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Comments (6)

davidwdan avatar davidwdan commented on August 11, 2024 3

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mbonneau avatar mbonneau commented on August 11, 2024

I believe throwing a WampErrorException will allow you to control the WAMP error details that get sent back to the caller.

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27leaves avatar 27leaves commented on August 11, 2024

Thank you, but for that I would need an InvocationMessage and for that some information that I do not have at that point.

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27leaves avatar 27leaves commented on August 11, 2024

OK, it works like that

throw new WampInvocationException(
      new InvocationMessage(InvocationMessage.MSG_INVOCATION, 0, {}, args, argskw),
      'my.error.trysomething',
      args,
      argskw,
      {
        error: `Some error description`
      }
    );

that on the other side this releases:

Error {
  error: 'my.error.trysomething',
  args: [],
  kwargs: { code: 'd126' }
}

Seems like the only information that I don't have at that point is the registrationId. Don't I need that? And the details are not passed, but I guess that's by design?

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davidwdan avatar davidwdan commented on August 11, 2024

This should work:

async function trysomething(a, b) {
    throw new WampErrorException('my.error.trysomething', ['arg1', 'arg2']);
}

How do you plan on using the registrationId?

We limit what is sent back with the error message to limit the possibility of accidentally leaking information. You have fine grain control over everything with WampErrorExeption.

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27leaves avatar 27leaves commented on August 11, 2024

Oh yes, that should be sufficient.

I do not need the registrationId. I just would have needed it on using WampInvocationException with InvocationMessage.

Thank you! :)

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