CODEWITHSUNDEEP
scalaplayframework-2.0
Hemant
Hemant

Reputation: 4616

Play 2.0 + Bootstrap3: Showing active navigation item

I am using the following syntax in the Playframework 2.0 to figure out the given Bootstrap nav li should be active or not, but it doesn't work! Any idea what I missed here?

The following is snippet from my playframework template where I am trying to make Home li as active if the current request path is same as routes index/home

@* Home page is active. *@
<li class=@{"active".when("@request.path" == "@routes.Application.index()")}>
  <a href="@routes.Application.index()">Home</a>
</li>

Upvotes: 3

Views: 924

Answers (2)

4lex1v
4lex1v

Reputation: 21557

I'm don't know play, but using their template engine in my projects.

How about this way:

@* Home page is active. *@
<li class=@{if (request.path == routes.Application.index().toString()) "active" else ""}>
  <a href="@routes.Application.index()">Home</a>
</li>

Upvotes: 2

Manuel D.
Manuel D.

Reputation: 76

Try this :

<li class="@("active".when(request.path == controllers.routes.Application.index().toString()))">
  <a href="@routes.Application.index()">Home</a>
</li>

Upvotes: 1

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