CODEWITHSUNDEEP
pythongoogle-app-enginewebapp2
Derek 朕會功夫
Derek 朕會功夫

Reputation: 94309

Responding with 404 as status code in Webapp2

My app is dynamically generating pages by reading the URL. For example, it will handle all URLs formatted like this:

[url]/word

If /word is a valid URL then the app will generate a page and return it back. When the app can't find anything useful it should return a 404 page.

How can I do that? More specifically how do I set the status code to 404?

Upvotes: 0

Views: 806

Answers (1)

erichiggins
erichiggins

Reputation: 596

From within your RequestHandler, you can simply call self.abort(404) or webapp2.abort(404) to set the error status code.

References:

  • webapp2.RequestHandler.abort():

    Raises an HTTPException.

    This stops code execution, leaving the HTTP exception to be handled by an exception handler.

    Parameters: 

    code – HTTP status code (e.g., 404).
args – Positional arguments to be passed to the exception class.
kwargs – Keyword arguments to be passed to the exception class.

  • webapp2.abort():

    Raises an HTTPException.

    Parameters: 

    code – An integer that represents a valid HTTP status code.
args – Positional arguments to instantiate the exception.
kwargs – Keyword arguments to instantiate the exception.

Upvotes: 2

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