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The "onreadystatechange" dosen't work, What I am doing wrong ?

Question
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I am have this wired problem, every time the onreadystatechange called its not returning anything. for example I remove the if and I tell it to return yes that the state of the ajax change its still not returning instead I will get underfed any Idea why this is happening ?
function load_page(php_handler, url ,searchString) { // Get the ajax object using our function above. window.ajax = makeAJAXObject(); // Tell the AJAX object what to do when it's loaded the page window.ajax.onreadystatechange = function () { if (window.ajax.readyState == 4) { // 4 means it's loaded ok. // For simplicity, I'll just return this
return ("yes"); } } // Set up the variables you want to sent to your PHP page (namely, the URL of the page to load) var queryString = "?url=" + url; // Load the PHP script that opens the page window.ajax.open("GET", php_handler + queryString, true); window.ajax.send(null); }
here is my makeAJAXObject function
function makeAJAXObject() { // seems to be working ok var ajaxRequest; // The variable that makes Ajax possible! try { // Opera 8.0+, Firefox, Safari ajaxRequest = new XMLHttpRequest(); } catch (e) { // Internet Explorer Browsers try { ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try { ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP"); } catch (e) { // Something went wrong return false; } } } return ajaxRequest; }
I also tried to copy a working version of the script from example I found here and the same thing happens, I am working on the split form that coming with the studio.
Please help !
Friday, May 31, 2013 5:20 PM
Answers
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Hi,
Why don't you use the WinJS native xhr? (see the link for full documentation)
WinJS.xhr({ url: "http://your_url" }).done(function completed(result){ if (result.status === 200) { //do what you need with the result } }, function error(result){ //handle error });
- Marked as answer by Daniel Iliya Friday, June 7, 2013 3:52 PM
Monday, June 3, 2013 2:07 PM
All replies
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that seems like a lot of code for a simple task. try this...
serverxmlhttp = new XMLHttpRequest(); serverxmlhttp.open("GET", "http://www.website.com/api.php?id=" + Math.random(), true); serverxmlhttp.onreadystatechange = function (oEvent) { if (serverxmlhttp.readyState === 4) { if (serverxmlhttp.status === 200) { console.log(serverxmlhttp.responseText); } } }; serverxmlhttp.send();
I always throw in a dummy chunk as to never use a cache when GET or POST is used, that's why you see the id=Math.random();. you can eliminate that if you want. This should work on your machine, just make sure you have internet under you app manifest selected.Monday, June 3, 2013 7:08 AM -
Hi,
Why don't you use the WinJS native xhr? (see the link for full documentation)
WinJS.xhr({ url: "http://your_url" }).done(function completed(result){ if (result.status === 200) { //do what you need with the result } }, function error(result){ //handle error });
- Marked as answer by Daniel Iliya Friday, June 7, 2013 3:52 PM
Monday, June 3, 2013 2:07 PM