使用Sql Query显示错误的变量值

$selectedItem = mysql_escape_string($_POST['select']);
$id= mysql_escape_string($_POST['id']);

if ($selectedItem == "ID")
    {
        echo 2;
        $nome = mysql_query("SELECT nome FROM `eventos` WHERE ID = '$id'");
        echo $nome;
    }

Ok, i want to show the name of one event of id that i insert in textbox. It shows me 2, so if statement is working but when i say to show nome show me: 2Resource id #5

I have one row in table eventos with id = 1 and i put always in textbox number 1.

Why it doesnt show me the text that i have in collumn nome?

try this

$n = mysql_fetch_array($nome);
echo $n['nome']; 

It's recommended to use mysqli_query instead of mysql_query
like this

 //
 $nome = mysqli_query("SELECT nome FROM `eventos` WHERE ID = '$id'");
 $n = mysql_fetch_array($nome);
    echo $n['nome'];