I have 2 collection.
Collection "users"
{
"_id" : ObjectId("54b00098e0fdb6634b1f54e6"),
"state" : "active",
"backends" : [
DBRef("backends", ObjectId("54b001ebe0fd853df1c93419")),
DBRef("backends", ObjectId("54b00284e0fd853df1c9341b"))
]
}
Collection "backends"
{
"_id" : ObjectId("54b001ebe0fd853df1c93419"),
"state" : "running"
}
I want to get a list of backend of a user where the backend's state is "running".
How can mongodb do this like join two table?
Is it any method to search backward from backend or have function the filter?
I can search like this
db.users.find({"backends.$id" : "distring"})
But what if I want to search the state inside backend object? like.
db.users.find({"backends.$state" : "running"})
But ofcoure it is not working.
MongoDB doesn't support joins so you need to do this in two steps. In the shell:
var ids = db.backends.find({state: 'running'}, {_id: 1}).map(function(backend) {
return backend._id;
});
var users = db.users.find({'backends.$id': {$in: ids}}).toArray();
On a side note, you're probably better off using a plain ObjectId instead of a DBRef for the backends array elements unless the ids in that array can actually refer to docs in multiple collections.
Related
I'm new to MongoDB and have to work on a legacy project that I didn't create... and I'm struggling!
I need to reset some documents in one of my collections, based on a particular field value. I have had some success with this so far, but some of the data I need to update is within nested arrays in that document, and I can't work that part out.
Below is an example of one document in the collection:
{
"_id" : ObjectId("1234567890"),
"currentStatus" : "approved",
"itemsInstalled" : [
{
"installDate" : ISODate("2017-04-18T00:00:00.000Z"),
"_id" : ObjectId("1234567890"),
"status" : "approved"
},
{
"installDate" : ISODate("2017-04-18T00:00:00.000Z"),
"_id" : ObjectId("0987654321"),
"status" : "approved"
}
],
"__v" : 5005,
"approvalDate" : ISODate("2017-12-04T10:40:01.580Z"),
"approvedBy" : "automatic"
}
I need to update every document in my collection where the approvedBy field is set to automatic, and leave the others untouched.
For the documents I am updating, I need to remove the approvalDate and approvedBy fields completely, change the currentStatus field to action and change every item in the itemsInstalled array to pending. Everything else can stay as it is.
This is something I would persist with solving myself if I had more time.
Unfortunately, to my knowledge, you can't update multiple array elements. Your best bet is probably to use forEach. You can accomplish what you want with something like this:
db.your_collection.find({
approvedBy: "automatic"
}).forEach(function(doc) {
for(var i = 0; i < doc.itemsInstalled.length; i++) {
doc.itemsInstalled[i].status = "pending";
}
doc.currentStatus = "action";
delete doc.approvedBy;
delete doc.approvalDate;
db.your_collection.update({_id: doc._id}, doc);
});
Using forEach, you can update all of the array elements at once. The downside is that you will be performing multiple update queries, so you should be careful about doing this on particularly large collections or as part of your application logic. Ideally this should be a one-time use scenario.
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;
})
After few months working with Mongo trying to understand if using sub-documents for nested data is good or not, especially in this example:
Assume users collection that each document into it have the following:
{
"_id" : ObjectId("some valid Object ID"),
"userName" : "xxxxx",
"email" : "xxxx#xxxx.xx"
}
Now, in my system there are also rooms (another collection) and i want to save for each user scores per room.
In my mind, to do that i have 2 major options, (1) create new collection call userScores that will hold: userId, roomId, scores fields like i did previously in MySql and other relational DB's (2) create a sub-document into the above user document:
{
"_id" : ObjectId("sdfdfdfdfdf"),
"userName" : "xxxxx",
"email" : "xxxx#xxxx.xx",
"scores": {
"roomIdX": 50,
"roomIdY": 50,
"roomIdZ": 50
}
}
What do you think is better way so later i can handle searches, aggregations and other data queries via the code (mongoose in my case)
Thanks.
I'm using passport.js to store my users into my mongodb. A user object looks like this
{
"_id" : ObjectId("54893faf0907a100006341ee"),
"local" : {
"password" : [encrypted password],
"email" : "johnsmith#domain.com"
},
"__v" : 0
}
In a mongodb shell how would I go about listing all the emails? I'm finding it difficult to do this as my data sits two level deep within the object. Cheers!
You can use distinct to get a list of a field's distinct values in the collection, using dot notation to reference the embedded field:
db.users.distinct('local.email')
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.