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Home/ Questions/Q 8542321
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T12:02:06+00:00 2026-06-11T12:02:06+00:00

In Play1.x it was possible to define some conditional routes using the following syntax:

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In Play1.x it was possible to define some conditional routes using the following syntax:

%{ if (play.id == "test") }%

GET     /test/derp website.nonproduction.DerpController.index
...

The %{…}% syntax does not compile on Play2 and I cannot seem to find anything in the documentation that discusses conditional routing. Any ideas?

Thanks in advance & good luck with your Play2 migrations!

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  1. Editorial Team
    Editorial Team
    2026-06-11T12:02:08+00:00Added an answer on June 11, 2026 at 12:02 pm

    Routing in Play 2.0 compiles to a class and cannot be dynamic. However, you can wrap the Action block of any development / sensitive endpoints.

    You can implement something like Zentasks’s Secured trait: https://github.com/playframework/Play20/blob/master/samples/scala/zentasks/app/controllers/Application.scala

    Specifically, to check if Play! is running in Production mode, you can check Play.application().isProd.

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