How to update a field value globally in mongodb - mongodb

Is there a easier way to update field text in mongodb globally?
I don't have a mongodb knowledge but some thing like this,
Update "all collections that has field1" set field1=some where field1=1234
Please take a look at mysql solution using db dumps:
Similar to Find and replace in entire mysql database
If the above is not possible what is the best way to go about writing a one time script to migrate data in mongodb? This is in linux environment.

To update multiple documents in the same collection, use the multi:true option when you do an update:
db.collection.update(
{ "field1": 1234 },
{ "$set": { "field1" : somevalue } },
{ multi: true }
);
MongoDB has no commands which affect more than one collection at a time, so you will have to execute this for every collection separately. When you want to do this in the shell, you can perform the command on every collection separately with a script like this:
db.getCollectionNames().forEach( function(name) {
db[name].update(
{ "field1": 1234 },
{ "$set": { "field1" : somevalue } },
{ multi: true }
);
});

Related

MongoDB: Updating a document with existing fields using aggregation

I am new to mongodb, I am learning from some Udemy courses and I want to know how I can update a document existing field without overwriting it.
I have the following collection with these documents:
enter image description here
I want to add new warehouses in the "item":"drafts" within the stock field.
What I am trying is:
enter image description here
And giving the output it seems that is working, but when I do again db.matrices.find(), what I get is the exactly same output that in the first image.
How can I update it? I have tried also the update method, but does not do what I want to do.
Thanks!
PD: I am using linux mint, with mongo v5.0.3, and mongosh v1.1.1
You are using the aggregate pipeline, this does not update the document in the DB, it just retrieves the result. starting in Mongo version 4.2+ you can now use an aggregation pipeline ( with some limitations ) to update a document, like so:
db.collection.updateOne({
item: "drafts"
},
[
{
$set: {
stock: {
$concatArrays: [
"$stock",
[
{
"warehouse": "A",
qty: 20
}
]
]
}
}
}
])
Mongo Playground
I will just say that this specific update is very simple, there is no need to use an aggregation pipeline for it. a simple use of the $push operator in the update field will suffice in a "normal" update:
db.collection.updateOne({
item: "drafts"
},
{
$push: {
stock: {
"warehouse": "A",
qty: 20
}
}
})
Mongo Playground

Can I update the exsisting record of mongodb by its id? [duplicate]

I want update an _id field of one document. I know it's not really good practice. But for some technical reason, I need to update it.
If I try to update it I get:
db.clients.update({ _id: ObjectId("123")}, { $set: { _id: ObjectId("456")}})
Performing an update on the path '_id' would modify the immutable field '_id'
And the update is rejected. How I can update it?
You cannot update it. You'll have to save the document using a new _id, and then remove the old document.
// store the document in a variable
doc = db.clients.findOne({_id: ObjectId("4cc45467c55f4d2d2a000002")})
// set a new _id on the document
doc._id = ObjectId("4c8a331bda76c559ef000004")
// insert the document, using the new _id
db.clients.insert(doc)
// remove the document with the old _id
db.clients.remove({_id: ObjectId("4cc45467c55f4d2d2a000002")})
To do it for your whole collection you can also use a loop (based on Niels example):
db.status.find().forEach(function(doc){
doc._id=doc.UserId; db.status_new.insert(doc);
});
db.status_new.renameCollection("status", true);
In this case UserId was the new ID I wanted to use
In case, you want to rename _id in same collection (for instance, if you want to prefix some _ids):
db.someCollection.find().snapshot().forEach(function(doc) {
if (doc._id.indexOf("2019:") != 0) {
print("Processing: " + doc._id);
var oldDocId = doc._id;
doc._id = "2019:" + doc._id;
db.someCollection.insert(doc);
db.someCollection.remove({_id: oldDocId});
}
});
if (doc._id.indexOf("2019:") != 0) {... needed to prevent infinite loop, since forEach picks the inserted docs, even throught .snapshot() method used.
Here I have a solution that avoid multiple requests, for loops and old document removal.
You can easily create a new idea manually using something like:_id:ObjectId()
But knowing Mongo will automatically assign an _id if missing, you can use aggregate to create a $project containing all the fields of your document, but omit the field _id. You can then save it with $out
So if your document is:
{
"_id":ObjectId("5b5ed345cfbce6787588e480"),
"title": "foo",
"description": "bar"
}
Then your query will be:
db.getCollection('myCollection').aggregate([
{$match:
{_id: ObjectId("5b5ed345cfbce6787588e480")}
}
{$project:
{
title: '$title',
description: '$description'
}
},
{$out: 'myCollection'}
])
You can also create a new document from MongoDB compass or using command and set the specific _id value that you want.
As a very small improvement to the above answers i would suggest using
let doc1 = {... doc};
then
db.dyn_user_metricFormulaDefinitions.deleteOne({_id: doc._id});
This way we don't need to create extra variable to hold old _id.
Slightly modified example of #Florent Arlandis above where we insert _id from a different field in a document:
> db.coll.insertOne({ "_id": 1, "item": { "product": { "id": 11 } }, "source": "Good Store" })
{ "acknowledged" : true, "insertedId" : 1 }
> db.coll.aggregate( [ { $set: { _id : "$item.product.id" }}, { $out: "coll" } ]) // inserting _id you want for the current collection
> db.coll.find() // check that _id is changed
{ "_id" : 11, "item" : { "product" : { "id" : 11 } }, "source" : "Good Store" }
Do not use $match filter + $out as in #Florent Arlandis's answer since $out fully remove data in collection before inserting aggregate result, so effectively you will loose all data that don't match to $match filter

