I'm doing a simple insert into Mongo...
db.notes.insert({ title: "title", details: "note details"})
After the note document is inserted, I need to get the object id immediately. The result that comes back from the insert has some basic info regarding connection and errors, but no document and field info.
I found some info about using the update() function with upsert=true, I'm just not sure if that's the right way to go, I have not yet tried it.
One of the cool things about MongoDB is that the ids are generated client side.
This means you don't even have to ask the server what the id was, because you told it what to save in the first place. Using pymongo the return value of an insert will be the object id. Check it out:
>>> import pymongo
>>> collection = pymongo.Connection()['test']['tyler']
>>> _id = collection.insert({"name": "tyler"})
>>> print _id.inserted_id
4f0b2f55096f7622f6000000
The answer from Tyler does not work for me.
Using _id.inserted_id works
>>> import pymongo
>>> collection = pymongo.Connection()['test']['tyler']
>>> _id = collection.insert({"name": "tyler"})
>>> print(_id)
<pymongo.results.InsertOneResult object at 0x0A7EABCD>
>>> print(_id.inserted_id)
5acf02400000000968ba447f
It's better to use insert_one() or insert_many() instead of insert(). Those two are for the newer version. You can use inserted_id to get the id.
myclient = pymongo.MongoClient("mongodb://localhost:27017/")
myDB = myclient["myDB"]
userTable = myDB["Users"]
userDict={"name": "tyler"}
_id = userTable.insert_one(userDict).inserted_id
print(_id)
Or
result = userTable.insert_one(userDict)
print(result.inserted_id)
print(result.acknowledged)
If you need to use insert(), you should write like the lines below
_id = userTable.insert(userDict)
print(_id)
Newer PyMongo versions depreciate insert, and instead insert_one or insert_many should be used. These functions return a pymongo.results.InsertOneResult or pymongo.results.InsertManyResult object.
With these objects you can use the .inserted_id and .inserted_ids properties respectively to get the inserted object ids.
See this link for more info on insert_one and insert_many and this link for more info on pymongo.results.InsertOneResult.
updated; removed previous because it wasn't correct
It looks like you can also do it with db.notes.save(...), which returns the _id after it performs the insert.
See for more info:
http://api.mongodb.org/python/current/api/pymongo/collection.html
some_var = db.notes.insert({ title: "title", details: "note details"})
print(some_var.inserted_id)
You just need to assigne it to some variable:
someVar = db.notes.insert({ title: "title", details: "note details"})
To get the ID after an Insert in Python, just do like this:
doc = db.notes.insert({ title: "title", details: "note details"})
return str(doc.inserted_id) # This is to convert the ObjectID (type of doc.inserted_id into string)
Related
I can query my MongoDB and I see the "_id" value like this:
"_id" : BinData(3,"sFgVQWMKzUiWl5dql62j2g==")
Using Flask 0.10.1 and PyMongo 3.0.3 I attempt to "find_one" like this:
record = db.collection.find_one({'_id': ObjectId("sFgVQWMKzUiWl5dql62j2g==")})
And I get this error:
bson.errors.InvalidId: 'sFgVQWMKzUiWl5dql62j2g==' is not a valid ObjectId, it must be a 12-byte input or a 24-character hex string
Any help would be appreciated.
Try this:
record = db.collection.find_one({'_id': BinData(3, "sFgVQWMKzUiWl5dql62j2g==")})
Ok here is how I got this working. From my PyMongo client I queried the result and got this:
_id: "497ffaf0-5ed3-3a4e-99ae-6b5c5f9b431e"
Note how that is not like the MongoDB console client which returns this:
_id : BinData(3,"sFgVQWMKzUiWl5dql62j2g==")
So seeing that the value looked like a GUID I started digging around and found that is a UUID and with that I did this:
import uuid
record = db.collection.find_one({'_id': uuid.UUID("497ffaf0-5ed3-3a4e-99ae-6b5c5f9b431e")})
This worked! Hazzah.
You are storing _id as bindata and trying to retrive it as ObjectId,
instead of ObjectId.
First you need to convert your base64string to binary and then try to search.
bi = binary.Binary("sFgVQWMKzUiWl5dql62j2g==");
record = db.collection.find_one({'_id': bi});
This will work for you.
It will convert your id to binary then compare to get result.
Using Flask, MongoEngine I also was receiving errors searching by _id: mongoengine.errors.ValidationError: "DBRef('m_turk', ObjectId('5966b478b929570647f51a5c'))" is not a valid ObjectId, it must be a 12-byte input or a 24-character hex string
In my case, the object id from MongoEngine.Document is of type bson.dbref.DBREF, whereas MongoEngine.Document expects type bson.objectid.ObjectId, despite returning bson.dbref.DBREF on MongoEngine.Document.objects(...).*. I'm not sure if this is an API inconsistency or simply my fault.
