两个mysql查询之一不显示结果

Question

I'm learning PHP and I need to select some media from database by id or slug. Can I ask you why this code works fine:

index.php?id=1

if( isset( $_GET['id'] ) ) {
    $id = $_GET['id'];
    $query = mysql_query("SELECT * FROM media WHERE id = $id");

    while($row=mysql_fetch_array($query)) {
        echo $row['slug'];
    }
}

And this not?

index.php?slug=first-post-slug

if( isset( $_GET['slug'] ) ) {
    $slug = $_GET['slug'];
    $query = mysql_query("SELECT * FROM media WHERE slug = $slug");

    while($row=mysql_fetch_array($query)) {
        echo $row['id'];
    }
}

The second piece of code does not return anything, and the first piece of code returns proper slug ;/

Answer 1

You should escape the variable:

$slug = mysql_real_escape_string($_GET['slug']);
$query = mysql_query("SELECT * FROM media WHERE slug = '". $slug ."'");

And also you should cast id to int:

$id = (int) $_GET['id'];