Mongo add timestamp field from existing date field - mongodb

I currently have a collection with documents like the following:
{ foo: 'bar', timeCreated: ISODate("2012-06-28T06:51:48.374Z") }
I would now like to add a timestampCreated key to the documents in this collection, to make querying by time easier.
I was able to add the new column with an update and $set operation, and set the timestamp value but I appears to be setting the current timestamp using this:
db.reports.update({}, {
$set : {
timestampCreated : new Timestamp(new Date('$.timeCreated'), 0)
}
}, false, true);
I however have not been able to figure out a way to add this column and set it's value to the timestamp of the existing 'timeCreated' field.

Do a find for all the documents, limiting to just the id and timeCreated fields. Then loop over that and generate the timestampCreated value, and do an update on each.

Use updateMany() which can accept aggregate pipelines (starting from MongoDB 4.2) and thus take advantage of the $toLong operator which converts a Date into the number of milliseconds since the epoch.
Also use the $type query in the update filter to limit only documents with the timeCreated field and of Date type:
db.reports.updateMany(
{ 'timeCreated': {
'$exists': true,
'$type': 9
} },
[
{ '$set': {
'timestampCreated': { '$toLong': '$timeCreated' }
} }
]
)

Related

Update field only if document updated in MongoDB

I am calling findAndModify() using the $max function to set the value of a field to the largest value.
For example, as shown in the MongoDB documentation.
db.scores.update( { _id: 1 }, { $max: { highScore: 950 } } )
I'd like to also set a lastUpdatedTimestamp only if the document is updated. I can't just perform a $set because that will always change the last updated timestamp. Is there a good mechanism within MongoDB to set another value only if the document is updated? Something similar to $setOnInsert but for any update.
If there isn't what might be a good approach here? Right now I'm thinking I could perform a regular find. Then do a local comparison. If the new value is greater than the old, then there is a good possibility that the update will update the document. So I just include the $set for the lastUpdatedTimestamp.
You can first make a query to find records having highScore less than your input value and then update. This will only set lastUpdatedTimestamp on updating the record.
db.scores.findAndModify({
query: { highScore: { $lt: 950 } },
update: { $set: { "highScore" : 950, "lastUpdatedTimestamp" : new Date() } },
})
as I see you wanted to . update your document only if your highScore can be updated .
only the documents's score is lower than new score value ,it will be updated with score field and lastUpdatedTimestamp
the best way is put your new score in the filter to find the documents match old score < new score
do it like this
db.scores.update(
{_id :4,highScore:{$lt:900}},
{$set:{highScore:900},
$currentDate: { lastModified: true }})
or set the modify time like
{$set:{highScore:900 ,lastupdatetime: new_time},

How to get the particuler object fields using ReativeMongo without a case class [duplicate]