The fix was:
from flask_mongoengine import MongoEngine
from bson.objectid import ObjectId
... set Flask up ...
db = MongoEnginee(app)
class MiniFoo(db.Document):
foo = db.StringField()
class MyDoc(db.Docment):
key = db.StringField()
myself = db.ReferenceField(MiniFoo)
other_foos = db.ReferenceField(MiniFoo)
myobj = MyDoc.objects.get(key="over the rainbow")
# want to find other foos!
# Will fail with exception above
try:
MiniFoo.objects.with_id(myobj.other_foos[0].id)
except:
MiniFoo.objects.with_id(ObjectId(obj.id))
with MongoEngine
from flask_mongoengine import MongoEngine
from bson.objectid import ObjectId
db = MongoEngine()
db.init_app(app)
class Model(db.Document):
_id = db.ObjectIdField()
name = db.StringField(required=True)
#classmethod
def findById(cls, _id: str):
return cls.objects(_id=ObjectId(_id)).first()
modelById = Model.findById('601ec2d13ad7559bf7ebad76')
When I use mongo-ruby-driver and I insert new document it returns generated '_id':
db = MongoClient.new('127.0.0.1', '27017').db('ruby-mongo-examples')
id = db['test'].insert({name: 'example'})
# BSON::ObjectId('54f88b01ab8bae12b2000001')
I'm trying to get the '_id' of a document after doing an insertion using Moped:
db = Moped::Session.new(['127.0.0.1:27017'])
db.use('ruby-mongo-examples')
id = db['coll'].insert({name: 'example'})
# {"connectionId"=>15, "n"=>0, "syncMillis"=>0, "writtenTo"=>nil, "err"=>nil, "ok"=>1.0}
How I get the id using Moped?
Update:
I also try use safe mode but it doesn't work:
db = Moped::Session.new(['127.0.0.1:27017'])
db.use('ruby-mongo-examples')
db.with(safe: true) do |safe|
id = safe['coll'].insert({name: 'example'})
# {"connectionId"=>5, "n"=>0, "syncMillis"=>0, "writtenTo"=>nil, "err"=>nil, "ok"=>1.0}
end
After inserting/saving, the returned object will have a property inserted_id which is a BSON::ObjectId:
# I'm using insert_one
result = safe['coll'].insert_one({name: 'example'})
result.methods.sort # see list of methods/properties
result.inserted_id
result.inserted_id.to_s # convert to string
From this issue:
It would be nice, but unfortunately Mongo doesn't give us anything
back when inserting (since it's fire and forget), and when in safe
mode it still doesn't give the id back if it generated it on the
server. So there really isn't any possible way for us to do this
unless it was a core feature in MongoDB.
Your best bet would be to generate the id before inserting the document:
document = { _id: Moped::BSON::ObjectId.new, name: "example" }
id = document[:_id]
I want to access a document in collection by 'name' attribute for getting its ObjectId so that i can insert that unique objectid to other document for reference.
cursorObject = db.collectionIngredient.find({'name': 'sugar'})
I want _id field of cursorObject.
cursorObject.'_id' or cursorObject._id not working.
I have tried __getitem__, __getattribute__ and so much internet surfing but couldn't able to find a way.
Please help
First, as #jjmartinez pointed out, find returns a cursor, which you need to iterate over, in order to get hold of the documents returned by your query. The _id field belongs to the documents, not the cursor.
I'm guessing that in your case the name is unique, so you can avoid cursor/iterating if you use find_one instead of find. Then you get the document directly.
Then, to access the _id, you just need a standard dict-item access:
id = doc['_id']
So we get:
ingredient = db.collectionIngredient.find_one({'name': 'sugar'})
if ingredient is not None:
id = ingredient['_id']
else:
id = None
When you do cursorObject = db.collectionIngredient.find({'name': 'sugar'}) you have a collection of documents, not a single element. So you need to explore all the collection. You need to iterate inside the cursor:
try:
cursorObject = db.collectionIngredient.find({'name': 'sugar'})
except:
print "Unexpected error:", sys.exc_info()[0]
for doc in cursorObject:
print doc
Here you have the Pymongo Tutorial
I have the following Mongoengine document:
class MyEmbed(EmbeddedDocument):
embedField = StringField(primary_key=True)
varField = StringField()
class TestDoc(Document):
myField = StringField()
embed_list = ListField(EmbeddedDocumentField(MyEmbed))
So I keep a list of embedded documents, to which I wish to add new documents if they don't exist already. The problem is that when I use the atomic update operator add_to_set things don't turn out the way I want them to.
