i'm using delayed.start().
the way my code works is that once it finishes generating, it sends a 200 status code. the problem is that delayed.start() also sends a response code therefore making the 200 sent invalid. is there any way to circumvent this?
i'm probably just using this package wrong, but i'd love to hear your thoughts.
i'm using delayed.start().
the way my code works is that once it finishes generating, it sends a 200 status code. the problem is that delayed.start() also sends a response code therefore making the 200 sent invalid. is there any way to circumvent this?
i'm probably just using this package wrong, but i'd love to hear your thoughts.