XF 2.1 Conditional statement not working

[C Allingham]

Trying to use this conditional statement in an advertisement in Container content: Above on my forum list page. I'm guessing you can't use these xf statements in an ad?

<xf:if is="$template == 'forum_list'">
Show test content...
</xf:if>

The page in question is https://tvwbb.com. My homepage is the forum list page.

 

[C Allingham]

Do xf conditional statements work in ad containers?

 

[Russ]

Try this out instead:

<xf:if is="$xf.reply.template == 'forum_list'">
ad code here
</xf:if>

 

[C Allingham]

That works. Thank you Russ!

 

[tonmo]

Try this out instead:

<xf:if is="$xf.reply.template == 'forum_list'">
ad code here
</xf:if>

This was helpful, what about excluding specific forums/nodes? The following does not seem to work:

<xf:if is="!in_array($forum.node_id, [1,2,3])">

 

[tonmo]

got it:

XF 2.1 - Widget display condition

If I enter in widget display position for example $forum.node_id == 270 to show this widget only in specific forum it does not work. But if I add this to forum_view template it works: What am I missing here?

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