CODEWITHSUNDEEP
phpjavascriptjqueryajaxjson
niczak
niczak

Reputation: 3917

jQuery JSON Problem

I want to populate form fields with values from a database immediately after the user enters a 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 gets 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 param is passed just fine to test.php and the JSON object comes back just fine:

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

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

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 stab at ajax with jQuery so any input would be much appreciated!

Thanks,

Upvotes: 3

Views: 7022

Answers (4)

Dancrumb
Dancrumb

Reputation: 27539

I don't believe the answers here are correct.

The problem here is that the PHP (and Perl) libraries encode the JSON that you provide them, and nothing else.

Thus, if you give the encoding function an array, it will output:

[ element1, element2, element3]

This is the correct encoding for an array, but it is not a correct JSON response. A correct JSON response must be an object as outlined at http://json.org/.

Thus, you need to make a wrapper object, around your array and then encode and respond with that.

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

echo json_encode($json);
?>

Upvotes: 0

Lior Cohen
Lior Cohen

Reputation: 9055

I agree with the previous repliers. That script is an SQL injection waiting to happen. You should probably use something like PDO with prepared statements or at least something like pg_escape_string.

Upvotes: 1

Crescent Fresh
Crescent Fresh

Reputation: 116980

Nice little injection attack waiting to happen there ;)

Try changing

$.each(data.items,

to:

$.each(data,

Edit: to answer your comment, I like to name my fields the same as the data key:

<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)
})

Upvotes: 7

Arron S
Arron S

Reputation: 5537

Not sure, but I do see a huge Sql Injection hole.

Upvotes: 0

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