在我的 sql 數據庫中插入外鍵值？

Question

我想在表中插入外鍵值。我有兩個故事員工（employee_id）和出勤。 這里employee_id 是考勤表中的外鍵。 我嘗試了很多，但沒有插入值。 這是我的代碼

if(isset($_POST['submit']))
{
$date   =    date('Y-m-d',strtotime($_POST['daily_date']));
$in     =    $_POST['daily_in'];
$l_out  =    $_POST['lunch_out'];
$l_in   =    $_POST['lunch_in'];
$out    =    $_POST['daily_out'];
$emp_remarks     =            $_POST['remarks'];
$sql = "INSERT INTO  attendance (atten_id,daily_date,daily_in,lunch_out,lunch_in,daily_out,remarks,employee_id)
VALUES('NULL','$date','$in','$l_out','$l_in','$out','$emp_remarks','".$_REQUEST['employee_id']."')";
   $res = mysql_query($sql);
        if ($res > 0) {
        echo "inserted";  
         }

如果我運行下面的代碼然后

            if(isset($_POST['submit']))
        {  
            $date            =          $_POST['daily_date'];
            $in              =            $_POST['daily_in'];
            $l_out           =            $_POST['lunch_out'];
            $l_in            =            $_POST['lunch_in'];
            $out             =            $_POST['daily_out'];
            $emp_remarks     =            $_POST['remarks'];

        if(isset($_REQUEST['employee_id']))
    {
     echo "Employee Id" .$_REQUEST['employee_id'];
    } 

    else {
     echo "Smoething went wrong";
    }
            $sql = "INSERT INTO  attendance (atten_id,daily_date,daily_in,lunch_out,lunch_in,daily_out,remarks,employee_id)
    VALUES
    ('NULL','$date','$in','$l_out','$l_in','$out','$emp_remarks','".$_REQUEST['employee_id']."')";



its gives 
Smoething went wrong not inserted error

Answer 1

在插入 try this 之前，try this 檢查值是否被傳遞。

isset($_REQUEST['employee_id'])
{ echo "Employee Id" .$_REQUEST['employee_id'];
} else {
 echo "Smoething went wrong";
}