simple SESSION problem

[plodos]

$sql="SELECT d_name,password FROM doctor WHERE d_name='$myusername' and password='$mypassword'";
    $result=mysql_query($sql);
    // Mysql_num_row is counting table row
    $count=mysql_num_rows($result);
    // If result matched table row must be 1 row
    if($count==1){
	$_SESSION['loggedInD'] = true;
	$_SESSION['UserName'] = $myusername;
	$_SESSION['PassWord'] = $mypassword;
	header('location:doctor.php?'.SID.'');
}

There is a problem

<?
session_start();
include("dbconfig.php");
if (!$_SESSION['loggedInD']) 
{
header("location:login.php"); 
die ();
}
else
{ 
$dd_name=$_SESSION['PassWord'];
$sql = "select d_id from doctor where password like  '%$dd_name%'  ";
echo mysql_query($sql);
echo "\n";
echo $sql;
}

output is like that

Resource id #5 select d_id from doctor where password like '%lukeW%' 

but LUKE ID is 1...

 

$dd_name=$_SESSION['PassWord']; //this is working..

but there is a problem here, $sql = "select d_id from doctor where password like  '%$dd_name%'  ";

im my opinion, we can use the session variables in the SQL querys..

 

What can be the error?

[DarkWater]

Because you're echoing the query results instead of assigning it to a var.  MySQL returns a result resource.

 

Basically, don't do:

echo mysql_query($sql);

[plodos]

sorry but

i didnt understand the solution way?

 

$dd_name=$_SESSION['PassWord'];
$sql = "select d_name from doctor where password='$dd_name'  ";
$id = mysql_query($sql);
echo $id;

$dd_name=$_SESSION['PassWord'];
$sql = "select d_id from doctor where password='$dd_name'  ";
$id = mysql_query($sql);
echo $id;

if I write like that....Output is only Resource id #4

 

What must I write for see the true result...??

 

[discomatt]

You need to FETCH the result out of the resource, using one of these

 

mysql_result()

mysql_fetch_row()

mysql_fetch_array()

mysql_fetch_assoc()

mysql_fetch_object()

 

 

[DarkWater]

$id = mysql_query($sql);
$result = mysql_fetch_assoc($id);
       foreach ($result as $k=>$v) {
         echo "$k - $v <br />";
       }