python - How to respond with 400 in Django on `MultiValueDictKeyError`? -

coming flask, request.args[key] , request.form[key] responds 400 if key doesn't exist. has nice property of failing fast.

in django, request.get[key] , request.form[key] raises httpresponsebadrequest if key doesn't exist, causes api respond 500 status if unhandled. i've seen other answers recommending request.get.get(key), inadvertently loosens api contract allow client-side bugs slip (e.g. query parameter typo, forgetting include param, etc).

i manually check required query params:

arg1 = request.post.get('arg1') if not arg1:     raise httpbadrequest('"arg1" not provided.')  arg2 = request.post.get('arg2') if not arg2:     raise httpbadrequest('"arg1" not provided.')  arg3 = request.post.get('arg3') if not arg3:     raise httpbadrequest('"arg3" not provided.') 

but leaves lot of room error. (did notice typo?)

i manually handle httpresponsebadrequest:

def my_view(request):     try:         arg1 = request.post['arg1']         arg2 = request.post['arg2']         arg3 = request.post['arg3']     except multivaluedictkeyerror ex:         raise httpbadrequest('"{}" not provided.'.format(ex)) 

but have thing remember when writing views.

compare above examples flask's equivalent:

arg1 = request.form['arg1'] arg2 = request.form['arg2'] arg3 = request.form['arg3'] 

how can write same thing in django without boilerplate in views , have respond 400 when param missing?

this should use django forms. validation framework checks user input , returns errors when not valid.

(note, it's not accessing of nonexistent key raises badrequest; raises multivaluedictkeyerror, 1 of snippet notes, , if unhandled response middleware return badrequest.)