Parse error sub queries [duplicate]

Question

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Reference - What does this error mean in PHP? (37 answers)

Closed 8 years ago.

I'm trying to show all the games related to the category of the current game played. But I'm getting a parse error in "if (!empty($_GET['gameId'])) $varGameId = $_GET['gameId'];"

How can I solve this?

Thank you

<?php
$varCategoria_GameData = "0";
$varGameId = 0
if (!empty($_GET['gameId'])) $varGameId  = $_GET['gameId'];
if (isset($_GET["cat"])) {
 $varCategoria_GameData = $_GET["cat"];
}

$sql_categoria = "SELECT * FROM jogos WHERE intCategoria =
( SELECT intCategoria FROM jogos WHERE idGames = $varGameId )";
$query_categoria = mysql_query($sql_categoria, $gameconnection) or die(mysql_error()); 
$categoria = mysql_fetch_assoc($query_categoria);

?> 

Answer 1

The error is here, semi-colon missing:

$varGameId = 0

Answer 2

You miss semi colon here: $varGameId = 0