Route pattern with the default path URL returning the Not found exception. about esmerald HOT 5 CLOSED

@harshalizode in the docs, it's explicitly said that if you want to do the include for all in one, in theory you should to a nested include, in your case you have 2 includes trying to reach the "/" + the the URLs from the namespace.

What is happening? So now you have the accounts and the employee apps (includes), both in the including without a path, which means it will default to "/".

Since Esmerald like everything, uses regex for the routes, as per docs, it will be a first found. Since both Includes have the same path, the accounts include is ignored.

Advice? If you want both accounts and employees to be under the "/", make those includes to be inside a main Include (nested includes) or give a different path to both includes like "/accounts" and "/employees".

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@tarsil can we includes more than 1 apps under "/" with the nested include?
As per the documentation They have includes for the same app urls.
Can you provide the example for this?

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@harshalizode literally from this section where the warning was also mentioned:

https://esmerald.dev/routing/routes/?h=nested+includes#application-routes

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@tarsil can we includes more than 1 apps under "/" with the nested include? As per the documentation They have includes for the same app urls. Can you provide the example for this?

As you can see, @harshalizode the routing behaves as expected, because internally Esmerald uses a regex to match the routes, you need to be careful the way the routes are assembled, usually, if you want to point to "/", that should be the last one. This is common in almost every framework.

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Okay, Thank you.

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