XF 2.2 Using style properties in a template conditional

[Lee]

If I wanted to use a style property in a template conditional what is the correct way to reference it?

As a side note, is using template conditionals in css/less templates possible and/or recommended practice?

 

[Brogan]

Standard XF conditional statements don't work in less templates.

What are you trying to do?

Style properties can be called in templates using {{ property('propertyName') }}.

 

[Lee]

Trying to apply some CSS to an element depending on what conditions are selected in style properties.

I want to enable / disable a menu.

I was trying to stop the CSS loading in the less templates if it isn't required.

I think what I will do is wrap the HTML in a conditional and then put the less in its own template and wrap that template include in a conditional.

Unless there is a better way to achieve this?

 

[Brogan]

Yes, the conditional statement would have to be in the html template, so that sounds about right.

 

[Lee]

Thanks Brogan.

What is the correct method to include a less template and have it parse as css?

 

[XDinc]

Thanks Brogan.

What is the correct method to include a less template and have it parse as css?

Hi @Lee
I can reply briefly. More details maybe will be adding by @Brogan
Create your.less template put it the your codes and include your less template with use the following;

<xf:css src="your.less"/>

or use the

<xf:css>your css codes</xf:css>

That's all.

 

[Lee]

Thank you, XDinc. Very helpful!

 

[Brogan]

<xf:css src="template_name.less" />