2 days old to Mongo, so bear with me.
I have a collection from which, I only want to retrieve specific values contingent to another key existing in the MongoDB environment.
Here is what I am doing:
db.results.find({'someKeyThatShouldExist':{$exists:true}}, {"parentKey.childKey.theKeyWoseValueIwant":1}
This yields data in the following format for me:
{ "_id" : ObjectId("532a2c2b6803fa486b8b456a"), "parentKey" : { "childKey" : { "theKeyWhoseValueIWant" : 102982577 }}}.....
Now, all I really want is the value 102982577, not everything else.
How can I do this ?
You can suppress the _id by adding _id:0 to the projection criteria.
db.results.find(
{"someKeyThatShouldExist":{$exists:true}},
{_id:0, "parentKey.childKey.theKeyWoseValueIwant":1}
)
To get just the value, you could do something like:
db.results.find(
{"someKeyThatShouldExist":{$exists:true}},
{_id:0, "parentKey.childKey.theKeyWoseValueIwant":1}
)[0].parentKey.childKey.theKeyWoseValueIwant
Related
I'm trying to order my collection using the following query:
db.getCollection('trip').find().sort({'itinerary.0.timestamp': 1})
The result is not being correctly sorted, however I exported the full collection to a local mongoDB database and the same query works like a charm. In order to perform that sort in cosmos DB I had to create the index 'itinerary.0.timestamp'.
data example:
{
"_id" : ObjectId("6087104ca68f171ce7715448"),
"tripId" : NumberLong(38533184),
"itinerary" : [
{
"transId" : NumberLong(39800097),
"timestamp" : NumberLong(1619372446291)
},
{
"transId" : NumberLong(39800576),
"timestamp" : NumberLong(1619372446321)
},
],
"results" : [],
"tripTimeSent" : ISODate("2021-04-29T14:44:53.253Z")
}
What am I missing?
Thanks!!
The solution was to create a new field, itiTimestamp, outside the array containing the value 'itinerary.0.timestamp'. Then just order by itiTimestamp
It's true that you need to create an index for the sort field. Here's the doc related:
To apply a sort to a query, you must create an index on the fields
used in the sort operation.
==========================================
I've tested in my side, after creating wildcard index on itinerary, sort query could be executed but has no luck. I also refer to this answer(new BasicDBObject("labels.0.value", 1)) and this one(db.testCollection.find().sort({"someArray.0": 1})), they all don't work for the date format Op provided.
But when I added a properity "score":[20,55,80] in each item in the collection, I found it can be sorted by the first item when sort by score directly.
I assume that this feature hasn't supported.
I'm struggling with a seemingly simple query in mongodb.
I have a job collection that has objects like:
{
"_id" : ObjectId("5995c1fc3c2a353a782ee51b"),
"stages" : [
{
"start" : ISODate("2017-02-02T22:06:26Z"),
"end" : ISODate("2017-02-03T22:06:26Z"),
"name" : "stage_one"
},
{
"start" : ISODate("2017-02-03T22:06:26Z"),
"end" : ISODate("2017-02-07T20:34:01Z"),
"name" : "stage_two"
}
]
}
I want to get a job whose second stage does not have an end time, i.e. end is null or not defined.
According to the mongo docs on querying for null and querying an array of embedded documents, it would seem the correct query should be:
db.job.findOne({'stages.1.end': null})
However, running that query returns me the job above which does have a non-null end date. In fact, if I run the query with count instead of findOne, I see that all jobs are returned - no filtering is done at all.
For completeness, here is the output from an example on a fresh mongo instance:
So in this example, I would expect db.job.findOne({'stages.1.end': null}) to return nothing since there is only one document and its second stage has a non-null end date.
This feels like the sort of issue where it's just me being an idiot and if so, I apologise profusely.
Thanks in advance for your help and let me know if you need any more details!
EDIT:
After some more experimentation, I think I can achieve what I want with the following:
db.job.find({$or: [{'stages.1.end': { $type: 10 }}, {'stages.1.end': {$exists: false}}]})
While this gets the job done, it doesn't feel like the simplest way and I still don't understand why the original query doesn't work. If anyone could shed some light on this it'd be much appreciated.
Our (edX) original Mongo persistence representation uses a bson dictionary (aka object or subdocument) as the _id value (see, mongo/base.py). This id is missing a field.
Can some documents' _id values have more subfields than others without totally screwing up indexing?
What's the best way to handle existing documents without the additional field? Remove and replace them? Try to query w/ new _id format and if fails fall over to query w/o the new field? Try to query with both new and old _id format in one query?
To be more specific, the current format is
{'_id': {
'tag': 'i4x', // yeah, it's always this fixed value
'org': your_school_x,
'course': a_catalog_number,
'category': the_xblock_type,
'name': uniquifier_within_course
}}
I need to add 'run': the_session_or_term_for_course_run or 'course_id': org/course/run.
