mysql-从表中选择PHP不起作用

Question

So I am trying to get the FirstName value based on the email and password provided by the user. But for some reason the code I have does not return anything. Can anybody help?

$sql_result = "SELECT FirstName FROM $tbl_name WHERE Email='$email' and Password='$mypassword' limit 1";
$res = mysql_query($sql_result);
$value = mysql_fetch_object($res);

$_SESSION['name']=$value['FirstName'];

I have also tried the code below but that didn't work either.

 $_SESSION['name']=$value->FirstName;

Answer 1

$value = mysql_fetch_object($res);
                     ^^^^^^^---- fetching an object

$_SESSION['name']=$value['FirstName'];
                   ^^^^^^^^^---treating it like an array

Try mysql_fetch_assoc() instead, so your $value really is an array, or use $value->FirstName and use it as the object it really is.