Json and Mysql problem

Question

here is my code for json php

include("connect.php"); 
$id = $_GET['lid'];
function countRec($fname,$tname) {
$sql = "SELECT * FROM `mail` WHERE confirmed = 'no' AND label_id = '". $id ."'";
$result = mysql_query($sql) or die ('test'); 
$num = mysql_num_rows($result);
return $num;
}

$page = $_POST['page'];
$rp = $_POST['rp'];
$sortname = $_POST['sortname'];
$sortorder = $_POST['sortorder'];

if (!$sortname) $sortname = 'ID';
if (!$sortorder) $sortorder = 'desc';

    $sort = "ORDER BY $sortname $sortorder";

if (!$page) $page = 1;
if (!$rp) $rp = 10;

$start = (($page-1) * $rp);

$limit = "LIMIT $start, $rp";

$sql = "SELECT * FROM `mail` WHERE confirmed = 'no' AND label_id = '". $id ."' $sort $limit";
$result = mysql_query($sql) or die ('test'); 

$total = countRec();

header("Expires: Mon, 26 Jul 1997 05:00:00 GMT" );
header("Last-Modified: " . gmdate( "D, d M Y H:i:s" ) . "GMT" );
header("Cache-Control: no-cache, must-revalidate" );
header("Pragma: no-cache" );
header("Content-type: text/x-json");
$json = "";
$json .= "{\n";
$json .= "page: $page,\n";
$json .= "total: $total,\n";
$json .= "rows: [";
$rc = false;
while ($row = mysql_fetch_array($result)) {
if ($rc) $json .= ",";
$json .= "\n{";
$json .= "id:'".$row['ID']."',";
$json .= "cell:['".$row['email']."'";
$json .= ",'".addslashes($row['name'])."'";
$json .= ",'".addslashes($row['country'])."'";
$json .= ",'".addslashes($row['bus'])."'";
$json .= ",'".addslashes($row['website'])."'";
$json .= ",'".addslashes($row['music'])."'";
$json .= ",'".addslashes($row['radio'])."']";
$json .= "}";
$rc = true;
}
$json .= "]\n";
$json .= "}";
echo $json;

i am posting data to this php like that "req.php?lid=3434"

and getting "lid" like $id = $_GET['lid']; as you can see

but in my mysql, when i write WHERE label_id = '$id' it doesnt work

any suggestions?

Thanks

Tom Haigh · Accepted Answer · 2009-10-25 16:35:40Z

3

You are referencing the global $id inside a function. You need to mark it as global:

function countRec($fname,$tname) {
    global $id;
    //etc
}

Or you could pass it to the function as a third parameter, which is probably a better solution.

Note that this code is vulnerable to SQL injection attacks. You should either quote $id (e.g. using mysql_real_escape_string()), or if it always an integer you could cast it, e.g. $id = (int) $id. Better still, you could use PDO and use prepared statments and bound parameters, which removes this problem.

edited Oct 25, 2009 at 16:35

answered Oct 25, 2009 at 15:17

Tom Haigh

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1 Comment

cdmckay Over a year ago

I'd put the $id as a third parameter instead of using global.

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