In my MongoDB, I have a student collection with 10 records having fields name and roll. One record of this collection is:
{
"_id" : ObjectId("53d9feff55d6b4dd1171dd9e"),
"name" : "Swati",
"roll" : "80",
}
I want to retrieve the field roll only for all 10 records in the collection as we would do in traditional database by using:
SELECT roll FROM student
I went through many blogs but all are resulting in a query which must have WHERE clause in it, for example:
db.students.find({ "roll": { $gt: 70 })
The query is equivalent to:
SELECT * FROM student WHERE roll > 70
My requirement is to find a single key only without any condition. So, what is the query operation for that.
From the MongoDB docs:
A projection can explicitly include several fields. In the following operation, find() method returns all documents that match the query. In the result set, only the item and qty fields and, by default, the _id field return in the matching documents.
db.inventory.find( { type: 'food' }, { item: 1, qty: 1 } )
In this example from the folks at Mongo, the returned documents will contain only the fields of item, qty, and _id.
Thus, you should be able to issue a statement such as:
db.students.find({}, {roll:1, _id:0})
The above statement will select all documents in the students collection, and the returned document will return only the roll field (and exclude the _id).
If we don't mention _id:0 the fields returned will be roll and _id. The '_id' field is always displayed by default. So we need to explicitly mention _id:0 along with roll.
get all data from table
db.student.find({})
SELECT * FROM student
get all data from table without _id
db.student.find({}, {_id:0})
SELECT name, roll FROM student
get all data from one field with _id
db.student.find({}, {roll:1})
SELECT id, roll FROM student
get all data from one field without _id
db.student.find({}, {roll:1, _id:0})
SELECT roll FROM student
find specified data using where clause
db.student.find({roll: 80})
SELECT * FROM students WHERE roll = '80'
find a data using where clause and greater than condition
db.student.find({ "roll": { $gt: 70 }}) // $gt is greater than
SELECT * FROM student WHERE roll > '70'
find a data using where clause and greater than or equal to condition
db.student.find({ "roll": { $gte: 70 }}) // $gte is greater than or equal
SELECT * FROM student WHERE roll >= '70'
find a data using where clause and less than or equal to condition
db.student.find({ "roll": { $lte: 70 }}) // $lte is less than or equal
SELECT * FROM student WHERE roll <= '70'
find a data using where clause and less than to condition
db.student.find({ "roll": { $lt: 70 }}) // $lt is less than
SELECT * FROM student WHERE roll < '70'
I think mattingly890 has the correct answer , here is another example along with the pattern/commmand
db.collection.find( {}, {your_key:1, _id:0})
> db.mycollection.find().pretty();
{
"_id": ObjectId("54ffca63cea5644e7cda8e1a"),
"host": "google",
"ip": "1.1.192.1"
}
db.mycollection.find({},{ "_id": 0, "host": 1 }).pretty();
Here you go , 3 ways of doing , Shortest to boring :
db.student.find({}, 'roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}).select('roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}, {'roll' : 1 , '_id' : 1 ); // <---- Old lengthy boring way
To remove specific field use - operator :
db.student.find({}).select('roll -_id') // <--- Will remove id from result
While gowtham's answer is complete, it is worth noting that those commands may differ from on API to another (for those not using mongo's shell).
Please refer to documentation link for detailed info.
Nodejs, for instance, have a method called `projection that you would append to your find function in order to project.
Following the same example set, commands like the following can be used with Node:
db.student.find({}).project({roll:1})
SELECT _id, roll FROM student
Or
db.student.find({}).project({roll:1, _id: 0})
SELECT roll FROM student
and so on.
Again for nodejs users, do not forget (what you should already be familiar with if you used this API before) to use toArray in order to append your .then command.
Try the following query:
db.student.find({}, {roll: 1, _id: 0});
And if you are using console you can add pretty() for making it easy to read.
db.student.find({}, {roll: 1, _id: 0}).pretty();
Hope this helps!!
Just for educational purposes you could also do it with any of the following ways:
1.
var query = {"roll": {$gt: 70};
var cursor = db.student.find(query);
cursor.project({"roll":1, "_id":0});
2.
var query = {"roll": {$gt: 70};
var projection = {"roll":1, "_id":0};
var cursor = db.student.find(query,projection);
`
db.<collection>.find({}, {field1: <value>, field2: <value> ...})
In your example, you can do something like:
db.students.find({}, {"roll":true, "_id":false})
Projection
The projection parameter determines which fields are returned in the
matching documents. The projection parameter takes a document of the
following form:
{ field1: <value>, field2: <value> ... }
The <value> can be any of the following:
1 or true to include the field in the return documents.
0 or false to exclude the field.
NOTE
For the _id field, you do not have to explicitly specify _id: 1 to
return the _id field. The find() method always returns the _id field
unless you specify _id: 0 to suppress the field.
READ MORE
For better understanding I have written similar MySQL query.
Selecting specific fields
MongoDB : db.collection_name.find({},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name;
Selecting specific fields with where clause
MongoDB : db.collection_name.find({email:'you#email.com'},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name WHERE email = 'you#email.com';
This works for me,
db.student.find({},{"roll":1})
no condition in where clause i.e., inside first curly braces.
