How do I return HTTP error code without default template in Tornado?

Question

I am currently using the following to raise a HTTP bad request:

raise tornado.web.HTTPError(400)

which returns a html output:

&         


        

Answer 1

Tornado calls RequestHandler.write_error to output errors, so an alternative to VisioN's approach would be override it as suggested by the Tornado docs. The advantage to this approach is that it will allow you to raise HTTPError as before.

The source for RequestHandler.write_error is here. Below you can see an example of a simple modification of write_error that will change the set the status code and change the output if you provide a reason in kwargs.

def write_error(self, status_code, **kwargs):
    if self.settings.get("serve_traceback") and "exc_info" in kwargs:
        # in debug mode, try to send a traceback
        self.set_header('Content-Type', 'text/plain')
        for line in traceback.format_exception(*kwargs["exc_info"]):
            self.write(line)
        self.finish()
    else:
        self.set_status(status_code)
        if kwargs['reason']:
            self.finish(kwargs['reason'])
        else: 
            self.finish(""
                "%(code)d: %(message)s" % {
                    "code": status_code,
                    "message": self._reason,
                })