Rename a sub-document field within an Array

Considering the document below how can I rename 'techId1' to 'techId'. I've tried different ways and can't get it to work.
{
"_id" : ObjectId("55840f49e0b"),
"__v" : 0,
"accessCard" : "123456789",
"checkouts" : [
{
"user" : ObjectId("5571e7619f"),
"_id" : ObjectId("55840f49e0bf"),
"date" : ISODate("2015-06-19T12:45:52.339Z"),
"techId1" : ObjectId("553d9cbcaf")
},
{
"user" : ObjectId("5571e7619f15"),
"_id" : ObjectId("55880e8ee0bf"),
"date" : ISODate("2015-06-22T13:01:51.672Z"),
"techId1" : ObjectId("55b7db39989")
}
],
"created" : ISODate("2015-06-19T12:47:05.422Z"),
"date" : ISODate("2015-06-19T12:45:52.339Z"),
"location" : ObjectId("55743c8ddbda"),
"model" : "model1",
"order" : ObjectId("55840f49e0bf"),
"rid" : "987654321",
"serialNumber" : "AHSJSHSKSK",
"user" : ObjectId("5571e7619f1"),
"techId" : ObjectId("55b7db399")
}
In mongo console I tried which gives me ok but nothing is actually updated.
collection.update({"checkouts._id":ObjectId("55840f49e0b")},{ $rename: { "techId1": "techId" } });
I also tried this which gives me an error. "cannot use the part (checkouts of checkouts.techId1) to traverse the element"
collection.update({"checkouts._id":ObjectId("55856609e0b")},{ $rename: { "checkouts.techId1": "checkouts.techId" } })
In mongoose I have tried the following.
collection.findByIdAndUpdate(id, { $rename: { "checkouts.techId1": "checkouts.techId" } }, function (err, data) {});
and
collection.update({'checkouts._id': n1._id}, { $rename: { "checkouts.$.techId1": "checkouts.$.techId" } }, function (err, data) {});
Thanks in advance.
You were close at the end, but there are a few things missing. You cannot $rename when using the positional operator, instead you need to $set the new name and $unset the old one. But there is another restriction here as they will both belong to "checkouts" as a parent path in that you cannot do both at the same time.
The other core line in your question is "traverse the element" and that is the one thing you cannot do in updating "all" of the array elements at once. Well, not safely and without possibly overwriting new data coming in anyway.
What you need to do is "iterate" each document and similarly iterate each array member in order to "safely" update. You cannot really iterate just the document and "save" the whole array back with alterations. Certainly not in the case where anything else is actively using the data.
I personally would run this sort of operation in the MongoDB shell if you can, as it is a "one off" ( hopefully ) thing and this saves the overhead of writing other API code. Also we're using the Bulk Operations API here to make this as efficient as possible. With mongoose it takes a bit more digging to implement, but still can be done. But here is the shell listing:
var bulk = db.collection.initializeOrderedBulkOp(),
count = 0;
db.collection.find({ "checkouts.techId1": { "$exists": true } }).forEach(function(doc) {
doc.checkouts.forEach(function(checkout) {
if ( checkout.hasOwnProperty("techId1") ) {
bulk.find({ "_id": doc._id, "checkouts._id": checkout._id }).updateOne({
"$set": { "checkouts.$.techId": checkout.techId1 }
});
bulk.find({ "_id": doc._id, "checkouts._id": checkout._id }).updateOne({
"$unset": { "checkouts.$.techId1": 1 }
});
count += 2;
if ( count % 500 == 0 ) {
bulk.execute();
bulk = db.collection.initializeOrderedBulkOp();
}
}
});
});
if ( count % 500 !== 0 )
bulk.execute();
Since the $set and $unset operations are happening in pairs, we are keeping the total batch size to 1000 operations per execution just to keep memory usage on the client down.
