[2.0] how to get current page's URL in view templates?

[Chris B.]

I want to customize the page's navigation menu if the page's URL is of a certain value: e.g. /, /blog, etc.

here is the pseudo code in view template, :

@if (current page's URL == "/") { "Home Menu Selected" }

how to implement this code? Is this the best way to approach this issue?

[Korrawit Yindeeyoungyeon]

You can pass request object to template view by using implicit request.

Request class : http://www.playframework.org/documentation/api/2.0/scala/index.html#play.api.mvc.Request

It is often useful to mark the request parameter as implicit so it can be implicitely used by other APIs that need it:

Action { implicit request =>
  Ok("Got request [" + request + "]")
}

Ref : http://www.playframework.org/documentation/2.0/ScalaActions

Then in your controller,

def index = Action { implicit request =>

    Ok(views.html.index())

}

In your view,

@(message: String)(implicit request: play.api.mvc.Request[Any])

@main("Index") {

<div>

                @if (request.uri == "/") { "Home Menu Selected" }

</div>

}

[Chris B.]

it works. thank you.

[Sergii Starodubtsev]

If I put:

<div>

 @if (request.uri == "/") { "Home Menu Selected" }

</div>

I got

'(' expected but ')' found.

[sunnykaka]

If you're using Play for Java, you can do something this:

<li @if(requestHeader.uri == routes.Help.index.url){class="active"}>

Just ignore the error in your ide, it compiles and run well.

The requestHeader method was import automatically by class play.TemplateImports:

javaImports.add("play.core.j.PlayMagicForJava._");