The Little Guy Posted March 24, 2007 Share Posted March 24, 2007 I don't think that this is Sending the request, because it never passes step 3 (I don't actually know what step it gets to, but its not four.) Basically it sends the request, and the PHP uploads the file to the server... That doesn't happen. Please Help. The AJAX: <script type="text/javascript"> <!-- //Browser Support Code function ajax(sourcetxt){ 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 alert("You Browser Doesn't support AJAX."); return false; } } } // Create a function that will receive data sent from the server ajaxRequest.onreadystatechange = function(){ if(ajaxRequest.readyState < 4){ document.getElementById('con').innerHTML = '<p><img alt="loading" src="loading.gif" /></p>'; } if(ajaxRequest.readyState == 4){ window.location = 'user.php'; //document.getElementById('con').innerHTML = ajaxRequest.responseText; } } ajaxRequest.open("POST", sourcetxt, true); ajaxRequest.send(null); } //--> </script> The HTML: <form enctype="multipart/form-data"> <table> <tr> <td> <input type="submit" name="submit" value="Upload"> </td> </tr> <tr> <td> <input type="radio" name="write" value="1" checked>Overwrite Older files<br> <input type="radio" name="write" value="2">Create with new file name </td> </tr> <tr> <td> <input class="uploadinputbutton" name="file[]" onchange="add_file('file', 1);" type="file"><br><span id="file_1"><input value="Add another" onclick="add_file('file', 1);" type="button"></span> <br> </td> </tr> <tr> <td> <input type="submit" onClick="javascript:ajax('upload_files.php')" value="Upload"> </td> </tr> </table> </form> Quote Link to comment Share on other sites More sharing options...
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