $socket = new IO::Socket::INET ( LocalHost => '$host', LocalPort => '$port', Proto => 'tcp', Listen => 5, Reuse => 1 ); die "Could not connect to $host";

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  • WSACONNREFUSED when connecting to server

    - by Robert Mason
    I'm currently working on a server. I know that the client side is working (I can connect to www.google.com on port 80, for example), but the server is not functioning correctly. The socket has socket()ed, bind()ed, and listen()ed successfully and is on an accept loop. The only problem is that accept() doesn't seem to work. netstat shows that the server connection is running fine, as it prints the PID of the server process as LISTENING on the correct port. However, accept never returns. Accept just keeps running, and running, and if i try to connect to the port on localhost, i get a 10061 WSACONNREFUSED. I tried looping the connection, and it just keeps refusing connections until i hit ctrl+c. I put a breakpoint directly after the call to accept(), and no matter how many times i try to connect to that port, the breakpoint never fires. Why is accept not accepting connections? Has anyone else had this problem before? Known: [breakpoint0] if ((new_fd = accept(sockint, NULL, NULL)) == -1) { throw netlib::error("Accept Error"); //netlib::error : public std::exception } else { [breakpoint1] code...; } breakpoint0 is reached (and then continued through), no exception is thrown, and breakpoint1 is never reached. The client code is proven to work. Netstat shows that the socket is listening. If it means anything, i'm connecting to 127.0.0.1 on port 5842 (random number). The server is configured to run on 5842, and netstat confirms that the port is correct.

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  • Post complex types to WebApi Client

    - by BumbleBee
    I am new to WebAPI. I have a MVC project and webApi project both reside under the **same solution**. Also, BLL Class library and DAL class library reside under the same solution. Earlier, my MVC project will talk to the BLL now I am trying to create a WebAPi project which stands in between MVC and BLL. Here is what I have come up with so far : I'm using the HTTPClient to post to a WebApi project. My post method on my controller accepts a single parameter (a model). StatsCriteria criteria = new StatsCriteria(); ...... var client = new HttpClient(); var response = client.PostAsJsonAsync("http://localhost:52765/api/reports", criteria).Result; ....... Here's the signature for my controller in Webapi [HttpPost] public CMAReportVM Reports([FromBody] StatsCriteria criteria) { var cmaReport = Service3.GetCMAReport(criteria.Mlsnums); //Create Map to enable mapping business object to View Model Mapper.CreateMap<CMAReport, CMAReportVM>(); // Maps model to VM model class var cmaVM = Mapper.Map<CMAReport, CMAReportVM>(cmaReport); reutn cmaVM; } // and here's my routing: config.Routes.MapHttpRoute( name: "DefaultApi", routeTemplate: "api/{controller}/{id}", defaults: new { id = RouteParameter.Optional } I am getting the following 405 : Method not allowed. As the WebAPI and MVC project both reside under same sloution I am not sure where/how to host my webapi.

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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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  • Memcache error: Failed reading line from stream (0) Array

    - by daviddripps
    I get some variation of the following error when our server gets put under any significant load. I've Googled for hours about it and tried everything (including upgrading to the latest versions and clean installs). I've read all the posts about it here on SA, but can't figure it out. A lot of people are having the same problem, but no one seems to have a definitive answer. Any help would be greatly appreciated. Thanks in advance. Fatal error: Uncaught exception 'Zend_Session_Exception' with message 'Zend_Session::start() - /var/www/trunk/library/Zend/Cache/Backend/Memcached.php(Line:180): Error #8 Memcache::get() [memcache.get]: Server localhost (tcp 11211) failed with: Failed reading line from stream (0) Array We have a copy of our production environment for testing and everything works great until we start load-testing. I think the biggest object stored is about 170KB, but it will probably be about 500KB when all is said and done (well below the 1MB limit). Just FYI: Memcache gets hit about 10-20 times per page load. Here's the memcached settings: PORT="11211" USER="memcached" MAXCONN="1024" CACHESIZE="64" OPTIONS="" I'm running Memcache 1.4.5 with version 2.2.6 of the PHP-memcache module. PHP is version 5.2.6. memcache details from php -i: memcache memcache support = enabled Active persistent connections = 0 Version = 2.2.6 Revision = $Revision: 303962 $ Directive = Local Value = Master Value memcache.allow_failover = 1 = 1 memcache.chunk_size = 8192 = 8192 memcache.default_port = 11211 = 11211 memcache.default_timeout_ms = 1000 = 1000 memcache.hash_function = crc32 = crc32 memcache.hash_strategy = standard = standard memcache.max_failover_attempts = 20 = 20 Thanks everyone

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  • Read quicktime movie from servlet in a webpage?