Documents within a collection need not have values for _id that are of the same structure. Hence, it is perfectly acceptable to have the following documents within a collection:
> db.foo.find()
{ "_id" : { "a" : 1 } }
{ "_id" : { "a" : 1, "b" : 2 } }
{ "_id" : { "c" : 1, "b" : 2 } }
Note that because the index is on only _id, only queries that specify a value for _id will use the index:
db.foo.find({_id:1}) // will use the index on _id
db.foo.find({_id:{state:"Alaska"}) // will use the index on _id
db.foo.find({"_id.a":1}) // will NOT use the index on _id
Note also that only a complete match of the "value" of _id will return a document. So this returns no documents for the collection above:
db.foo.find({_id:{c:1}})
Hence, for your case, you are welcome to add fields to the object that is the value for the _id key. And it does not matter that all documents have a different structure. But if you are hoping to query the collection by_id and have it be efficient, you are going to need to add indexes for all relevant sub parts that might be used in isolation. That is not super efficient.
_id is no different than any other key in this regard.
It seems that i can go further than one subdocument if i want to add it dynamicaly, here is the code:
db.users.update({"pup.cmn.id":id}, {"$addToSet":{"pup.cmn":{"abus":email}}})
this give error:
OperationFailure: can't append to array using string field name: cmn
then, if i add positional element i get this:
db.users.update({"pup.cmn.id":id}, {"$addToSet":{"pup.$.cmn":{"abus":email}}})
"cmn" :
[
{
"fto" : ObjectId("5190e8a53a5f3a0c102af045")
"id" : "14.05.2013 12:29:53"
},
{
"abus" : "u...#example.com"
}
]
so as you can see, it will add it in the same level, and i dont want that, because the application will get errors.
It seems that Mongodb for the time of writing (2.4.x) have not this feature, there is a ticket:
https://jira.mongodb.org/browse/SERVER-831
How to get position (index) of selected document in mongo collection?
E.g.
this document: db.myCollection.find({"id":12345})
has index 3 in myCollection
myCollection:
id: 12340, name: 'G'
id: 12343, name: 'V'
id: 12345, name: 'A'
id: 12348, name: 'N'
If your requirement is to find the position of the document irrespective of any order, that is not
possible as MongoDb does not store the documents in specific order.
However,if you want to know the index based on some field, say _id , you can use this method.
If you are strictly following auto increments in your _id field. You can count all the documents
that have value less than that _id, say n , then n + 1 would be index of the document based on _id.
n = db.myCollection.find({"id": { "$lt" : 12345}}).count() ;
This would also be valid if documents are deleted from the collection.
As far as I know, there is no single command to do this, and this is impossible in general case (see Derick's answer). However, using count() for a query done on an ordered id value field seems to work. Warning: this assumes that there is a reliably ordered field, which is difficult to achieve in a concurrent writer case. In this example _id is used, however this will only work with a single writer case.:
MongoDB shell version: 2.0.1
connecting to: test
> use so_test
switched to db so_test
> db.example.insert({name: 'A'})
> db.example.insert({name: 'B'})
> db.example.insert({name: 'C'})
> db.example.insert({name: 'D'})
> db.example.insert({name: 'E'})
> db.example.insert({name: 'F'})
> db.example.find()
{ "_id" : ObjectId("4fc5f040fb359c680edf1a7b"), "name" : "A" }
{ "_id" : ObjectId("4fc5f046fb359c680edf1a7c"), "name" : "B" }
{ "_id" : ObjectId("4fc5f04afb359c680edf1a7d"), "name" : "C" }
{ "_id" : ObjectId("4fc5f04dfb359c680edf1a7e"), "name" : "D" }
{ "_id" : ObjectId("4fc5f050fb359c680edf1a7f"), "name" : "E" }
{ "_id" : ObjectId("4fc5f053fb359c680edf1a80"), "name" : "F" }
> db.example.find({_id: ObjectId("4fc5f050fb359c680edf1a7f")})
{ "_id" : ObjectId("4fc5f050fb359c680edf1a7f"), "name" : "E" }
> db.example.find({_id: {$lte: ObjectId("4fc5f050fb359c680edf1a7f")}}).count()
5
>
This should also be fairly fast if the queried field is indexed. The example is in mongo shell, but count() should be available in all driver libs as well.
This might be very slow but straightforward method. Here you can pass as usual query. Just I am looping all the documents and checking if condition to match the record. Here I am checking with _id field. You can use any other single field or multiple fields to check it.
var docIndex = 0;
db.url_list.find({},{"_id":1}).forEach(function(doc){
docIndex++;
if("5801ed58a8242ba30e8b46fa"==doc["_id"]){
print('document position is...' + docIndex);
return false;
}
});
There is no way that MongoDB can return this as it does not keep documents in order in the database, just like MySQL f.e. doesn't name row numbers.
The ObjectID trick from jhonkola will only work if only one client creates new elements, as the ObjectIDs are generated on the client side, with the first part being a timestamp. There is no guaranteed order if different clients talk to the same server. Still, I would not rely on this.
I also don't quite understand what you are trying to do though, so perhaps mention that in your question? I can then update the answer.
Restructure your collection to include the position of any entry i.e {'id': 12340, 'name': 'G', 'position': 1} then when searching the database collection(myCollection) using the desired position as a query
The queries I use that return the entire collection all use sort to get a reproducible order, find.sort.forEach works with the script above to get the correct index.