inside next curly braces: list of projection field names to be needed in the result and 1 indicates particular field is the part of the query result
getting name of the student
student-details = db.students.find({{ "roll": {$gt: 70} },{"name": 1, "_id": False})
getting name & roll of the student
student-details = db.students.find({{ "roll": {$gt: 70}},{"name": 1,"roll":1,"_id": False})
I just want to add to the answers that if you want to display a field that is nested in another object, you can use the following syntax
db.collection.find( {}, {{'object.key': true}})
Here key is present inside the object named object
{ "_id" : ObjectId("5d2ef0702385"), "object" : { "key" : "value" } }
var collection = db.collection('appuser');
collection.aggregate(
{ $project : { firstName : 1, lastName : 1 } },function(err, res){
res.toArray(function(err, realRes){
console.log("response roo==>",realRes);
});
});
it's working
Use the Query like this in the shell:
1. Use database_name
e.g: use database_name
2. Which returns only assets particular field information when matched , _id:0 specifies not to display ID in the result
db.collection_name.find( { "Search_Field": "value" },
{ "Field_to_display": 1,_id:0 } )
If u want to retrieve the field "roll" only for all 10 records in the collections.
Then try this.
In MongoDb :
db.students.find( { } , { " roll " : { " $roll " })
In Sql :
select roll from students
The query for MongoDB here fees is collection and description is a field.
db.getCollection('fees').find({},{description:1,_id:0})
Apart from what people have already mentioned I am just introducing indexes to the mix.
So imagine a large collection, with let's say over 1 million documents and you have to run a query like this.
The WiredTiger Internal cache will have to keep all that data in the cache if you have to run this query on it, if not that data will be fed into the WT Internal Cache either from FS Cache or Disk before the retrieval from DB is done (in batches if being called for from a driver connected to database & given that 1 million documents are not returned in 1 go, cursor comes into play)
Covered query can be an alternative. Copying the text from docs directly.
When an index covers a query, MongoDB can both match the query conditions and return the results using only the index keys; i.e. MongoDB does not need to examine documents from the collection to return the results.
When an index covers a query, the explain result has an IXSCAN stage that is not a descendant of a FETCH stage, and in the executionStats, the totalDocsExamined is 0.
Query : db.getCollection('qaa').find({roll_no : {$gte : 0}},{_id : 0, roll_no : 1})
Index : db.getCollection('qaa').createIndex({roll_no : 1})
If the index here is in WT Internal Cache then it would be a straight forward process to get the values. An index has impact on the write performance of the system thus this would make more sense if the reads are a plenty compared to the writes.
If you are using the MongoDB driver in NodeJs then the above-mentioned answers might not work for you. You will have to do something like this to get only selected properties as a response.
import { MongoClient } from "mongodb";
// Replace the uri string with your MongoDB deployment's connection string.
const uri = "<connection string uri>";
const client = new MongoClient(uri);
async function run() {
try {
await client.connect();
const database = client.db("sample_mflix");
const movies = database.collection("movies");
// Query for a movie that has the title 'The Room'
const query = { title: "The Room" };
const options = {
// sort matched documents in descending order by rating
sort: { "imdb.rating": -1 },
// Include only the `title` and `imdb` fields in the returned document
projection: { _id: 0, title: 1, imdb: 1 },
};
const movie = await movies.findOne(query, options);
/** since this method returns the matched document, not a cursor,
* print it directly
*/
console.log(movie);
} finally {
await client.close();
}
}
run().catch(console.dir);
This code is copied from the actual MongoDB doc you can check here.
https://docs.mongodb.com/drivers/node/current/usage-examples/findOne/
db.student.find({}, {"roll":1, "_id":0})
This is equivalent to -
Select roll from student
db.student.find({}, {"roll":1, "name":1, "_id":0})
This is equivalent to -
Select roll, name from student
In mongodb 3.4 we can use below logic, i am not sure about previous versions
select roll from student ==> db.student.find(!{}, {roll:1})
the above logic helps to define some columns (if they are less)
Using Studio 3T for MongoDB, if I use .find({}, { _id: 0, roll: true }) it still return an array of objects with an empty _id property.
Using JavaScript map helped me to only retrieve the desired roll property as an array of string:
var rolls = db.student
.find({ roll: { $gt: 70 } }) // query where role > 70
.map(x => x.roll); // return an array of role
Not sure this answers the question but I believe it's worth mentioning here.
There is one more way for selecting single field (and not multiple) using db.collection_name.distinct();
e.g.,db.student.distinct('roll',{});
Or, 2nd way: Using db.collection_name.find().forEach(); (multiple fields can be selected here by concatenation)
e.g., db.collection_name.find().forEach(function(c1){print(c1.roll);});
_id = "123321"; _user = await likes.find({liker_id: _id},{liked_id:"$liked_id"}); ;
let suppose you have liker_id and liked_id field in the document so by putting "$liked_id" it will return _id and liked_id only.
For Single Update :
db.collection_name.update({ field_name_1: ("value")}, { $set: { field_name_2 : "new_value" }});
For MultiUpdate :
db.collection_name.updateMany({ field_name_1: ("value")}, { $set: {field_name_2 : "new_value" }});
Make sure indexes are proper.

