PHP Code Problem...

Question

function check_login($array_val)
   {


  $strQury = "Select * from  tblsignup where usr_email ='".$array_val[0]."' and usr_password  = '".$array_val[1]."'" ;

    $result  = mysql_query($strQury);
    $row_user = mysql_fetch_array($result);
    if(mysql_num_rows($result)>0)
     {
      $msg = "true";

     }
    else
     {
      $msg = "false";
     }
    return $msg ;
   }

The return value is Object id #1true???? what is object id#1?

Answer 1

Try this:

function check_login($array_val)
{
    $strQury = "Select * from  tblsignup where usr_email ='".$array_val[0]."' and usr_password  = '".$array_val[1]."'" ;

    $result  = mysql_query($strQury);
    $row_user = mysql_fetch_array($result);

    if(mysql_num_rows($result)>0)
    {
        return true;
    }
    else
    {
        return false;
    }
}

Let us know what result you get when using that code.

Answer 2

You're returning the strings "true" or "false" when you probably mean the boolean values true and false.

Oh, and your code is wide open to a visit from Little Bobby Tables. You really should use mysqli and proper prepared statements instead.

Answer 3

user single quotes and things will start to work better. also check your query for sql injection bug as it does have it.

Answer 4

Change from:

echo $objUser.check_login($array_login);

to:

echo $objUser->check_login($array_login);

The . operator in PHP does string concatenation, while the arrow allows you to access object methods and attributes.

Answer 5

Change

echo $objUser.check_login($array_login);

to

echo $objUser;
echo check_login($array_login);

You should end up with the following result:

Object id #1
true

My guess is that $objUser was set earlier with something along these lines:

$objUser = new User;

As a result, it is an object (the first one declared) and will return Object id #1 when you just echo it. You will need to read up on classes to understand that more.