The loop simply looks for documents where the field to be renamed "exists" and then iterates each array element of each document and commits the two changes. As Bulk Operations, these are not sent to the server until the .execute() is called, where also a single response is returned for each call. This saves a lot of traffic.
If you insist on coding with mongoose. Be aware that a .collection acessor is required to get to the Bulk API methods from the core driver, like this:
var bulk = Model.collection.inititializeOrderedBulkOp();
And the only thing that sends to the server is the .execute() method, so this is your only execution callback:
bulk.exectute(function(err,response) {
// code body and async iterator callback here
});
And use async flow control instead of .forEach() such as async.each.
Also, if you do that, then be aware that as a raw driver method not governed by mongoose, you do not get the same database connection awareness as you do with mongoose methods. Unless you know for sure the database connection is already established, it is safter to put this code within an event callback for the server connection:
mongoose.connection.on("connect",function(err) {
// body of code
});
But otherwise those are the only real ( apart from call syntax ) alterations you really need.
This worked for me, I created this query to perform this procedure and I share it, (although I know it is not the most optimized way):
First, make an aggregate that (1) $match the documents that have the checkouts array field with techId1 as one of the keys of each sub-document. (2) $unwind the checkouts field (that deconstructs the array field from the input documents to output a document for each element), (3) adds the techId field (with $addFields), (4) $unset the old techId1 field, (5) $group the documents by _id to have again the checkout sub-documents grouped by its _id, and (6) write the result of these aggregation in a temporal collection (with $out).
const collection = 'yourCollection'
db[collection].aggregate([
{
$match: {
'checkouts.techId1': { '$exists': true }
}
},
{
$unwind: {
path: '$checkouts'
}
},
{
$addFields: {
'checkouts.techId': '$checkouts.techId1'
}
},
{
$project: {
'checkouts.techId1': 0
}
},
{
$group: {
'_id': '$_id',
'checkouts': { $push: { 'techId': '$checkouts.techId' } }
}
},
{
$out: 'temporal'
}
])
Then, you can make another aggregate from this temporal collection to $merge the documents with the modified checkouts field to your original collection.
db.temporal.aggregate([
{
$merge: {
into: collection,
on: "_id",
whenMatched:"merge",
whenNotMatched: "insert"
}
}
])

Spring Data with Mongo: Subdocument name as query parameter

I have the following document structure:
{
"_id": "project1",
"customer": "someDefaultCustomer",
"users": {
"user1": {
"projectRoles": ["CUSTOMER"]
}
}
}
Now i'm going to query all projects with user contain 'user1' with Spring Data Mongo:
#Query("{'users.?1': {$exists : true} }")
Project findUserProject(String login);
The problem is that Spring Data escapes replacement in queries, so i've got the following 'real' query:
o.s.d.m.r.q.StringBasedMongoQuery - Created query { "users.\"user1\"" : { "$exists" : true}}
Is it possible to avoid escaping? Of course, i can create custom query using spring criteria, but i'd like to keep '#Query' approach.
Pass in the full users.user1 as the variable, i.e.:
#Query("{?1: {$exists : true} }")
If you really want to avoid people having to do that then have a helper function that adds the "users." at the start of the string then calls this method.

