In Flask convert form POST object into a represent

问题:

I am using Flask and MongoDB. I am trying to convert the content of request.form into something suitable for saving via PyMongo. It seems like something that should come up often enough to have a ready-made solution.

So what Flask give me is something like:

ImmutableMultiDict([('default', u''), ('required': u'on'), ('name', u'short_text'), ('name', u'another'), ('submit', u'Submit')])

And what I am looking to get is something close to this:

{
  'default': '',
  'name': ['short_text', 'another'],
  'required': true
}

回答1:

>>> from werkzeug.datastructures import ImmutableMultiDict
>>> imd = ImmutableMultiDict([('default', u''), ('required', u'on'), ('name', u'short_text'), ('name', u'another'), ('submit', u'Submit')])
>>> imd.to_dict(flat=False)
>>> {'default': [''], 
'name': ['short_text', 'another'],
'required': ['on'],
'submit': ['Submit']}

.to_dict(flat=False) is the thing to keep in mind. See the relevant documentation

回答2:

The Flask ImmutableMultiDict data structure has a built in to_dict method.

This knowledge in addition to the Flask request object form property being an ImmutableMultiDict allows for simple handling of a form POST request to MongoDB.

See below for a naive example:

from flask import request

@app.route('/api/v1/account', methods=['POST'])
def create_account():
    """Create user account"""
    account_dict = request.form.to_dict()

    db.account.insert_one(account_dict)

回答3:

You can use werkzeug's getlist to write code like this

data = dict((key, request.form.getlist(key)) for key in request.form.keys())

Now each key of data would be a list which would contain 1 more element. To get results exactly in your format do this

data = dict((key, request.form.getlist(key) if len(request.form.getlist(key)) > 1 else request.form.getlist(key)[0]) for key in request.form.keys())

Now this is inefficient because for each key there are 3 calls to request.form.getlist(key). You can write a loop and get around it.

回答4:

request.form.to_dict() would yield what you need

回答5:

>>> from werkzeug.datastructures import ImmutableMultiDict
>>> so = ImmutableMultiDict([('default', u''), ('required', u'on'), ('name', u'short_text'), ('name', u'another'), ('submit', u'Submit')])

# Most earlier answers have comments suggesting so.to_dict()
# It doesn't work, duplicates are lost like in a normal dict
>>> so.to_dict()
{'default': '', 'required': 'on', 'name': 'short_text', 'submit': 'Submit'}

# The response by Vb407 is better but litters lists everywhere
>>> dso = dict(so)
{'default': [''], 'required': ['on'], 'name': ['short_text', 'another'], 'submit': ['Submit']}

# We can achieve the requested state by cleaning this up
>>> { k: dso[k][0] if len(dso[k]) <= 1 else dso[k] for k in dso }
{'default': '', 'required': 'on', 'name': ['short_text', 'another'], 'submit': 'Submit'}

回答6:

Comparison of dict() and .to_dict() method before and after python version 3.6.

from werkzeug.datastructures import ImmutableMultiDict
imd = ImmutableMultiDict([('default', u''), ('required', u'on'), ('name', u'short_text'), ('name', u'another'), ('submit', u'Submit')])

Till python3.5

dict(imd)
#output: {'default': [''], 'required': ['on'], 'name': ['short_text', 'another'], 'submit': ['Submit']}

imd.to_dict(flat=false)
#output: {'default': [''], 'required': ['on'], 'name': ['short_text', 'another'], 'submit': ['Submit']}

imd.to_dict(flat=True) # or imd.to_dict() 
#output: {'default': '', 'required': 'on', 'name': 'short_text', 'submit': 'Submit'}

Thus,

dict(imd) == imd.to_dict(flat=False)
#output: True

From python3.6 onwards

dict(imd)
#output: {'default': '', 'required': 'on', 'name': 'short_text', 'submit': 'Submit'}

imd.to_dict(flat=false)
#output: {'default': [''], 'required': ['on'], 'name': ['short_text', 'another'], 'submit': ['Submit']}

imd.to_dict(flat=True) # or imd.to_dict() 
#output: {'default': '', 'required': 'on', 'name': 'short_text', 'submit': 'Submit'}

Thus,

dict(imd) == imd.to_dict(flat=False)
#output: False

Using .to_dict() with flat=True/False is a safer option.

回答7:

Once you have a Python dictionary, it is relatively simple to transform it to JSON. Supposing your dictionary is mdict:

import json

mdict = {'default' : u'', 'required': u'on', 'name': u'short_text', 'name': u'another', 'submit': u'Submit'}
json_value = json.dumps(mdict, separators=(',',':'))

PS: I think mutable dicts have a to_dict() method to convert them to a regular dictionary.