Validating $_GET - php

Question

How do i validate $_GET thats the number coming from correct source.
My url look like : index.php?page=items&catID=5
When users put something like 3 which is doesn't exist on catID. I want it to display error message.
$catID = intval($_GET["catID"]);
if($catID) {
$checkSQL = mysql_query("SELECT * FROM category WHERE category_type='2'");
while($checkROW = mysql_fetch_array($checkSQL)) {
$checkCAT != $checkROW["categoryID"];
echo "err msg";
}
This i can come up so far but it doesn't working as it fire error msg even in correct page.
Thank you

Answer 1

wallk makes a good point, there is a missing if. but if i read this correctly, wouldn't something along the lines of this be more what you are going for? Right now the line:
if($catID) {
is actually only checking if catID (or, catID from the $_GET) is non-zero (not false). My guess if you are looking to check if catID is the categoryID returned from SQL?
$catID = intval($_GET["catID"]);
checkcat($catID);
function checkcat($check_category) {
$checkSQL = mysql_query("SELECT * FROM category WHERE category_type='2'");
while($checkROW = mysql_fetch_array($checkSQL)) {
if ( $check_category != $checkROW["categoryID"] ) {
echo "err msg";
} else {
echo "not an error message";
}
}
}
Expounding on what you are looking for, how about something like this then?
$catID = ($_GET["catID");
if ( !is_numeric($catID) ) {
echo "Not a numeric category!"
} else {
$checkSQLQuery = "SELECT * FROM category WHERE categoryID = '{$catID}' AND category_type='2'"
$resultSQL = mysql_query($checkSQLQuery, $db);
/* NOTE!: Guessing on what your database resouce
pointer is - it isn't included in the origin snippet.
Although, the last opened should be used by default if
this is left out. */
if ( mysql_num_rows($resultSQL) < 1 ) {
echo "Error message, category ID not found"
} else {
echo "Found it!"
}
}

Answer 2

Oh, I see. The first line inside the while loop should have an "if":
while ($checkROW = mysql_fetch_array($checkSQL)) {
if ($checkCAT != $checkROW["categoryID"])
echo "err msg";

Answer 3

It looks like you'll be wanting to use mysql_fetch_assoc(), rather than mysql_fetch_array().