Can't use function return value in write context

Question

Ok, here is my code. What it is supposed to do is retrieve the referer that sent you to the page, the user will type someurl.com/refcreate.php?ref=username

<?php
session_start();

$referer = $_GET['ref'];
$_SESSION['referer'] = $referer;

if (!isset($referer))
{
    echo 'You did not specify a referer, please correct this to continue';
    die;
}
elseif($referer == "")
{
    echo 'You did not specify a referer, please correct this to continue';
    die;
}

The above part works fine if they forgot to specify the referer. The below half is to check if the current referer specified is an actual user in the database.

if(refcheck($referer) = false)
    {
        echo 'that referer is not in our database,please double chek the spelling and try again.';
        die;
    }

    function refcheck($ref)
    {
        require('mysql_con.php');

        $query="SELECT username FROM jos_users WHERE username='". $ref ."'";
        echo $query;
        $result = mysql_query($query, $con);
        $exists =mysql_fetch_assoc($result);
        if ($exists != false)
        {
            //return true;      
            echo 'true';    
            return true;

        }

        require('mysql_close.php');
    }
    ?>

OK, I figured out the problem (or problems rather). 1 was it needed to look like this if(refcheck($referer) == false){} instead of if(refcheck($referer) = false);. So it was a missing equal sign and a misplaced colon :P thanks guys

Throwing in a pile of source code along with "it does not work, why is that?" isn't how this site works. Please post the relevant (!) bit of code here directly, explain what it is supposed to do, say what it actually does instead and ask a an actual question. — Tomalak
– Tomalak, Commented Mar 16, 2011 at 5:52
@ycs no this helped me - the question is a verbatim quote of the unhelpful php error message i was getting because of missing = — ErichBSchulz
– ErichBSchulz, Commented Aug 3, 2012 at 14:16

JohnP · Accepted Answer · 2011-03-16 05:50:37Z

9

You're assigning a variable and not comparing it

This refcheck($referer) = false

Should be refcheck($referer) == false

Also, your method should have a default return if your IF condition fails.

answered Mar 16, 2011 at 5:50

JohnP

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Msquared86 · Accepted Answer · 2011-03-16 16:34:34Z

0

ok i figured out hte problem or problems rather 1 was it need to look like this if(refcheck($referer) == false){} instead of if(refcheck($referer) = false); so a missing equal sign and a misplaced colon :P thanks guys

answered Mar 16, 2011 at 16:34

Msquared86

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nv39 · Accepted Answer · 2012-05-22 14:38:03Z

0

you misspelled "check" in the third line

answered May 22, 2012 at 14:38

nv39

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