Android Uncaught Error Network_err Xmlhttprequest Exception 101
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Network_err Xmlhttprequest Exception 101 Chrome
'email':$('#email').val(), 'password':$('#password').val(), }, cache: false, async: false, crossDomain: "true", beforeSend: function() { }, complete: function() { }, success: function(resp) { alert(resp); var result = $.parseJSON(resp); if (result.result == "Success") { alert(emailID,password); } else { alert(result.msg); } }, error: function(request, status, error) { //$("#LoadingImage").hide(); alert("Result = " + error); } }); return false; }); It is returning with this error: NETWORK_ERR: XMLHttpRequest Exception 101 on android tablet. I have read a bunch of SO posts on this error, most suggest network_err: xmlhttprequest exception 101 jquery that I set async to true. This DOES remove the error message- but it is still an error, and I never get valid data. It just seems to remove the error message which is not helpful. Please help me. javascript android jquery ajax share|improve this question edited Nov 6 '13 at 10:02 Rory McCrossan 153k22129179 asked Nov 6 '13 at 10:00 peter 565520 You should definitely remove the async: false. What is the error you get then? Also, check the status in the error handler. –Rory McCrossan Nov 6 '13 at 10:02 @RoryMcCrossan If you see my last comment then just ignore me. Not had enough coffee yet :) –Archer Nov 6 '13 at 10:08 @Archer you're lucky, I missed it :D –Rory McCrossan Nov 6 '13 at 10:09 @RoryMcCrossan i tried but now alert showing empty window –peter Nov 6 '13 at 10:13 android 4.0 is working but problem with android 4.1.x –peter Nov 6 '13 at 11:08 add a comment| 2 Answers 2 active oldest votes up vote 4 down vote accepted Finally i got solution.. Android 4.1 and 4.2 introduce this new method: getAllowUniversalAccessFromFileURLs Since it's not working on API below 16 the solution needs some few more lines, to assure that this inexistent method do not cause errors in previous API. int currentapiVersion = android.os.Build.VERSION.SDK_INT; if (currentapiVersion >= android.os.Build.VERSION_
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Unfortunately, Casperjs Cannot Make Cross Domain Ajax Requests
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Cross Origin Requests Are Only Supported For Http.
you, helping each other. Join them; it only takes a minute: Sign up NETWORK_ERROR: XMLHttpRequest Exception 101 up vote 27 down vote favorite 5 I am getting this Error NETWORK_ERROR: XMLHttpRequest Exception 101 when trying to get http://stackoverflow.com/questions/19808991/ajax-post-error-network-err-xmlhttprequest-exception-101-on-android-device XML content from one site. Here is my code var xmlhttp; if(window.XMLHttpRequest) { xmlhttp = new XMLHttpRequest(); } if (xmlhttp==null) { alert ("Your browser does not support XMLHTTP!"); return; } xmlhttp.onReadyStateChange=function() { if(xmlhttp.readyState==4) { var value =xmlhttp.responseXML; alert(value); } } xmlhttp.open("GET",url,false); xmlhttp.send(); //alert(xmlhttp.responseXML); } xmlhttp.open("GET",url,false); xmlhttp.send(null); Does any one have a solution? ajax xmlhttprequest share|improve this question edited Jun 7 '13 at 21:14 Ribo 1,7261424 asked Feb 10 '10 at 10:09 pawan Mangal 136123 http://stackoverflow.com/questions/2235929/network-error-xmlhttprequest-exception-101 this looks like a cross domain problem –Moataz Elmasry Jul 24 '12 at 11:01 Please mention what browser and/or platform you see the problem on (I got the exception on Android 2.3.7 with WebKit 533.1 –Ribo Jun 7 '13 at 21:15 add a comment| 7 Answers 7 active oldest votes up vote 21 down vote If the url you provide is located externally to your server, and the server has not allowed you to send requests, you have permission problems. You cannot access data from another server with a XMLHttpRequest, without the server explicitly allowing you to do so. See this related question share|improve this answer edited Oct 8 '15 at 22:30 Undo♦ 18.3k147298 answered Apr 7 '10 at 10:54 Frederik Wordenskjold 5,55352649 [This post] is a dead link. –Léon Pelletier Jan 25 '13 at 4:46 1 Thanks! Dead link removed. –Frederik Wordenskjold Jan 26 '13 at 1:01 URLs located on another domain/site are not necessarily a permission problem. Cross-Site requests can be made using XMLHttpRequest if the target server permits it. –Ribo Jun 7 '13 at 21:07 8 I find it moderately amusing that 5 years later (as of 2015) this is the first hit on this subject on google and this answer says that google has a lot of information on it ;). –j03
here for a quick overview of the site Help Center Detailed answers to http://stackoverflow.com/questions/6965942/network-err-xmlhttprequest-exception-101 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 xmlhttprequest exception 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 NETWORK_ERR: XMLHttpRequest Exception 101 up vote 8 down vote favorite 3 I'm having an AJAX problem in Chrome, giving the following error: Uncaught Error: NETWORK_ERR: XMLHttpRequest xmlhttprequest exception 101 Exception 101 This is my code: function IO(filename) { if (window.XMLHttpRequest) { // Mozilla, Safari,... xmlhttp = new XMLHttpRequest(); } else if (window.ActiveXObject) { // IE try { xmlhttp = new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try { xmlhttp = new ActiveXObject("Microsoft.XMLHTTP"); } catch (e) { } } } xmlhttp.open("GET", filename+"?random="+Math.floor(Math.random()*100000001), false); xmlhttp.send(); if(xmlhttp.readyState==4) return xmlhttp.responseXML; } javascript ajax google-chrome xmlhttprequest share|improve this question edited Aug 6 '11 at 21:41 pimvdb 90.5k40218308 asked Aug 6 '11 at 9:34 Pawan Goswami 41112 What is filename you're using? –pimvdb Aug 6 '11 at 9:41 add a comment| 2 Answers 2 active oldest votes up vote 11 down vote The solution is setting the async parameter to true: xmlhttp.open("GET", filename+"?random="+Math.floor(Math.random()*100000001), true); share|improve this answer edited Jun 14 '12 at 16:48 Tim Cooper 86.3k21162181 answered Jun 14 '12 at 16:26 headmax 11113 1 W00t. this worked for me. –gnarbarian Sep 1