Showing wrong value of variable using Sql Query

Question

$selectedItem = mysql_escape_string($_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?

mysql_query function returns a resource on success, or FALSE on error according to the documentation. in your case, it returns success. try user3091574's suggestion. — Barış Akkurt
– Barış Akkurt, Commented May 8, 2014 at 9:18

ashishmaurya · Accepted Answer · 2014-05-08 09:15:47Z

1

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'];

answered May 8, 2014 at 9:15

ashishmaurya

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