jayzz Posted October 31, 2007 Share Posted October 31, 2007 hi guys, i am new to Ajax and I've hit a little snag with the code that i am writing. i have student information that is retrieved from a database and displayed in forms on a page i then have two checkbox's where when they are checked the information is submitted into the database the problem that i am having is only the first form is being submitted. I am thinking that it is to do with the form name needing to to be submitted to Ajax, but I am not sure how to do that. could someone give me a sample of how this is done. i have been searching this for a few days but cant find anything and i would like to get this sorted so i can move on to the next part of the site. thanks. here is the code <html> <body> <script language="javascript" type="text/javascript"> //Browser Support Code function ajaxFunction(){ 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("Your browser broke!"); return false; } } } // Create a function that will receive data sent from the server ajaxRequest.onreadystatechange = function(){ if(ajaxRequest.readyState == 4){ var ajaxDisplay = document.getElementById('ajaxDiv'); ajaxDisplay.innerHTML = "<img src='images/absent.gif' width='20' height='20'/>"; } } var stu_id = document.getElementById('stu_id').value; var class_id = document.getElementById('class_id').value; var absent = document.getElementById('absent').value; var uniform = document.getElementById('uniform').value; var queryString = "?stu_id=" + stu_id + "&class_id=" + class_id + "&absent=" + absent + "&uniform=" + uniform; ajaxRequest.open("GET", "insert_data.php" + queryString, true); ajaxRequest.send(null); } </script> <?php echo'<link rel="stylesheet" type="text/css" href="body.css">'; $class_id=$_POST['class_id']; include 'sql_config.php'; $student_id = $rc->Fields("StudentID"); $first_name = $rc->Fields("FirstName"); $last_name = $rc->Fields("LastName"); $yr_level = $rc->Fields("YearLevel"); echo '<table>'; $counter = 0; $cells_per_row = 7; $tables = 0; while (!$rc->EOF) { $student = $student_id->value; $fname = $first_name->value; $lname = $last_name->value; $counter++; if(($counter % $cells_per_row) == 1) { echo '<tr>'; } echo '<td><table border="0" cellspacing="0" cellpadding="0" id="student"> <tr> <td colspan="3" width="150px" align="center">';echo $fname, $lname; echo'</td> </tr> <tr> <td colspan="3" width="150px" align="center">';echo $student; echo'</td> </tr> <tr> <td width="40px"><div id="ajaxDiv" value="';echo $student; echo'" > </div></td> <td width="70px" align="center"><img src="'; echo $images, $student; echo'.jpg" width="70" height="83" /></td> <td width="40px"> </td> </tr> <tr> <td id="forms" colspan="3" width="150px" align="left"><form name="';echo $student; echo'" method="post" >'; ?> <input name="stu_id" id="stu_id" type="hidden" readonly="true" value="<?php echo $student; ?>"/> <input type="checkbox" id="absent" name="<?php echo $student; ?>" onclick="ajaxFunction();" value="90" />Absent.<br> <input name="class_id" id="class_id" type="hidden" readonly="true" value="<?php echo $class_id; ?>"/> <input type="checkbox" id="uniform" name="<?php echo $student; ?>" onclick="ajaxFunction();" value="65" />Out of Uniform. </form> </td> </tr> </table> <?php $rc->MoveNext(); echo'</td>'; if(($counter % $cells_per_row) == 0) { echo '</tr>'; } } $rc->Close(); echo' </body> </html>'; ?> Link to comment https://forums.phpfreaks.com/topic/75543-urgent-help-form-submitting/ Share on other sites More sharing options...
phpQuestioner Posted November 15, 2007 Share Posted November 15, 2007 you have not set-up you xmlHttp.readyState==2 in your ajax routine. you need this this query your php page. take a closer look at xmlHttpRequest. take a look at this tutorial: http://www.tizag.com/ajaxTutorial/ajaxxmlhttprequest.php look were it says "Ajax - Sending a Request for Information" Link to comment https://forums.phpfreaks.com/topic/75543-urgent-help-form-submitting/#findComment-391945 Share on other sites More sharing options...
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