How to make a query without nested document in MongoDB? [duplicate]

In my MongoDB, I have a student collection with 10 records having fields name and roll. One record of this collection is:
{
"_id" : ObjectId("53d9feff55d6b4dd1171dd9e"),
"name" : "Swati",
"roll" : "80",
}
I want to retrieve the field roll only for all 10 records in the collection as we would do in traditional database by using:
SELECT roll FROM student
I went through many blogs but all are resulting in a query which must have WHERE clause in it, for example:
db.students.find({ "roll": { $gt: 70 })
The query is equivalent to:
SELECT * FROM student WHERE roll > 70
My requirement is to find a single key only without any condition. So, what is the query operation for that.
From the MongoDB docs:
A projection can explicitly include several fields. In the following operation, find() method returns all documents that match the query. In the result set, only the item and qty fields and, by default, the _id field return in the matching documents.
db.inventory.find( { type: 'food' }, { item: 1, qty: 1 } )
In this example from the folks at Mongo, the returned documents will contain only the fields of item, qty, and _id.
Thus, you should be able to issue a statement such as:
db.students.find({}, {roll:1, _id:0})
The above statement will select all documents in the students collection, and the returned document will return only the roll field (and exclude the _id).
If we don't mention _id:0 the fields returned will be roll and _id. The '_id' field is always displayed by default. So we need to explicitly mention _id:0 along with roll.
get all data from table
db.student.find({})
SELECT * FROM student
get all data from table without _id
db.student.find({}, {_id:0})
SELECT name, roll FROM student
get all data from one field with _id
db.student.find({}, {roll:1})
SELECT id, roll FROM student
get all data from one field without _id
db.student.find({}, {roll:1, _id:0})
SELECT roll FROM student
find specified data using where clause
db.student.find({roll: 80})
SELECT * FROM students WHERE roll = '80'
find a data using where clause and greater than condition
db.student.find({ "roll": { $gt: 70 }}) // $gt is greater than
SELECT * FROM student WHERE roll > '70'
find a data using where clause and greater than or equal to condition
db.student.find({ "roll": { $gte: 70 }}) // $gte is greater than or equal
SELECT * FROM student WHERE roll >= '70'
find a data using where clause and less than or equal to condition
db.student.find({ "roll": { $lte: 70 }}) // $lte is less than or equal
SELECT * FROM student WHERE roll <= '70'
find a data using where clause and less than to condition
db.student.find({ "roll": { $lt: 70 }}) // $lt is less than
SELECT * FROM student WHERE roll < '70'
I think mattingly890 has the correct answer , here is another example along with the pattern/commmand
db.collection.find( {}, {your_key:1, _id:0})
> db.mycollection.find().pretty();
{
"_id": ObjectId("54ffca63cea5644e7cda8e1a"),
"host": "google",
"ip": "1.1.192.1"
}
db.mycollection.find({},{ "_id": 0, "host": 1 }).pretty();
Here you go , 3 ways of doing , Shortest to boring :
db.student.find({}, 'roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}).select('roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}, {'roll' : 1 , '_id' : 1 ); // <---- Old lengthy boring way
To remove specific field use - operator :
db.student.find({}).select('roll -_id') // <--- Will remove id from result
While gowtham's answer is complete, it is worth noting that those commands may differ from on API to another (for those not using mongo's shell).
Please refer to documentation link for detailed info.
Nodejs, for instance, have a method called `projection that you would append to your find function in order to project.
Following the same example set, commands like the following can be used with Node:
db.student.find({}).project({roll:1})
