neverforget98 Posted November 15, 2013 Share Posted November 15, 2013 Hello, On one of the pages of the system I am coding there is a form to update the destination of one of the rows. Every row that has no destination has a form to update the destination. However, when you update the destination it updates it for every row that doesn't have a destination. This is the portion of the code, thanks: if($destination=='') { echo "<form method='post'> <td width='25%'><center><input type='text' name='destination_save2' placeholder='Destination Location' /><input type='submit' name='submit2' value='Update Call' /></center></td>"; if(isset($_POST['submit2'])) { $destination_save2=$_POST['destination_save2']; $result8=mysql_query("UPDATE uc_calls SET destination='$destination_save2' WHERE id=$id"); if(!$result8) { die('Failed to update call. Please contact your Technical Analyst with this error:<br />' .mysql_error()); } else { header("Location: ?"); } } echo "</form>"; } else { echo "<td width='25%'><center>$destination</center></td>"; } Quote Link to comment Share on other sites More sharing options...
Solution Ch0cu3r Posted November 15, 2013 Solution Share Posted November 15, 2013 (edited) Is this code being ran in a loop? if it is then it'll update every row to the same destination value. You will need to the records id to a hidden form field. You will also want to move the code for updating the record outside of the loop. Code for displaying the destination value or edit form if($destination=='') { echo "<form method='post'> <td width='25%'><center> <input type='hidden' name='id' value='$id' /> <input type='text' name='destination_save2' placeholder='Destination Location' /> <input type='submit' name='submit2' value='Update Call' /></center> </td>"; echo "</form>"; } else { echo "<td width='25%'><center>$destination</center></td>"; } Then outside of the loop you'll have the update code if(isset($_POST['submit2'])) { $id = (int) $_POST['id']; $destination_save2 = mysql_real_escape_string($_POST['destination_save2']); $result8=mysql_query("UPDATE uc_calls SET destination='$destination_save2' WHERE id=$id"); if(!$result8) { die('Failed to update call. Please contact your Technical Analyst with this error:<br />' .mysql_error()); } else { header("Location: ?"); } } Edited November 15, 2013 by Ch0cu3r Quote Link to comment Share on other sites More sharing options...
neverforget98 Posted November 15, 2013 Author Share Posted November 15, 2013 Is this code being ran in a loop? if it is then it'll update every row to the same destination value. You will need to the records id to a hidden form field. You will also want to move the code for updating the record outside of the loop. Code for displaying the destination value or edit form if($destination=='') { echo "<form method='post'> <td width='25%'><center> <input type='hidden' name='id' value='$id' /> <input type='text' name='destination_save2' placeholder='Destination Location' /> <input type='submit' name='submit2' value='Update Call' /></center> </td>"; echo "</form>"; } else { echo "<td width='25%'><center>$destination</center></td>"; } Then outside of the loop you'll have the update code if(isset($_POST['submit2'])) { $id = (int) $_POST['id']; $destination_save2 = mysql_real_escape_string($_POST['destination_save2']); $result8=mysql_query("UPDATE uc_calls SET destination='$destination_save2' WHERE id=$id"); if(!$result8) { die('Failed to update call. Please contact your Technical Analyst with this error:<br />' .mysql_error()); } else { header("Location: ?"); } } Thanks so much Ch0cu3r! Quote Link to comment Share on other sites More sharing options...
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