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Pirata Nervo · 09/09/2007 06:09 AM

well the member.php has the $usefields varaible, and i can use the $userfields variable there without doing anything i just need to do what i did in my code.

Like this per example:
Replace this:

PHP Code:

$formattedname = format_name($memprofile['username'], $memprofile['usergroup'], $memprofile['displaygroup']);

with this:

PHP Code:

    if($userfields['fid5'])
    {
        $formattedname = "<span style=\"color:{$userfields['fid5']}\">{$memprofile['username']}</span>";
    }
    else
    {
        $formattedname = format_name($memprofile['username'], $memprofile['usergroup'], $memprofile['displaygroup']);
    }

So this is my function now:

PHP Code:

function format_namePN($username, $usergroup, $displaygroup="")
{
    global $groupscache, $cache, $db;
	
	$query = $db->simple_select("SELECT * FROM userfields");
    $userfields = $db->fetch_field($query);

    if(!is_array($groupscache))
    {
        $groupscache = $cache->read("usergroups");
    }

    if($displaygroup != 0)
    {
        $usergroup = $displaygroup;
    }

    if($userfields['fid5')
    {
        $format = "<span style=\"${userfields['fid5'};\">{username}</span>";
    }
    else
    {
        $ugroup = $groupscache[$usergroup];
        $format = $ugroup['namestyle'];
    }
    $userin = substr_count($format, "{username}"); // function that counts how many times {username} substring is in the $formats string
    if($userin == 0) // if {username} substring wasnt found in the $format string, then
    {
		$format = "{username}"; // the $format string = "{username}"
    }
    $format = stripslashes($format); // The stripslashes() function removes backslashes from the $format string
    return str_replace("{username}", $username, $format); // replaces {username} with $username, so the $format string will contain $username instead of {username}
}

Still not working.

And its $format = "{username}"; not with &#092

ZiNgA BuRgA · 09/09/2007 06:12 PM

Uhh.... the two functions kinda contradict each other... Try working through your logic first.

Pirata Nervo · 10/09/2007 02:38 AM

but i know what im doing and i know what to edit but i don't know what i need to acess userfields.

Also what functions contradict each other?

ZiNgA BuRgA · 10/09/2007 03:07 AM

Well, if you know what you're doing, you obviously don't need help! :P

*sigh* okay, I'll give you a quick 1 minute function (too lazy to test if it works). It _should_ work, but do note that it's _very_ slow:

PHP Code:

function format_name($username, $usergroup, $displaygroup="")
{
	global $db;
	$query = $db->query("SELECT fid5 FROM ".TABLE_PREFIX."userfields f LEFT JOIN ".TABLE_PREFIX."users u ON u.uid = f.uid WHERE u.username = '".$db->escape_string($username)."'");
	
	if(!strpos($format = $db->fetch_field($query, 'fid5'), '{username}'))
	{
		global $groupscache, $cache;

		if(!is_array($groupscache))
			$groupscache = $cache->read("usergroups");

		if($displaygroup) $usergroup = $displaygroup;
		
		if(!strpos(($format = $groupscache[$usergroup]['namestyle']), '{username}'))
			$format = "{username}";
	}
	return str_replace("{username}", $username, stripslashes($format));
}

Pirata Nervo · 10/09/2007 03:24 AM

i don't think that's what i want, i just want users to insert the hex color, not the {username}

I tested that:
MySQL error: 1054
Unknown column 'f.uid' in 'on clause'
Query: SELECT fid5 FROM mybb_userfields f LEFT JOIN mybb_users u ON u.uid = f.uid WHERE u.username = 'GoogleBot'

Pirata Nervo · 10/09/2007 02:53 PM

also i tried that in the function format_name instead of format_namePN
and i get this error:
MySQL error: 1054
Unknown column 'f.uid' in 'on clause'
Query: SELECT fid5 FROM mybb_userfields f LEFT JOIN mybb_users u ON u.uid = f.uid WHERE u.username = 'Pirata Nervo'

ZiNgA BuRgA · 10/09/2007 04:12 PM

Well, if it's not called "uid" replace it with the proper name. I'm too lazy to look up what the correct name is.

If you don't want people entering a then just remove it...

Pirata Nervo · 11/09/2007 01:41 AM

Ok this is how it is:
f.ufid

No errors shown, BUT, nothing changed.
Also remove what? the span???
Doesn't your function requires people to insert {username} instead of just #hexcolor?

ZiNgA BuRgA · 11/09/2007 03:54 AM

Urgh, okay, tell me what do you have in fid5?

Try setting it to something like:

Code:

<span style="color: red;">{$username}</span>

Let's take this one step at a time.

Pirata Nervo · 11/09/2007 04:00 AM

I have #ff0000.

ok let me try with that.

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