User Function

[Kieran-R]

Hello

Well I was trying to make a function where it would make the URL, staff colours, and donator image appear. Although ive ran into a problem.

Ive never really attempted to work with functions before, but this is what I came up with:

function username_formatter($username)
{
$USER=mysql_query("SELECT * FROM users WHERE username=$username");
$FETCH=mysql_fetch_array($USER);
if($FETCH['donatordays']&&$FETCH['user_level']==1) { $FETCH['username'] = "<font color=white>[b]{$FETCH['username']}[/b]</font>";$d="[img=donator.gif]"; }
if($FETCH['user_level']==2) { $FETCH['username'] = "<font color=red>[b]{$FETCH['username']}[/b]</font>";$d="[img=donator.gif]"; }
if($FETCH['user_level']==5) { $FETCH['username'] = "<font color=#00FF00>[b]{$FETCH['username']}[/b]</font>";$d="[img=donator.gif]"; }
if($FETCH['gang']>0){
$gang=mysql_query("SELECT * FROM gangs WHERE gangID={$FETCH['gang']}");
$gn=mysql_fetch_array($gang);
$gangtag="[url='gangs.php?action=gang_view&gang_id={$FETCH['][{$gn['gangPREF']}][/url]";
}
else
{
$gangtag="";
}
$user="$gangtag <a href=viewuser?u={$FETCH['userid']} />{$FETCH['username']}</a> $d";

return $user;
}

 

to use this function, I would do: username_formatter(***USERNAME***). But it just displays nothing. Yes. I am printing it correctly lol

If someone could assist me then it would be much appreciated!

[Danny696]

Try this:

function username_formatter($username)
{
$USER=mysql_query("SELECT * FROM users WHERE username=$username");
$FETCH=mysql_fetch_array($USER);
if($FETCH['donatordays']&&$FETCH['user_level']==1) { $FETCH['username'] = "<font color=white>[b]{$FETCH['username']}[/b]</font>";$d="[img=donator.gif]"; }
if($FETCH['user_level']==2) { $FETCH['username'] = "<font color=red>[b]{$FETCH['username']}[/b]</font>";$d="[img=donator.gif]"; }
if($FETCH['user_level']==5) { $FETCH['username'] = "<font color=#00FF00>[b]{$FETCH['username']}[/b]</font>";$d="[img=donator.gif]"; }
if($FETCH['gang']>0){
$gang=mysql_query("SELECT * FROM gangs WHERE gangID={$FETCH['gang']}");
$gn=mysql_fetch_array($gang);
$gangtag="[url='gangs.php?action=gang_view&gang_id={$FETCH['][{$gn['gangPREF']}][/url]";
}
else
{
$gangtag="";
}
echo "$gangtag <a href=viewuser?u={$FETCH['userid']} />{$FETCH['username']}</a> $d";
}

[Kieran-R]

Hmm Im getting an error from: $FETCH=mysql_fetch_array($USER);

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

Cant seem to se why that would be erroring though...

[Danny696]

Show me the part where your using the function

[Kieran-R]

echo "Welcome Back ".username_formatter($ir['username'])."!";

[Kieran-R]

Ahh.

I found the error. Its because in teh MySQL query, i put a string on its own instead of surrounding it with ' '

[SilvaTungDevil]

Line 3 and line 9.

You are fetching EVERYTHING from the users and gang tables, only having briefly scanned your function I suspect you dont need every column.

My suggestions, optimize it a bit and only grab the columns that you need:

So maybe [mysql]SELECT `userid`, `username`, `user_level`, `donatordays`, `gang` FROM `users` WHERE `username` = $username[/mysql]

And also [mysql]SELECT `gangID`, `gangPREF` FROM `gangs` WHERE `gangID` = {$FETCH['gang']} [/mysql]