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Add kwargs into insert method #2169 #2170

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Add kwargs into insert method #2169 #2170

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5268bf5
Add kwargs into insert method #2169
Sep 25, 2019
4460e51
Fix tests
Sep 25, 2019
68695ce
Fix feedback
Sep 26, 2019
eaf65ca
Fix test
Sep 26, 2019
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1 change: 1 addition & 0 deletions docs/changelog.rst
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Original file line number Diff line number Diff line change
Expand Up @@ -19,6 +19,7 @@ Development
- ``ListField`` now accepts an optional ``max_length`` parameter. #2110
- The codebase is now formatted using ``black``. #2109
- In bulk write insert, the detailed error message would raise in exception.
- ``Doc.objects.insert`` accepts kwargs to pass ``ordered``, ``bypass_document_validation``, etc. #2169

Changes in 0.18.2
=================
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9 changes: 7 additions & 2 deletions mongoengine/queryset/base.py
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Expand Up @@ -288,7 +288,8 @@ def first(self):
return result

def insert(
self, doc_or_docs, load_bulk=True, write_concern=None, signal_kwargs=None
self, doc_or_docs, load_bulk=True, write_concern=None, signal_kwargs=None,
**kwargs
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):
"""bulk insert documents

Expand Down Expand Up @@ -323,6 +324,10 @@ def insert(
return_one = True
docs = [docs]

if return_one and 'ordered' in kwargs:
# insert_one does not accept `ordered` argument
kwargs.pop('ordered')

for doc in docs:
if not isinstance(doc, self._document):
msg = "Some documents inserted aren't instances of %s" % str(
Expand All @@ -345,7 +350,7 @@ def insert(
insert_func = collection.insert_one

try:
inserted_result = insert_func(raw)
inserted_result = insert_func(raw, **kwargs)
ids = (
[inserted_result.inserted_id]
if return_one
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I see a potential bug related to the following code (that is a bit below in this method):

        # Apply inserted_ids to documents
        for doc, doc_id in zip(docs, ids):
            doc.pk = doc_id

In case you provide ordered=False and it raises an error (e.g if you try to insert 2 times a document with same pk), that piece of code won't be reached, leaving the mongoengine documents without id's. Instead of relying on zip to get the corresponding id, we could rely on the raw dict that actually gets updated with the id by pymongo (see this post). Now since it will raise an error anyway, perhaps it could be acceptable to not set the id on the Documents instances. What do you think?

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@bagerard bagerard Sep 25, 2019 •
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btw in case of successful unordered insert, I'm also unsure if we could rely on zip , as I don't know if we can rely on the order of inserted_result.inserted_ids

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6 changes: 6 additions & 0 deletions tests/queryset/test_queryset.py
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Expand Up @@ -945,6 +945,12 @@ class Blog(Document):
Blog.objects.insert(Blog(title=blog2.title))

self.assertEqual(Blog.objects.count(), 2)
# pass ordered=False
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Blog.drop_collection()
Blog.objects.insert([
Blog(title="foo", posts=[post1, post2]),
Blog(title="bar", posts=[post2, post3])
], ordered=False)

def test_bulk_insert_different_class_fails(self):
class Blog(Document):
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