SELECT _id, roll FROM student
Or
db.student.find({}).project({roll:1, _id: 0})
SELECT roll FROM student
and so on.
Again for nodejs users, do not forget (what you should already be familiar with if you used this API before) to use toArray in order to append your .then command.
Try the following query:
db.student.find({}, {roll: 1, _id: 0});
And if you are using console you can add pretty() for making it easy to read.
db.student.find({}, {roll: 1, _id: 0}).pretty();
Hope this helps!!
Just for educational purposes you could also do it with any of the following ways:
1.
var query = {"roll": {$gt: 70};
var cursor = db.student.find(query);
cursor.project({"roll":1, "_id":0});
2.
var query = {"roll": {$gt: 70};
var projection = {"roll":1, "_id":0};
var cursor = db.student.find(query,projection);
`
db.<collection>.find({}, {field1: <value>, field2: <value> ...})
In your example, you can do something like:
db.students.find({}, {"roll":true, "_id":false})
Projection
The projection parameter determines which fields are returned in the
matching documents. The projection parameter takes a document of the
following form:
{ field1: <value>, field2: <value> ... }
The <value> can be any of the following:
1 or true to include the field in the return documents.
0 or false to exclude the field.
NOTE
For the _id field, you do not have to explicitly specify _id: 1 to
return the _id field. The find() method always returns the _id field
unless you specify _id: 0 to suppress the field.
READ MORE
For better understanding I have written similar MySQL query.
Selecting specific fields
MongoDB : db.collection_name.find({},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name;
Selecting specific fields with where clause
MongoDB : db.collection_name.find({email:'you#email.com'},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name WHERE email = 'you#email.com';
This works for me,
db.student.find({},{"roll":1})
no condition in where clause i.e., inside first curly braces.
inside next curly braces: list of projection field names to be needed in the result and 1 indicates particular field is the part of the query result
getting name of the student
student-details = db.students.find({{ "roll": {$gt: 70} },{"name": 1, "_id": False})
getting name & roll of the student
student-details = db.students.find({{ "roll": {$gt: 70}},{"name": 1,"roll":1,"_id": False})
I just want to add to the answers that if you want to display a field that is nested in another object, you can use the following syntax
db.collection.find( {}, {{'object.key': true}})
Here key is present inside the object named object
{ "_id" : ObjectId("5d2ef0702385"), "object" : { "key" : "value" } }
var collection = db.collection('appuser');
collection.aggregate(
{ $project : { firstName : 1, lastName : 1 } },function(err, res){
res.toArray(function(err, realRes){
console.log("response roo==>",realRes);
});
});
it's working
Use the Query like this in the shell:
1. Use database_name
e.g: use database_name
2. Which returns only assets particular field information when matched , _id:0 specifies not to display ID in the result
db.collection_name.find( { "Search_Field": "value" },
{ "Field_to_display": 1,_id:0 } )
If u want to retrieve the field "roll" only for all 10 records in the collections.
Then try this.
In MongoDb :
db.students.find( { } , { " roll " : { " $roll " })
In Sql :
select roll from students
The query for MongoDB here fees is collection and description is a field.
db.getCollection('fees').find({},{description:1,_id:0})
Apart from what people have already mentioned I am just introducing indexes to the mix.
So imagine a large collection, with let's say over 1 million documents and you have to run a query like this.
