Update and insert together in mongodb using pymongo? - mongodb

I have a list of dictionaries as follows:
data=[{
"uni_id": '101',
"name":"abc",
"age": 10,
},
{
"uni_id": '102',
"name":"def",
"age": 12,
}]
I want to use this dictionary to update my database only if the "uni_id" exists in the database, if it doesn't exist in my database then I want to insert that particular "uni_id" and also its corresponding "name" and "age":
My update statement is as follows, however I would also like to use insert with it to satisfy the above mentioned conditions. How to write the insert statement? Or is there a way to write some if-else statements to do update and insert?
db.students.update_one(
{"uni_id":data[0]['uni_id']},
{
"$set":
{
"name":"abc_1",
}})

Set upsert to true... It creates a new document if the updated document is not existed in the collection
db.students.update_one(
{ "uni_id": data[0]['uni_id']},
{ "$set": { "name":"abc_1" }},
True
)

Related

Is there a way in mongoDB using db.runCommand, where i can insert multiple doc in collection only it's not present. if present, ignore that document

I am writing a migration script, which uses a db.runCommand to apply migration in mongoDb.
Something like this:
db.runCommand(
{
insert: "countries",
documents: [
{
"name": "Algeria"
},
{
"name": "Andorra"
},
{
"name" : "Angola"
}
]
}
)
let's say if Algeria is already present and Andorra along with Angola is not present in countries collection.
My requirement is we should be able to only insert only Andorra and Angola in countries collection as they are not present. Algeria should be ignored and no exception should be thrown.
How can we achieve this?
You have two options, I'll start with my preferred option:
I would start by creating a unique index on the name field.
Now when you try to insert a document that violates the uniqueness constraint it fails:
Inserting a duplicate value for any key that is part of a unique index, ... With ordered to false, the insert operation would continue with any remaining documents.
Let's understand how that affects the insert command via the ordered option:
Optional. If true, then when an insert of a document fails, return without inserting any remaining documents listed in the inserts array. If false, then when an insert of a document fails, continue to insert the remaining documents. Defaults to true.
This means that if you use the ordered: false option for the insert command, all inserts will "try" to execute even if one fails (in our case due to unique index violation).
To summarise what you have to do is 1. build a unique index on the name field, 2. add ordered: false option to the insert command like so:
db.runCommand(
{
insert: "countries",
documents: [
{
"name": "Algeria"
},
{
"name": "Andorra"
},
{
"name": "Angola"
}
],
ordered: false
}
)
The other option you have is to use an update command with the upsert option instead of the insert command, I would personally not choose this as it has more overhead:
db.runCommand(
{
update: "countries",
updates: [
{
q: {"name": "Algeria"},
u: {"name": "Algeria"},
upsert: true
},
{
q: {"name": "Andorra"},
u: {"name": "Andorra"},
upsert: true
},
{
q: {"name": "Angola"},
u: {"name": "Angola"},
upsert: true
},
]
}
)

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

Mongodb how to insert ONLY if does not exists (no update if exist)?

How can I insert a document if it does not exist while not updating existing document if it exists?
let's say I have a document as follows:
{
"company":"test",
"name":"nameVal"
}
I want to check whether the collection contains company test, if it doesn't exist I want to create a new document. If it exists I want to do NOTHING. I tried update with upsert = true. But it updates the existing document if it exists.
This is what I tried:
db.getCollection('companies').update(
{
"company": "test"
},
{
"company": "test",
"name": "nameVal2"
},
{upsert:true}
)
Appreciate any help to resolve this using one query.
You can use $setOnInsert like,
db.companies.updateOne(
{"company": "test"},
{ $setOnInsert: { "name": "nameVal2", ... } },
{ upsert: true }
)
If this update operation does not do insert, $setOnInsert won't have any effect. So, the name will be updated only on insert.
Edited: I was informed that my solution only works for an array which is not what the original post asked for.
If you are using an array you can use the $addToSet: operator instead of $SetOnInsert.
db.companies.updateOne(
{"company": "test"},
{ $addToSet: { ["name": "nameValue"]} },
{ new: true })
)

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

How to update the _id of one MongoDB Document?

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