[SOLVED] a function to check logins! doesn't return false!

[samoi]

Hello guys

 

I have this function

function phonebook($u, $p)
{
    $query = "SELECT * FROM eid WHERE phone = '" . $p . "' OR user = '" .
        $u . "'";
$result = mysql_query($query) or die(mysql_error("Error result"));
    $row = mysql_fetch_row($result) or die(mysql_error("Error row"));
    return $row;
}

 

and this if statement

 

if (isset($_POST['book'])) {

    $user = $_POST['user'];
    $phone = $_POST['phone'];
phonebook($user, $phone);
    if (phonebook($user, $phone)) {
        $_SESSION["user"] = phonebook($user, $phone);

        echo "The contact id is " . $_SESSION["record"][0] . " And the number is " . $_SESSION["record"][1];
    } else {
        echo 'No record'; // To tell the user there was no record
    }


}
else {
// Get the user out to search.php page again
}

 

Please help me with debug the problem!

 

Thank you all!

 

 

 

Please note, I am doing this to learn only! no purpose behind all of this! I wanted to do some functions and learn with it.

[Garethp]

You don't use die("Error") to return 1; (I'm pretty sure, but it might be return 0;)

[samoi]

You don't use die("Error") to return 1; (I'm pretty sure, but it might be return 0;)

 

wow !

 

thank you for the speed reply!

But as I stated above, that I want to learn doing the functions!

I did not understand what you meant!

also I noticed return with 1 and 0 values in many scripts! what does that mean?

 

Thank you and sorry for being asking too much!

[Garethp]

Ok, return will stop the function as soon as it's executed, and whatever is in the return (like, return $var or return "This") gets passed outside. I'll show you an example

 

function Test()
{
$Var = "Hello World";
return $Var;
}

echo Test();

 

That would echo "Hello World" Or you could do this

 

function Test()
{
return "Hello World";
}

$Var = Test();
echo $Var;

 

That would also echo Hello World.

 

Now, if you were to use return 1, that wouldn't return a string, but you could use an if function. I'll show you

 

function Test()
{
$Var = "On";
if($Var == "On")
	{
	return 1;
	}
else
	{
	return 0;
	}
}

if(Test())
{
echo "Function is Working";
}
else
{
echo "Error";
}

 

Try it, it should work. 1 is for success, 0 is for error