Anyone?

[Hex_legend]

I have set:

PHP Code:

<if condition="$referrals == 0"><font color="red">
Referrals: $referrals</font><else /> Referrals: $referrals</if> 


Now..

I need it so that if:

0 Referrals = That message Above
1+ Referrals = Shows their number (as shown in else statement)
20+ = Shows in blue.

How would I do that? I need an If condition x2, but not sure how to do it.

[SEOvB]

<a href="https://vborg.vbsupport.ru/showthread.php?t=188930" target="_blank">https://vborg.vbsupport.ru/showthread.php?t=188930</a>

Also font codes are bad coding use CSS

[Hex_legend]

Im not sure how this would help me use and / or conditions.

If they have 0 referrals, it will say 0 referrals in red.
If they have 1-19 it will say it in default
If they have 20+ it will say it in blue.

This is what I can't work out the code for.

[Dismounted]

The number of users that have be referred by a certain user is not stored in the database, it requires an additional query to count.

[Hex_legend]

$referrals does work. It comes up with XX referrals. I am using a referral mod to show this.
But what I am asking is how to specify how to use the command

If 0 = red font
and 1 - 19 = normal
or 20+ = blue font

Thus using IF AND OR commands.. How would I do this?

[Lynne]

<if condition="$referrals == 0">
<span style="color:red;"> Referrals: 0</span>
<else /> 
<if condition="$referrals <= 19">
Referrals: $referrals
<else /> 
<span style="color:blue;"> Referrals: $referrals</span>
</if> 
</if>

Not tested.

[Hex_legend]

Perfect, Thats what I was asking!!

Thanks Lynne.

I was using

<if condition"$referrals == <20"> Which was what was giving me the error.

[Dismounted]

Originally Posted by Hex_legend

$referrals does work. It comes up with XX referrals. I am using a referral mod to show this.

Ah OK, did not know that .