JSON JSON problem

Question

I want to fill in the form fields with the values ​​from the database right after the user enters the value in the #sid field. Here is my jQuery / HTML example:

<script src="jquery-1.3.1.min.js"></script>
<script type="text/JavaScript">
$(document).ready(function()
{
  $('#sid').bind("change", function(){
    $.getJSON("test.php?sid=" + $("#sid").val(), 
    function(data)
    {
      $.each(data.items, 
      function(i, item)
      {
        if (item.field == "saffil")
        {
              $("#saffil").val(item.value);
        }
        else if (item.field == "sfirst")
        {
              $("#sfirst").val(item.value);
        }
      });
      });
   });
});
</script>

Here is my processing script (test.php, which is called by the .getJSON method)

<?
require_once("db_pers.inc");

$ssql = "SELECT * FROM contacts_mview WHERE sempid = '".$_GET['sid']."'";

$rres = pg_query($hdb, $ssql);
pg_close($hdb);

$ares = pg_fetch_assoc($rres);

$json = array(array('field' =>  'saffil',
            'value' =>  $ares['saffil']),
          array('field' =>  'sfirst',
            'value' =>  $ares['sfirst']));

echo json_encode($json);
?>

According to firebug, the GET parameter is passed just fine for test.php, and the JSON object is returned just fine:

[{"field":"saffil","value":"Admin"},{"field":"sfirst","value":"Nicholas"}]

however, nothing happens on the page, and I get the following error message:

G is undefined
init()()jquery-1....1.min.js (line 12)
(?)()()test.html (line 15)
I()jquery-1....1.min.js (line 19)
F()()jquery-1....1.min.js (line 19)
[Break on this error] (function(){var l=this,g,y=l.jQuery,p=l.....each(function(){o.dequeue(this,E)})}});

This is my first hit in ajax with jQuery, so any input would be much appreciated!

Thanks,

• Nikolay

Answer 1

Nice little injection waiting there there;)

Try to change

$.each(data.items,

in

$.each(data,

: , , :

<input type="text" name="saffil" value="" />
<input type="text" name="sfirst" value="" />

var data = {saffil:'foo', sfirst:'bar'};
$.each(data, function(key, value) {
   $('[name='+key+']').val(value)
})

Answer 2

. script - SQL-, . , - PDO , , - pg_escape_string.

Answer 3

, Sql.

Answer 4

, .

, PHP ( Perl) JSON, , .

, , :

[ element1, element2, element3]

, JSON. JSON , http://json.org/.

, -, , .

<?php
$json = array('items' =>
              array(
                    array('field' =>  'saffil',
                          'value' =>      $ares['saffil']),
                    array('field' =>      'sfirst',
                          'value' =>      $ares['sfirst'])
              )
        );

echo json_encode($json);
?>