XF 2.2 Widget Conditional Statement

[Itworx4me]

I want to add a widget to a particular page. What would the conditional statement look like?

Thanks,
Itworx4me

 

[Paul B]

This is the syntax: $var == 'thing' .

From here: https://xenforo.com/community/resou...nd-widgets.8041/updates#resource-update-37750

What is the actual page?

 

[Itworx4me]

This is the syntax: $var == 'thing' .

From here: https://xenforo.com/community/resou...nd-widgets.8041/updates#resource-update-37750

What is the actual page?

Its a custom addon that I had developed for our site.

 

[Itworx4me]

This is the syntax: $var == 'thing' .

From here: https://xenforo.com/community/resou...nd-widgets.8041/updates#resource-update-37750

What is the actual page?

How can I find the $var the page is using?

Thanks,
Itworx4me

 

[Paul B]

The main part of the guide explains how.

 

[Itworx4me]

The main part of the guide explains how.

Is there a conditional statement that I could use to select a template instead of a $var ???

 

[Itworx4me]

Ok I have figured out how to get this to work. Now just need to expand it so it includes all sub-url. This is what I have so far:

$xf.uri == '/my-page/'

How can I include everything that my-page url uses. Like:
/my-page/page-1
/my-page/page-2

without naming all sub-pages that are associated with url?

Thanks,
Itworx4me

 

[Paul B]

If there isn't a var which is common to all pages, you will have to use an array.