You need to return $x out of the function.

So do this:

function hello() {
        print 'hello ' . $name;
        $x = 1;
        return ($x);
}

etc

Then call it assigned to a variable:

$x = hello();
echo $x;
bye();

Kyle Babich wrote:
I think this is short example of my problem...

<?php

$name = 'kyle';

function hello() {
    print 'hello ' . $name;
    $x = 1;
}

function bye() {
    if ($x == 1) print 'x = 1';
    else print 'x != 1';
}

hello();
bye();

?>

Right now this returns:  hello x != 1
What do I have to do to get bye() to return 'x = 1'? I tried declaring
the x = 1 in hello() global and I tried declaring it static.



On Tue, 01 Jul 2003 17:27:48 -0500, "Kyle Babich" <[EMAIL PROTECTED]>
said:

Any php programmers out there have a little free time? I've been trying
to find the bug in my logging system forever and I've all but given up. If anyone else wants to try their luck then...


http://babich.us/log/source/log.php.txt
http://babich.us/log/source/config.inc.php.txt
http://babich.us/log/source/test.php.txt
http://babich.us/log/source/clearLogs.inc.php.txt
http://babich.us/log/source/logIpData.inc.php.txt
http://babich.us/log/source/logAgentData.inc.php.txt
http://babich.us/log/source/logLangData.inc.php.txt

I have a feeling that it is an obvious, simple error that I am missing.
Anyway, back to the bug hunting...
--
Kyle

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