drutjkpsr67393592 2009-08-12 06:51
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I am using php / mysql and protype.js to delete record from a table. The problem is that the record in the database is not deleted.


       <a href="javascript: deleteId('<?php echo $studentVo->id?>')">Delete</a></td>

Script is

   function deleteId(id)
       alert("ID : "+id);
       new Ajax.Request('delete.php?action=Delete&id='+id,{method:'post'});
       $(id).remove(); // because <tr id='".$row[id]."'> :)



      /* Database connection */
      echo "hello,...";
          $ID = $_POST['id'];
          $sql = 'DELETE FROM student where id="'.$ID.'"';
      else { echo '0'; }

alert("ID : "+id); is working properly but the code after that is not.

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3条回答 默认 最新

  • dongzan9069 2009-08-12 06:58

    You are using a GET request, from JS :


    And your PHP code uses data he thinks arrives as POST :

    $ID = $_POST['id'];

    You should use the same method on both sides.

    (As you are modifying / deleting data, you should probably use POST)

    As a sidenote, you should definitly escape/protected/check the data you are using in the SQL query, to avoid SQL injections, using, for instance, intval as you are working with an integer ; you'd use mysql_real_escape_string if you were working with a string.

    Another way would be to stop using the old mysql extension, and start using mysli or PDO, which means you could use prepared statements (mysqli, pdo)

    EDIT after the comment : you also, now that the request is made in POST, need to change the way parameters are passed : they should not be passed in the URL anymore.

    I suppose that something like this should work :

    var myAjax = new Ajax.Request(
        method: 'post',
        parameters: {action: id}

    Or you could also use something like this, building the parameters string yourself :

    var myAjax = new Ajax.Request(
        method: 'post',
        parameters: 'action=' + id

    (Not tested, so you might have to change a few things ;-) )

    For more informations, take a look at Ajax.Request and Ajax options :-)

    本回答被题主选为最佳回答 , 对您是否有帮助呢?



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