Drewser33 Posted July 29, 2014 Share Posted July 29, 2014 What I am trying to do is to submit as POST values to database_write.php, from within the while statement. What is happening is I am getting the second row of data every time I change the primary button. Currently database_write.php is just doing print_r($_POST), And my array is always the same, no matter which select box I choose from. How can I get the values to be associated with the row I am currently changing? Any help would be great, thanks. What I have so far: <table class="table table-bordered table-hover"> <thead> <th>Room Number</th> <th>Primary Caregiver</th> <th>Seconday Caregiver</th> </thead> <tbody class="list"> <?php $sql = 'SELECT alarm_device_id, alarm_description, alarm_device_type, notes FROM alarm_device where notes in (\'MSU\') ORDER BY alarm_description'; $retval = mysql_query( $sql, $con ); if(! $retval ) { die('Could not get data: ' . mysql_error()); } $x=0; while($row = mysql_fetch_array($retval, MYSQL_ASSOC)) { $id = $row['alarm_device_id']; $alarm_description = $row['alarm_description']; echo '<form id="msu_form">'; echo "<tr><td>{$row['alarm_description']}</td>"; echo "<td>"; $query2 = "SELECT alert_device_id,alert_description FROM alert_device WHERE notes = 'MSU'"; $result2 = mysql_query($query2) or die("Error in alarm_device select:" . mysql_error()); $count2 = mysql_num_rows($result2); if($count2 > 0) { //echo '<select name='.$x.'>'; echo '<select id="Primary" name="primary" onchange="doAjaxPost(this)">'; while($row2 = mysql_fetch_array($result2)) { echo "<option value=".$row2['alert_device_id'].">".$row2['alert_description']."</option>"; } echo "</select>"; }else { echo "Please update alert device to this area"; } echo "</td>"; echo "<td>"; $query3 = "SELECT alert_device_id,alert_description FROM alert_device WHERE notes = 'MSU2'"; $result3 = mysql_query($query3) or die("Error in alarm_device select:" . mysql_error()); $count3 = mysql_num_rows($result3); if($count3 > 0) { echo '<select id="Secondary" name="secondary">'; while($row3 = mysql_fetch_array($result3)) { echo "<option value=".$row3['alert_device_id'].">".$row3['alert_description']."</option>"; } echo "</select>"; }else { echo "Please update alert device to this area"; } echo "</td>"; $aid = $id + $x; //echo $aid; //$ad = $alarm_description + $x; echo '<input type="hidden" id="ID" name="ID" value="'.$id.'"/>'; //echo '<input type="hidden" id="desc" name="desc" value="'.$ad.'"/>'; //echo '<td>'."<input type='submit' name='btnupdate' value='UPDATE' /></td>"; //echo '<td><input type="button" value="Ajax Request" onClick="doAjaxPost()"></td>'; echo '</form>'; $x = $x+1; } ?> <script> function doAjaxPost() { // get the form values var primary = $('#Primary').val(); var secondary = $('#Secondary').val(); var hidden = $('#ID').val(); //var desc = $(sel).parent().nextAll('#desc').val(); $.ajax({ type: "POST", url: "functions/database_write.php", data: $('#msu_form').serialize(), //data: "Primary="+primary+"&Hidden="+hidden+"&Secondary="+secondary, success: function(resp){ //we have the response alert("'" + resp + "'"); }, error: function(e){ alert('Error: ' + e); } }); } </script> </tr> </tbody> </table> Quote Link to comment https://forums.phpfreaks.com/topic/290183-my-first-attempt-at-ajax-empty-array/ Share on other sites More sharing options...
Zane Posted July 31, 2014 Share Posted July 31, 2014 There's no need to create a form in a loop, making several forms. I'm surprised that you get the "second" value, and not the last one. Maybe you only have two loop iterations. Put the form tag before the while loop, and the end form tag after the loop. Quote Link to comment https://forums.phpfreaks.com/topic/290183-my-first-attempt-at-ajax-empty-array/#findComment-1486527 Share on other sites More sharing options...
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