Jquery .post() return array - php

Question

Sorry for the bad title, but I don't know how to name this. My problem is that whenever I pass a value from a select box I trigger this jquery event in order to check on the check boxes. Bassically I echo $res[]; at selecctedgr.php. Do I need to use json? and how can I do this?
Mainpage:
$("#group_name").change(function(){
var groupname = $("#group_name").val();
var selectedGroup = 'gr_name='+ groupname;
$.post("selectedgr.php", {data: selectedGroup}, function(data){
$.each(data, function(){
$("#" + this).attr("checked","checked");
});
},"json");
});
PHP (selectedgr.php):
<?php
include_once '../include/lib.php';
$gr_name=mysql_real_escape_string($_POST['gr_name']);
$sqlgr = "SELECT * FROM PRIVILLAGE WHERE MAINGR_ID=".$gr_name;
$resultgr = sql($sqlgr);
while($rowgr = mysql_fetch_array($resultgr)){
$res[] = $rowgr['ACT_ID'];
}
echo $res[];
?>

Answer 1

Change the last line in your PHP sample (echo $res[];) to:
echo json_encode($res);
json_encode() manual page will tell you more.
Also as #Unicron says you need to validate the $gr_name variable before passing it to your SQL statement.
You could use:
if(isset($_POST['gr_name'])) {
$gr_name = mysql_real_escape_string($_POST['gr_name']);
}
See: http://php.net/manual/en/function.mysql-real-escape-string.php for more information in the PHP manual.

Answer 2

You can use json_encode function to convert arbitrary data into JSON. Assuming that you want to return an array of strings, here is how you will use json_encode:
<?php
include_once '../include/lib.php';
$res = array(); // initialize variables
$sqlgr = sprintf("
SELECT ACT_ID
FROM PRIVILLAGE
WHERE MAINGR_ID=%d
",
$_POST['gr_name']
); // only select those columns that you need
// and do not trust user input
$resultgr = sql($sqlgr);
while($rowgr = mysql_fetch_array($resultgr)){
$res[] = $rowgr['ACT_ID'];
}
echo json_encode($res); // use json_encode to convert the PHP array into a JSON object
// this will output something like ['foo', 'bar', 'blah', 'baz'] as a string
?>
On the client side you can use jQuery.post method, like this:
<script type="text/javascript">
$("#group_name").change(function () {
$.post("selectedgr.php", {
gr_name: $(this).val()
}, function (data) {
// console.log(data);
// jQuery will convert the string "['foo', 'bar', 'blah', 'baz']" into a JavaScript object
// (an array in this case) and pass as the first parameter
for(var i = 0; i < data.length; i++) {
$("#" + data[i]).attr("checked", "checked");
}
}, "json");
});
</script>

Answer 3

If you want to use JSON then just use echo json_encode($res);
But I don't really understand what you'll gain if your code is working now, since you'll still have to do some processing in the Javascript to handle the result.

Answer 4

I found my major problem as below
instead of (before):
$.post("selectedgr.php", {data: selectedGroup}, function(data){
do this (after):
$.post("selectedgr.php", selectedGroup, function(data){
Forgive my bad. Ahh ya guys, regarding the escaping on mysql actually #group_name is not any input field but a select box. Appreciate for every comment, suggestion and guide.
Eric.