set current menu item active

Notice: This thread is very old.
11 years ago
Garcha
Member | 9
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In the @layout.latte I have a menu, and I want to set current presenter (page) as active, but after searching all day long, could not get help.
I'm sure it's possible in very easy way.
Thanks

11 years ago
Azathoth
Member | 495
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I am not sure if it is what you need, but if you need to test if current page is some action,
there is small trick,
$presenter->isLinkCurrent(‘:Presenter:action’);
it returns boolean.

Last edited by Azathoth (2015-04-10 21:48)

11 years ago
F.Vesely
Member | 369
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In latte

<a {ifCurrent Presenter:action $params}class="active"{/ifCurrent}>...</a>
11 years ago
duke
Member | 650
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Or more elegantly:

<a n:href="Presenter:action" n:class="$presenter->isLinkCurrent() ? active">...</a>

Or if used with non-linking tag:

<li n:class="$presenter->isLinkCurrent("Presenter:action") ? active">...</li>
11 years ago
Garcha
Member | 9
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Thanks, and what if I want to add some additional classes?

11 years ago
BigCharlie
Member | 285
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divide by commas, see documentation. I'm not sureif you don't need to use quotes to wrap “regular classes”. Try this example: n:class="$iterator->odd ? alt, 'my-class other-class'".

Last edited by BigCharlie (2015-04-30 14:30)

11 years ago
Garcha
Member | 9
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Thanks but I need to have one class name always but also active when active, an this case you said you can switch two versions :/

11 years ago
BigCharlie
Member | 285
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I do not understand your comment. Have you tried the proposal and its functionality?

11 years ago
tpr
Member | 55
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It worked for me, thanks!