The WiredTiger Internal cache will have to keep all that data in the cache if you have to run this query on it, if not that data will be fed into the WT Internal Cache either from FS Cache or Disk before the retrieval from DB is done (in batches if being called for from a driver connected to database & given that 1 million documents are not returned in 1 go, cursor comes into play)
Covered query can be an alternative. Copying the text from docs directly.
When an index covers a query, MongoDB can both match the query conditions and return the results using only the index keys; i.e. MongoDB does not need to examine documents from the collection to return the results.
When an index covers a query, the explain result has an IXSCAN stage that is not a descendant of a FETCH stage, and in the executionStats, the totalDocsExamined is 0.
Query : db.getCollection('qaa').find({roll_no : {$gte : 0}},{_id : 0, roll_no : 1})
Index : db.getCollection('qaa').createIndex({roll_no : 1})
If the index here is in WT Internal Cache then it would be a straight forward process to get the values. An index has impact on the write performance of the system thus this would make more sense if the reads are a plenty compared to the writes.
If you are using the MongoDB driver in NodeJs then the above-mentioned answers might not work for you. You will have to do something like this to get only selected properties as a response.
import { MongoClient } from "mongodb";
// Replace the uri string with your MongoDB deployment's connection string.
const uri = "<connection string uri>";
const client = new MongoClient(uri);
async function run() {
try {
await client.connect();
const database = client.db("sample_mflix");
const movies = database.collection("movies");
// Query for a movie that has the title 'The Room'
const query = { title: "The Room" };
const options = {
// sort matched documents in descending order by rating
sort: { "imdb.rating": -1 },
// Include only the `title` and `imdb` fields in the returned document
projection: { _id: 0, title: 1, imdb: 1 },
};
const movie = await movies.findOne(query, options);
/** since this method returns the matched document, not a cursor,
* print it directly
*/
console.log(movie);
} finally {
await client.close();
}
}
run().catch(console.dir);
This code is copied from the actual MongoDB doc you can check here.
https://docs.mongodb.com/drivers/node/current/usage-examples/findOne/
db.student.find({}, {"roll":1, "_id":0})
This is equivalent to -
Select roll from student
db.student.find({}, {"roll":1, "name":1, "_id":0})
This is equivalent to -
Select roll, name from student
In mongodb 3.4 we can use below logic, i am not sure about previous versions
select roll from student ==> db.student.find(!{}, {roll:1})
the above logic helps to define some columns (if they are less)
Using Studio 3T for MongoDB, if I use .find({}, { _id: 0, roll: true }) it still return an array of objects with an empty _id property.
Using JavaScript map helped me to only retrieve the desired roll property as an array of string:
var rolls = db.student
.find({ roll: { $gt: 70 } }) // query where role > 70
.map(x => x.roll); // return an array of role
Not sure this answers the question but I believe it's worth mentioning here.
There is one more way for selecting single field (and not multiple) using db.collection_name.distinct();
e.g.,db.student.distinct('roll',{});
Or, 2nd way: Using db.collection_name.find().forEach(); (multiple fields can be selected here by concatenation)
e.g., db.collection_name.find().forEach(function(c1){print(c1.roll);});
_id = "123321"; _user = await likes.find({liker_id: _id},{liked_id:"$liked_id"}); ;
let suppose you have liker_id and liked_id field in the document so by putting "$liked_id" it will return _id and liked_id only.
For Single Update :
db.collection_name.update({ field_name_1: ("value")}, { $set: { field_name_2 : "new_value" }});
For MultiUpdate :
db.collection_name.updateMany({ field_name_1: ("value")}, { $set: {field_name_2 : "new_value" }});
Make sure indexes are proper.

