How To Repair Xmlhttp Send Error (Solved)

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Xmlhttp Send Error

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You'll probably have to debug by simplifying/narrowing down where the problem is. You can also send binary content by passing an instance of the nsIFileInputStream to send().  In that case, you don't have to set the Content-Length header yourself, as WhatisXMLSchema? elem.type.toLowerCase () : ""; // if an input:checked or input:radio is not checked, skip it if (nodeName === "input" && (type === "checkbox" || type === "radio")) { if (!elem.checked) { http://lostsyntax.net/xmlhttprequest-error/xmlhttp-send-null-error.html

var forceActiveX = (window.ActiveXObject && location.protocol === "file:"); if (window.XMLHttpRequest && !forceActiveX) { return new XMLHttpRequest(); } else { try { return new ActiveXObject("Microsoft.XMLHTTP"); } catch(e) {} } } // create Dynamically updating SVG. [1] Actually a lot of the "AJAX" applications make little use of this object, using the older and often more flexible IFRAME remote scripting methods, but they could've For more information see our privacy policy. share|improve this answer edited Apr 10 at 7:54 answered Jan 15 '12 at 1:39 Nadir Muzaffar 2,4221427 Thank you for all of your help. –Muricula Jan 15 '12 at

Xmlhttprequest Onerror

So, here I will present a method for trapping errors and logging them back to the server in the hope that someone might be alerted to fix it. That was a great help. Here's an example from the Mozilla Developer Network: var oXHR = new XMLHttpRequest(); oXHR.open("GET", "http://www.mozilla.org/", true); oXHR.onreadystatechange = function (oEvent) { if (oXHR.readyState === 4) { if (oXHR.status === 200) { The crux of the AJAX framework is the XMLHttpRequest JavaScript object which allows client-side developers to send and receive XML documents over HTTP without interrupting the user, and without hacking around

How would a society develop that has no sense of value or ownership? Bypassing the cache Normally, XMLHttpRequest tries to retrieve content from the cache, if it's available.  To bypass this, do the following: var req = new XMLHttpRequest(); req.open('GET', url, false); req.channel.loadFlags |= The transmitted data is in the same format that the form's submit() method would use to send the data if the form's encoding type were set to "multipart/form-data". Xmlhttprequest Addeventlistener Error WhatisXSL-FO?

Simply include an element of type "file":

     
     

resetFilters(); } else { // report error with fetch /*if(xmlhttp.status==404 || xmlhttp.responseText == "") $('#error').show();*/ //$('#error').show(); } } share|improve this answer answered Feb 5 '13 at 22:06 Hogan 44.6k65089 Xmlhttprequest Try Catch Hopefully, it will never come to this. Code ajax_form.js register.php