    - by khue
    Hi, I have a servlet that construct response to a media file request by reading the file from server: File uploadFile = new File("C:\\TEMP\\movie.mov"); FileInputStream in = new FileInputStream(uploadFile); Then write that stream to the response stream. My question is how do I play the media file in the webpage using embed or object tag to read the media stream from the response? Here is my code in the servlet: public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { request.getParameter("location"); uploadFile(response); } private void uploadFile(HttpServletResponse response) { File transferFile = new File("C:/TEMP/captured.mov"); FileInputStream in = null; try { in = new FileInputStream(transferFile); } catch (FileNotFoundException e) { System.out.println("File not found"); } try { System.out.println("in byes i s" + in.available()); } catch (IOException e) { } DataOutputStream responseStream = null; try { responseStream = new DataOutputStream(response.getOutputStream()); } catch (IOException e) { System.out.println("Io exception"); } try { Util.copyStream(in, responseStream); } catch (CopyStreamException e) { System.out.println("copy Stream exception"); } try { responseStream.flush(); } catch (IOException e) { } try { responseStream.close(); } catch (IOException e) { } } And here is html page as Ryan suggested: <embed SRC="http://localhost:7101/movies/transferservlet" WIDTH=100 HEIGHT=196 AUTOPLAY=true CONTROLLER=true LOOP=false PLUGINSPAGE="http://www.apple.com/quicktime/"> Any ideas?

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  • Ajax div can't access address bar variable

    - by Elaine Adams
    Can someone please advise me on how my Ajax div can get an address bar variable. The usual way just doesn't work. My address bar currently looks like this: http://localhost/social3/browse/?locationName=Cambridge Usually, I would use a little php and do this: $searchResult = $_POST['locationName']; echo $searchResult; But because I'm in an Ajax div, I can't seem to get to the variable. Do I need to add some JavaScript wizardry to my Ajax coding? (I have little knowledge of this) My Ajax: <script> window.onload = function () { var everyone = document.getElementById('everyone'), searching = document.getElementById('searching'), searchingSubmit = document.getElementById('searchingSubmit'); everyone.onclick = function() { loadXMLDoc('indexEveryone'); everyone.className = 'filterOptionActive'; searching.className = 'filterOption'; } searching.onclick = function() { loadXMLDoc('indexSearching'); searching.className = 'filterOptionActive'; everyone.className = 'filterOption'; } searchingSubmit.onclick = function() { loadXMLDoc('indexSearchingSubmit'); } function loadXMLDoc(pageName) { var xmlhttp; if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("leftCont").innerHTML=xmlhttp.responseText; } } function get_query(){ var url = location.href; var qs = url.substring(url.indexOf('?') + 1).split('&'); for(var i = 0, result = {}; i < qs.length; i++){ qs[i] = qs[i].split('='); result[qs[i][0]] = decodeURIComponent(qs[i][1]); } return result; } xmlhttp.open("GET","../browse/" + pageName + ".php?user=" + get_query()['user'],true); xmlhttp.send(); } } </script> <!-- ends ajax script -->

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  • Creating a RESTful service in CakePHP

    - by NathanGaskin
    I'm attempting to create a RESTful service in CakePHP but I've hit a bit of a brick wall. I've enabled the default RESTful routing using Router::mapResources('users') and Router::parseExtensions(). This works well if I make a GET request, and returns some nicely formatted XML. So far so good. The problem is if I want to make a POST or PUT request. CakePHP doesn't seem to be able to read the data from the request. At the moment my add(), edit() and delete() actions don't contain any logic, they're simply setting $this-data to the view. I'm testing with the following cURL command: curl -v -d "<user><username>blahblah</username><password>blahblah</password>" http://localhost/users.xml --header 'content-type: text/xml' Which only returns a 404 header. If I remove the --header parameter then it returns the view but no data is set. It feels like I'm missing something obvious here. Any ideas?

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  • javascript :Object doesn't support this property or method

    - by Kaushik
    In my jsp page, I have in the tag, the following code: <script type="text/javascript" src="<%=request.getContextPath()%>/static/js/common/common.js"></script> <script type="text/javascript"> // Function for Suppressing the JS Error function silentErrorHandler() {return true;} window.onerror=silentErrorHandler; </script> If there is some javascript executing on the jsp page after this, then I guess silentErrorHandler() will have no effect. i.e. the error will still show on page. IS this correct? Because the error is showing and am not sure why. The second part of the question is this: The error is Webpage error details User Agent: Mozilla/4.0 (compatible; MSIE 8.0; Windows NT 6.1; Trident/4.0; SLCC2; .NET CLR 2.0.50727; .NET CLR 3.5.30729; .NET CLR 3.0.30729; Media Center PC 6.0; InfoPath.2; AskTbFXTV5/5.9.1.14019) Timestamp: Fri, 7 Jan 2011 21:26:23 UTC Message: Object doesn't support this property or method Line: 613 Char: 1 Code: 0 URI: http://localhost:9080/Claris/static/js/common/common.js And finally, line 613 states document.captureEvents(Event.MOUSEUP); There is error on IE8. Runs fine on Mozilla and IE7. Any suggestions will be very helpful

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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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  • 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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