I am posting multiple checkbox values via ajax to a php file which processes the data sent and updates a database. The post works as does the database updates based on the checkbox's values sent.
The problem I have is that it does not return to the calling php file where the ajax call is made from. I have to click on the previous button in the browser and refresh the browser page to see the results from the updated database.
The code for the checkboxes.
echo "<form id='assign' method ='post' action='../../assign_engineer_post_code.php'>";
echo "<input type='checkbox' checked name='select-all' id='select_all' />Select All<br/><br/>";
echo "<p style='font-size: 10px; '>";
while($row = mysqli_fetch_assoc($res)){
echo "<input type='checkbox' class='checkbox' name='engineers[]' value='".$row['Engineer_id']."' checked><label for='" .$row['Engineer'] ."'>" .$row['Engineer'] ."</label>";
echo " ";
$counter++;
if ($counter == 8){
//echo "<br/>";
}
}
echo "</p>";
?>
<table><tr>
<td>
<button type="button" onclick="clear_sap_data()">Clear Data</button>
</td>
<td>
<button type="button" onclick="import_sap_data()">Import SAP Data</button>
</td>
<td>
<!-- <input name="confirm" type="button" value="Assign Engineers" onclick="doit();" />
<input id="submit" type="submit" name="submit" value="Submit" onclick="submitForm()" /> -->
<input id="submit" type="submit" name="submit" value="Submit" onclick="submitForm()" />
</td>
</tr>
</table>
</form>
The script called submitForm is as follows
<script>
function submitForm() {
var form = document.assign;
var dataString = $(form).serialize();
$.ajax({
contentType: false,
processData: false
type:'POST',
url:'../../assign_engineer_post_code.php',
data: dataString,
success: function(data){
window.location.reload(); // This is not jQuery but simple plain ol' JS
}
});
return false;
}
</script>
The php file called does what is is supposed to do but it does not return to the calling page and the ajax window.location.reload() is not invoked.
Regards
Alan
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