This is what I am trying to do:
embed1 = models.MyEmbed(embedField="F1")
parent = models.TestDoc(myField="ParentField")
embed_list = []
embed_list.append(embed1)
parent.embed_list = embed_list
parent.save()
embed2 = models.MyEmbed(embedField="F1", varField="varField")
TestDoc.objects(id=parent.id).update_one(add_to_set__embed_list=embed2)
The problem is that after doing this, I have in the DB a list of embedded documents with 2 elements. And what I want is to decide upon one field (embedField in this case) whether 2 EmbeddedDocuments are equal or not, and not by taking into account all the properties. My questions are:
What are the default criteria according to which Mongoengine decides whether 2 EmbeddedDocuments are equal or not?
How can I redefine the function that makes Mongoengine decide when 2 EmbeddedDocuments are equal or not?
Thanks!
The actual checking is done inside MongoDB and not mongoengine.
The object sent to mongodb should be the same, but this is where it gets tricky as with BSON order is important and in python with dictionaries its not. When converting to send to mongodb mongoengine just passes a dictionary. This is a bug - so I've added #296 and will fix for 0.8
See https://github.com/MongoEngine/mongoengine/blob/master/mongoengine/document.py#L51 and https://github.com/MongoEngine/mongoengine/blob/master/mongoengine/base/document.py#L52:
def __eq__(self, other):
if isinstance(other, self.__class__):
return self._data == other._data
return False
It compare dicts of Embedded documents data. So you can override this method.
If you look at Document update that calls QuerySet update (find add_to_set and addToSet) you can find that mongoengine doesnt't check exists document in list and just call mongo $addToSet operation: https://github.com/MongoEngine/mongoengine/blob/master/mongoengine/queryset/transform.py#L156.
In your code you have document MyEmbed(embedField="F1") and try add another document MyEmbed(embedField="F1", varField="varField") so logic right: it add new document. If you try next code:
embed1 = models.MyEmbed(embedField="F1")
parent = models.TestDoc(myField="ParentField")
embed_list = []
embed_list.append(embed1)
parent.embed_list = embed_list
parent.save()
embed2 = models.MyEmbed(embedField="F1", varField="varField")
TestDoc.objects(id=parent.id).update_one(add_to_set__embed_list=embed2)
embed3 = models.MyEmbed(embedField="F1")
TestDoc.objects(id=parent.id).update_one(add_to_set__embed_list=embed3)
embed4 = models.MyEmbed(embedField="F1", varField="varField")
TestDoc.objects(id=parent.id).update_one(add_to_set__embed_list=embed4)
you can find that parent contains only embed1 and embed2.
So, to resolve you problem you can override __eq__ method and check document in list, but you must find another solution for update document list, because it have direct call of mongo method.
I have not clearly issue during query using two criterials like Id and Other. I use a Repository storing some data like id,iso,value. I have created an index("_id","Iso") to performs queries but queries are only returning my cursor if i use only one criterial like _id, but is returning nothing if a use two (_id, Iso) (commented code).
Are the index affecting the response or the query method are failing?
use :v1.6.5 and C# official.
Sample.
//Getting Data
public List<BsonObject> Get_object(string ID, string Iso)
{
using (var helper = BsonHelper.Create())
{
//helper.Db.Repository.EnsureIndex("_Id","Iso");
var query = Query.EQ("_Id", ID);
//if (!String.IsNullOrEmpty(Iso))
// query = Query.And(query, Query.EQ("Iso", Iso));
var cursor = helper.Db.Repository.FindAs<BsonObject>(query);
return cursor.ToList();
}
}
Data:
{
"_id": "2345019",
"Iso": "UK",
"Data": "Some data"
}
After that I have Updated my data using Update.Set() methods. I can see the changed data using MongoView. The new data are correct but the query is always returning the sames olds values. To see these values i use a page that can eventually cached, but if add a timestamp at end are not changing anything, page is always returning the same olds data. Your comments are welcome, thanks.
I do not recall offhand how the C# driver creates indexes, but the shell command for creating an index is like this:
db.things.ensureIndex({j:1});
Notice the '1' which is like saying 'true'.
In your code, you have:
helper.Db.Repository.EnsureIndex("_Id","Iso");
Perhaps it should be:
helper.Db.Repository.EnsureIndex("_Id", 1);
helper.Db.Repository.EnsureIndex("Iso", 1);
It could also be related to the fact that you are creating indexes on "_Id" and the actual id field is called "_id" ... MongoDB is case sensitive.
Have a quick look through the index documentation: http://www.mongodb.org/display/DOCS/Indexes