MongoDb timestamp

i have created and want to now import a dummy collection. one of the fields in each item are "created" and "updated" fields. what can i put in the source/json file so that MongoDb will use the current date and time as the value on import?
this wont work
"created" : Date()
mongoimport is intended for importing data existing data in CSV, TSV, or JSON format. If you want to insert new fields (such as a created timestamp) you will have to set a value for them.
For example, if you want to set the created timestamp to the current time, you could get a unix timestamp from the command line (which will be seconds since the epoch):
$ date +%s
1349960286
The JSON <date> representation that mongoimport expects is a 64-bit signed integer representing milliseconds since the epoch. You'll need to multiply the unixtime seconds value by 1000 and include in your JSON file:
{ "created": Date(1349960286000) }
An alternative approach would be to add the created timestamps to documents after they have been inserted.
For example:
db.mycoll.update(
{created: { $exists : false }}, // Query criteria
{ $set : { created: new Date() }}, // Add 'created' timestamp
false, // upsert
true // update all matching documents
)
As Stennie correctly pointed out, you can not do this with just mongoimport or mongorestore: they are just for restoring your previously dumped data. Correct way of doing this is to restore the data and then to make update on the restored data.
With a new mongo 2.6 you can do this easily using $currentDate operation, which was created to update time to a current timestamp.
In your case you need something like
db.users.update(
{},
{
$currentDate: {
created: true,
updated: true
},
}
)

MongoDB: update every document on one field

I have a collected named foo hypothetically.
Each instance of foo has a field called lastLookedAt which is a UNIX timestamp since epoch. I'd like to be able to go through the MongoDB client and set that timestamp for all existing documents (about 20,000 of them) to the current timestamp.
What's the best way of handling this?
Regardless of the version, for your example, the <update> is:
{ $set: { lastLookedAt: Date.now() / 1000 } }
However, depending on your version of MongoDB, the query will look different. Regardless of version, the key is that the empty condition {} will match any document. In the Mongo shell, or with any MongoDB client:
$version >= 3.2:
db.foo.updateMany( {}, <update> )
{} is the condition (the empty condition matches any document)
3.2 > $version >= 2.2:
db.foo.update( {}, <update>, { multi: true } )
{} is the condition (the empty condition matches any document)
{multi: true} is the "update multiple documents" option
$version < 2.2:
db.foo.update( {}, <update>, false, true )
{} is the condition (the empty condition matches any document)
false is for the "upsert" parameter
true is for the "multi" parameter (update multiple records)
This code will be helpful for you
Model.update({
'type': "newuser"
}, {
$set: {
email: "abc#gmail.com",
phoneNumber:"0123456789"
}
}, {
multi: true
},
function(err, result) {
console.log(result);
console.log(err);
})
I have been using MongoDB .NET driver for a little over a month now. If I were to do it using .NET driver, I would use Update method on the collection object. First, I will construct a query that will get me all the documents I am interested in and do an Update on the fields I want to change. Update in Mongo only affects the first document and to update all documents resulting from the query one needs to use 'Multi' update flag. Sample code follows...
var collection = db.GetCollection("Foo");
var query = Query.GTE("No", 1); // need to construct in such a way that it will give all 20K //docs.
var update = Update.Set("timestamp", datetime.UtcNow);
collection.Update(query, update, UpdateFlags.Multi);
You can use updateMany() methods of mongodb to update multiple document
Simple query is like this
db.collection.updateMany(filter, update, options)
For more doc of uppdateMany read here
As per your requirement the update code will be like this:
User.updateMany({"created": false}, {"$set":{"created": true}});
here you need to use $set because you just want to change created from true to false. For ref. If you want to change entire doc then you don't need to use $set