My Scenario : I need to join 2 below tables in Mongo DB and condition is
testScenarioId(table 1) = _id (table 2)
Table 1:
{
"_id" : ObjectId("58516a6838fdb54d744ba070"),
"_class" : "com.TestResults",
"testScenarioId" : "581cef861892ad1eb7d124dd",
"runId" : 314,
"status" : "passed"
}
Table 2:
{
"_id" : ObjectId("57f41cb9319ed34079df8a2d"),
"environment" : "STAGE",
"component" : "platform",
"scenarioName" : "ABC-1234",
}
i am able to do if i am joining with same local field and foreign field but not on the above case.
Mongodb does not support type coercion in $lookup. So field of type ObjectId can not be looked up with a string type foreign field.
What you need to do is while saving the testScenarioId, you need to store as objectId.
I tried using $type in aggregation but its not supported. So currently here is no way to do it directly in aggregation pipeline.
If you want to implement join in 2 collections, then you will be insert "testScenarioId" in ObjectId form.
At that time,you have insert id in string form and "lookup" Aggregation does not support this form of Id.
Reason why ObjectId: when query finds Id from first table (table 1), they will get id in ObjectId form, and after then they will compare id with second table parameter "testScenarioId" which was store in string form, and they will not match and query return null data.
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 am new to querying dbs and especially mongodb.If I run :
db.<customers>.find({"contact_name: Anny Hatte"})
I get:
{
"_id" : ObjectId("55f7076079cebe83d0b3cffd"),
"company_name" : "Gap",
"contact_name" : "Anny Hatte",
"email" : "ahatte#gmail.com"
}
I wish to get the value of the "_id" attribute from this query result. How do I achieve that?
Similarly, if I have another collection, named items, with the following data:
{
"_id" : ObjectId("55f7076079cebe83d0b3d009"),
"_customer" : ObjectId("55f7076079cebe83d0b3cfda"),
"school" : "St. Patrick's"
}
Here, the "_customer" field is the "_id" of the customer collection (the previous collection). I wish to get the "_id", the "_customer" and the "school" field values for the record where "_customer" of items-collection equals "_id" of customers-collection.
How do I go about this?
I wish to get the value of the "_id" attribute from this query result.
How do I achieve that?
The find() method returns a cursor to the results, which you can iterate and retrieve the documents in the result set. You can do this using forEach().
var cursor = db.customers.find({"contact_name: Anny Hatte"});
cursor.forEach(function(customer){
//access all the attributes of the document here
var id = customer._id;
})
You could make use of the aggregation pipeline's $lookup stage that has been introduced as part of 3.2, to look up and fetch the matching rows in some other related collection.
db.customers.aggregate([
{$match:{"contact_name":"Anny Hatte"}},
{$lookup:{
"from":"items",
"localField":"_id",
"foreignField":"_customer",
"as":"items"
}}
])
In case you are using a previous version of mongodb where the stage is not supported, then, you would need to fire an extra query to lookup the items collection, for each customer.
db.customers.find(
{"contact_name":"Anny Hatte"}).map(function(customer){
customer["items"] = [];
db.items.find({"_customer":customer._id}).forEach(function(item){
customer.items.push(item);
})
return customer;
})
Say I run a Mongo aggregate query and get a result set back, containing documents in the form,
{ user_id : 1234, value : 678 }
Is it possible to take the returned documents from the aggregation query and match the user_id to documents in a user collection then merge the result. i.e Match the above document to a user record with the form,
{_id : 1234, name : bob knight, email : bob#bob }
and return
{_id : 1234, name : bob knight, email : bob#bob, value : 678 }
Assuming either there is one value doc per user or you want to add them if there are multiple, you can do this:
db.collection.aggregate(
{$project:{user_id:{$ifNull:["$user_id","$_id"]}, email:1,name:1,value:1}},
{$group:{_id:"$user_id",name:{$max:"$name"}, email:{$max:"$email"},value:{$sum:"$value"}}}
)
$project needs to add user_id field into the documents with _id, hence the $ifNull.
$group just groups them, summing value works because null is 0, and $max works for the other fields because any value is > null.
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.
TENANT
{ "_ID" : 11, NAME : "ruben", OPERATION :[{OPERATION_ID: 100, NAME : "Check"}] }
how to set the OPERATION_ID has unique to avoid duplicate values and to avoid null values like primary key?
When you want the OPERATION_IDs to be unique for all tenants, then you can do it like that:
db.tenants.ensureIndex( { operation.OPERATION_ID : 1 }, { unique:true, sparse:true } );
When you want the OPERATION_IDs to be unique per tenant, so that two tenants can both have the operation_ID:100 but no tenant can have operation_id:100 twice, you have to add the _id of the tenant to the index so that any given combination of _id and operation_id is unique.
db.tenants.ensureIndex( { _id: 1, operation.OPERATION_ID : 1 }, { unique:true, sparse:true } );
Adding a unique index on OPERATION.OPERATION_ID will ensure that no two distinct documents will contain an element in OPERATION with the same OPERATION_ID.
If you want to prevent a single document from having two elements in OPERATION with the same OPERATION_ID, you can't use unique indexes; you will have to use set update operators (such as $set and $addToSet). You could turn OPERATION into a subdocument keyed by OPERATION_ID, like so:
{ "_ID" : 11, NAME : "ruben", OPERATION : {"100" : {NAME : "Check"} }}
Then you can enforce uniqueness by issuing updates with $set; for example:
db.<collection>.update({NAME: "ruben"}, {$set: {"OPERATION.100.NAME": "Uncheck"}})
Regarding null values: MongoDB doesn't feature non-null constraints on fields (it doesn't even force a given field to have a single type), so you will have to ensure in your application that null values aren't inserted.