Pymongo find by _id in subdocuments

Question

Assuming that this one item of my database:

{"_id" : ObjectID("526fdde0ef501a7b0a51270e"),
  "info": "foo",
  "status": true,
  "subitems : [ {"subitem_id" : ObjectID("65sfdde0ef501a7b0a51e270"),
                 //more},
                {....}
              ],
  //more
}

I want to find (or find_one, doesn't matter) the document(s) with "subitems.subitem_id" : xxx.

I have tried the following. All of them return an empty list.

from pymongo import MongoClient,errors
from bson.objectid import ObjectId

id = '65sfdde0ef501a7b0a51e270'

db.col.find({"subitems.subitem_id" : id } ) #obviously wrong
db.col.find({"subitems.subitem_id" : Objectid(id) })
db.col.find({"subitems.subitem_id" : {"$oid":id} })
db.col.find({"subitems.subitem_id.$oid" : id })
db.col.find({"subitems.$.subitem_id" : Objectid(id) })

In mongoshell this one works however:

find({"subitems.subitem_id" : { "$oid" : "65sfdde0ef501a7b0a51e270" } })

Rafa Viotti · Answer 1 · 2013-11-02T02:17:27.687

4

The literal 65sfdde0ef501a7b0a51e270 is not hexadecimal, hence, not a valid ObjectId.

Also, id is a Python built-in function. Avoid reseting it.

Finally, you execute a find but do not evaluate it, so you do not see any results. Remember that pymongo cursors are lazy.

Try this.

from pymongo import MongoClient
from bson.objectid import ObjectId

db = MongoClient().database
oid = '65cfdde0ef501a7b0a51e270'

x = db.col.find({"subitems.subitem_id" : ObjectId(oid)})

print list(x)

Notice I adjusted oid to a valid hexadecimal string.

Same query in the Mongo JavaScript shell.

db.col.find({"subitems.subitem_id" : new ObjectId("65cfdde0ef501a7b0a51e270")})

edited Nov 02 '13 at 02:17

answered Nov 01 '13 at 17:36

Rafa Viotti

9,998
4
42
62

There are big difference between id and id(). I don't see any problems specifyin `id = 5` – nickzam Nov 01 '13 at 17:51
If you do `id = 5`, the `id()` built-in is effectively hidden in that scope. See http://stackoverflow.com/questions/77552. – Rafa Viotti Nov 02 '13 at 02:04

nickzam · Accepted Answer · 2013-11-01T21:14:24.153

2

Double checked. Right answer is db.col.find({"subitems.subitem_id" : Objectid(id)}) Be aware that this query will return full record, not just matching part of sub-array.

Mongo shell:

a = ObjectId("5273e7d989800e7f4959526a")
db.m.insert({"subitems": [{"subitem_id":a},
                          {"subitem_id":ObjectId()}]})
db.m.insert({"subitems": [{"subitem_id":ObjectId()},
                          {"subitem_id":ObjectId()}]})
db.m.find({"subitems.subitem_id" : a })

>>> { "_id" : ObjectId("5273e8e189800e7f4959526d"), 
"subitems" : 
[
 {"subitem_id" : ObjectId("5273e7d989800e7f4959526a") },    
 {"subitem_id" : ObjectId("5273e8e189800e7f4959526c")} 
]}

edited Nov 01 '13 at 21:14

answered Nov 01 '13 at 16:43

nickzam

793
4
8

The quality of answers gets improved a lot when given additional explanations. – matthias krull Nov 01 '13 at 17:43
1

Improved and added examples. – nickzam Nov 01 '13 at 21:15
Notice the typographical errors in this answer, and on the question. Where you see Objectid, or ObjectID, it should read ObjectId. Capital letter I and lowed case letter d. Both in JavaScript and Python. – Rafa Viotti Nov 02 '13 at 03:16
There are no errors in answer. It's tested in Mongo shell and copy-pasted here. Fixed typos in question. – nickzam Nov 02 '13 at 11:34

Pymongo find by _id in subdocuments

2 Answers2