Is there any equivalent in MongoDB for MS-SQL command 'SET IDENTITY_INSERT tablename OFF'? [duplicate]

I want update an _id field of one document. I know it's not really good practice. But for some technical reason, I need to update it.
If I try to update it I get:
db.clients.update({ _id: ObjectId("123")}, { $set: { _id: ObjectId("456")}})
Performing an update on the path '_id' would modify the immutable field '_id'
And the update is rejected. How I can update it?
You cannot update it. You'll have to save the document using a new _id, and then remove the old document.
// store the document in a variable
doc = db.clients.findOne({_id: ObjectId("4cc45467c55f4d2d2a000002")})
// set a new _id on the document
doc._id = ObjectId("4c8a331bda76c559ef000004")
// insert the document, using the new _id
db.clients.insert(doc)
// remove the document with the old _id
db.clients.remove({_id: ObjectId("4cc45467c55f4d2d2a000002")})
To do it for your whole collection you can also use a loop (based on Niels example):
db.status.find().forEach(function(doc){
doc._id=doc.UserId; db.status_new.insert(doc);
});
db.status_new.renameCollection("status", true);
In this case UserId was the new ID I wanted to use
In case, you want to rename _id in same collection (for instance, if you want to prefix some _ids):
db.someCollection.find().snapshot().forEach(function(doc) {
if (doc._id.indexOf("2019:") != 0) {
print("Processing: " + doc._id);
var oldDocId = doc._id;
doc._id = "2019:" + doc._id;
db.someCollection.insert(doc);
db.someCollection.remove({_id: oldDocId});
}
});
if (doc._id.indexOf("2019:") != 0) {... needed to prevent infinite loop, since forEach picks the inserted docs, even throught .snapshot() method used.
Here I have a solution that avoid multiple requests, for loops and old document removal.
You can easily create a new idea manually using something like:_id:ObjectId()
But knowing Mongo will automatically assign an _id if missing, you can use aggregate to create a $project containing all the fields of your document, but omit the field _id. You can then save it with $out
So if your document is:
{
"_id":ObjectId("5b5ed345cfbce6787588e480"),
"title": "foo",
"description": "bar"
}
Then your query will be:
db.getCollection('myCollection').aggregate([
{$match:
{_id: ObjectId("5b5ed345cfbce6787588e480")}
}
{$project:
{
title: '$title',
description: '$description'
}
},
{$out: 'myCollection'}
])
You can also create a new document from MongoDB compass or using command and set the specific _id value that you want.
As a very small improvement to the above answers i would suggest using
let doc1 = {... doc};
then
db.dyn_user_metricFormulaDefinitions.deleteOne({_id: doc._id});
This way we don't need to create extra variable to hold old _id.
Slightly modified example of #Florent Arlandis above where we insert _id from a different field in a document:
> db.coll.insertOne({ "_id": 1, "item": { "product": { "id": 11 } }, "source": "Good Store" })
{ "acknowledged" : true, "insertedId" : 1 }
> db.coll.aggregate( [ { $set: { _id : "$item.product.id" }}, { $out: "coll" } ]) // inserting _id you want for the current collection
> db.coll.find() // check that _id is changed
{ "_id" : 11, "item" : { "product" : { "id" : 11 } }, "source" : "Good Store" }
Do not use $match filter + $out as in #Florent Arlandis's answer since $out fully remove data in collection before inserting aggregate result, so effectively you will loose all data that don't match to $match filter