Error Getxmlhttpobject Is Not Defined

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Uncaught Referenceerror: Xmlhttprequest Is Not Defined

Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping xmlhttprequest is not defined node js each other. Join them; it only takes a minute: Sign up XMLHttpRequest is not defined, in a chrome extension options page up vote 1 down vote favorite I'm trying to make an XMLHttpRequest in the options page of xmlhttprequest is not defined chrome an extension. In my options.js file I simply have the following code : if (window.XMLHttpRequest){ var xhr = new getXMLHttpRequest(); } But I have this error in the console Uncaught ReferenceError: getXMLHttpRequest is not defined I saw here that getXMLHttpRequests are a problem for hosted apps, but in this case, it's a simple extension, so I don't understand. javascript google-chrome-extension xmlhttprequest share|improve this question asked Apr 8 '13 at 8:29 emma sculateur 161315 add a comment| 1

Install Xmlhttprequest

Answer 1 active oldest votes up vote 4 down vote accepted To construct an XHR object you use new XMLHttpRequest();. getXMLHttpRequest is not a standard function. I saw here that getXMLHttpRequests are a problem… The question at the other end of the link doesn't use a function with a name starting with get. share|improve this answer answered Apr 8 '13 at 8:31 Quentin 489k56638807 Perfect ! But how could I miss that ?? I spent hours searching the reason why it was not working, and I did not even notice that I was not using the right function... –emma sculateur Apr 8 '13 at 8:51 add a comment| Your Answer draft saved draft discarded Sign up or log in Sign up using Google Sign up using Facebook Sign up using Email and Password Post as a guest Name Email Post as a guest Name Email discard By posting your answer, you agree to the privacy policy and terms of service. Not the answer you're looking for? Browse other questions tagged javascript google-chrome-extension xmlhttprequest or ask your own question. asked 3 years ago viewed 8084 times active 3 years ago Linked 7 Cross-Origin XMLHttpRequest in chrome extensions Related 1Parse XML Output with Javascript for a Chrome Extension?0XHR forbidden in Chrome extension, despite adding the correct permissions?0Error in Chrome: “Origin file:// is not allowed by Access-Control-Allow-Origin”1XMLHttpRequest with modifi

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Node Xmlhttprequest

Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping http://stackoverflow.com/questions/15874535/xmlhttprequest-is-not-defined-in-a-chrome-extension-options-page each other. Join them; it only takes a minute: Sign up XMLHttpRequest module not defined/found up vote 4 down vote favorite This is my code: var XMLHttpRequest = require("xmlhttprequest").XMLHttpRequest; var xhr = new XMLHttpRequest(); xhr.open("GET", "//URL") xhr.setRequestHeader("Content-Type: application/json", "Authorization: Basic //AuthKey"); xhr.send(); I am getting the error: Cannot find module 'xmlhttprequest' When I remove the first line, I http://stackoverflow.com/questions/32604460/xmlhttprequest-module-not-defined-found am getting: XMLHttpRequest is not defined I have searched all over and people have mentioned a problem with Node.js here and there but my installation of Node was correct so I'm not sure what the issue is. I'm using IntelliJ IDE for this project. Any help appreciated. javascript node.js intellij-idea xmlhttprequest share|improve this question asked Sep 16 '15 at 9:19 wmash 598219 Use http instead –Tushar Sep 16 '15 at 9:20 add a comment| 1 Answer 1 active oldest votes up vote 4 down vote accepted XMLHttpRequest is a built-in object in web browsers. It is not distributed with Node; you have to install it separately using npm. That said, Node comes with the http module which is the normal tool for choice for making HTTP requests from Node. share|improve this answer answered Sep 16 '15 at 9:23 Quentin 489k56638807 This has worked. Thank you –wmash Sep 16 '15 at 9:44 add a comment| Your Answer draft saved draft discarded Sign up or log in Sign up

here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more http://stackoverflow.com/questions/13585807/xmlhttp-is-undefined-javascript-ajax about Stack Overflow the company Business Learn more about hiring developers or posting ads http://stackoverflow.com/questions/13291901/referenceerror-ajaxcall-is-not-defined with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up xmlhttp is undefined. Javascript Ajax up vote 0 down vote favorite I is not am new to javascript and especially to ajax..And just trying to figure it out.. i wrote this code from one tutorial and can't find what i did wrong. Here you can see it live The error i am getting from Firebug: "TypeError: xmlhttp is undefined [Break On This Error] if (xmlhttp.readyState == 4){" my code is // JavaScript Document var xmlhttp; var url; function ajaxFunction(){ if (window.ActiveXObject){//if the window is not defined is InternetExplorer xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); }else if(window.XMLHttpRequest){// if Window is Firefox etc.. xmmlhttp= new XMLHttpRequest(); }else{ alert ("Get a New Browser") } }//end of ajaxFunction() function getInfo(){ ajaxFunction(); var entryInfo= document.getElementById("entry").value; function stateChanged(){ if (xmlhttp.readyState == 4){ document.getElementById("results").innerHTML = xmlhttp.responseText; }//if (xmlhttp.readyState == 4) }//end of function stateChanged() url = "info.php?user="+entryInfo; xmlhttp.onreadystateshange=stateChanged(); xmlhttp.open("GET",url,true); xmlhttp.send(null); }// end of function getInfo javascript ajax share|improve this question asked Nov 27 '12 at 13:56 Alex 1415 1 Your onreadystatechange property should be a reference to the function without (), not the function call with () as in xmlhttp.onreadystateshange = stateChanged; –Michael Berkowski Nov 27 '12 at 13:59 add a comment| 2 Answers 2 active oldest votes up vote 3 down vote accepted You have a typo here: xmmlhttp= new XMLHttpRequest(); ^ Change to xmlhttp= new XMLHttpRequest(); Also as Michael pointed out, you have parenthesis when assigning your onreadystatechange function: xmlhttp.onreadystateshange=stateChanged(); ^ remove the () If you don't remove the parenthesis, the stateChange() function will be called, and the return value will be given to xmlhttp.onreadystateshange which you don't want. share|improve this answer answered Nov 27 '12 at 13:58 MrCode 45.9k73875 1 aaa...i hate when it is so stupid mistakes..))thanks –Alex Nov 27 '12 at 14:03 add a co

here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up ReferenceError: ajaxCall is not defined up vote 0 down vote favorite I'm a little bit lost... I have a site, which has a lot AJAX calls. I tried with some JavaScript textfield formatting, but that was too difficult to handle, so I decided to go back to normal without it. But right now none of my Ajax calls are working and I don't know why... All I got is this error: ReferenceError: name-of-the-call is not defined The calls are locating in the "scripts.js" file and this is included in the header from the very begining. This is my header right now with an uploadify script and an arrow key navigation between the input fields.