-2

When I try to run this query

$SQL = " SELECT * FROM info WHERE username = '$led_nem'";
$result = mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)) {
        $tsk = $db_field['group_task'];
}

I get this error

Notice: Undefined variable: led_nem in C:\xampp\htdocs\online\Online Task Management System\task_mem.php on line 71

CRABOLO
  • 8,605
  • 39
  • 41
  • 68

2 Answers2

0

Test if the variable $led_nem exists before you use it:

if (isset($led_nem)) {
    $SQL = " SELECT * FROM info WHERE username = '$led_nem'";
    $result = mysql_query($SQL);
    while ($db_field = mysql_fetch_assoc($result)) {
        $tsk = $db_field['group_task'];
    }
}
pavel
  • 26,538
  • 10
  • 45
  • 61
0

Use this

if (isset($led_nem)) {
$SQL = " SELECT * FROM info WHERE username = '".$led_nem."'";
$result = mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)) {
        $tsk = $db_field['group_task'];
}
}
I'm Geeker
  • 4,601
  • 5
  • 22
  • 41