PHP MongoDB exception (Error handling) - php

Question

I have a PHP platform where user write mongodb query like picture below
and following code print result as a table
$m = new MongoClient();
$db = $m->Forensic;
$coll= $db->mobile_data;
if (isset($_POST['txt_area']) && !empty($_POST['txt_area'])) {
$d = ($_POST['txt_area']);
$p = json_decode($d);
$user_code = $coll->find($p);
When I type correct code system able to ouput all my result but if I write query wrong I am getting error message like
Warning: MongoCollection::find(): expects parameter 1 to be an array or object, null given in C:\xampp\htdocs\reports5.php on line 126
to catch that error i have tried following try catch code but no luck
try {
if (isset($_POST['txt_area']) && !empty($_POST['txt_area'])) {
$d = ($_POST['txt_area']);
$p = json_decode($d);
$user_code = $coll->find($p);
$NumberOfRow2 = $user_code->count();
$user_code->sort(array('Chat_group' => -1, 'Instant_Message' => 1 ));
}
}
catch (MongoCursorException $e) {
echo "error message: ".$e->getMessage()."\n";
echo "error code: ".$e->getCode()."\n";
}
catch (MongoException $e)
{
echo $e->getMessage();
}
catch(MongoResultException $e) {
echo $e->getMessage(), "\n";
$res = $e->getDocument();
var_dump($res);
}
what would be the best way to catch above error

Answer 1

The warning you're seeing is PHP complaining that $coll->find($p); expects $p to be either array or object while it's null. Quoting json_decode documentation:
NULL is returned if the json cannot be decoded or if the encoded data is deeper than the recursion limit.
So a proper way to defend against warning would be:
$p = json_decode($d);
if ($p === null) {
echo "Provided query is invalid!";
} else {
// do your logic
}