Related
How do I use the value of another field in $inc to define how much to increase?
Example-
Some doc
{_id: obi, length: 256, delta: 6}
Here, I want to increase the length by delta increment.
Pseudocode would be
db.collection.update({}, $inc: {length: $delta});
Thanks for helping...
Simple update operation can't allow to use internal fields in another fields, either its any of the operators,
For the solution you can use update with aggregation pipeline starting from MongoDB 4.2,
use $add/$sum any operator from both to sum the both fields number
db.collection.update({},
[{
$set: {
length: {
$add: ["$length", "$delta"]
}
}
}]
)
Playground
{
name: 'book',
tags: {
words: ['abc','123'],
lat: 33,
long: 22
}
}
Suppose this is a document. How do I remove "words" completely from all the documents in this collection? I want all documents to be without "words":
{
name: 'book',
tags: {
lat: 33,
long: 22
}
}
Try this: If your collection was 'example'
db.example.update({}, {$unset: {words:1}}, false, true);
Refer this:
http://www.mongodb.org/display/DOCS/Updating#Updating-%24unset
UPDATE:
The above link no longer covers '$unset'ing. Be sure to add {multi: true} if you want to remove this field from all of the documents in the collection; otherwise, it will only remove it from the first document it finds that matches. See this for updated documentation:
https://docs.mongodb.com/manual/reference/operator/update/unset/
Example:
db.example.update({}, {$unset: {words:1}} , {multi: true});
In the beginning, I did not get why the question has a bounty (I thought that the question has a nice answer and there is nothing to add), but then I noticed that the answer which was accepted and upvoted 15 times was actually wrong!
Yes, you have to use $unset operator, but this unset is going to remove the words key which does not exist for a document for a collection. So basically it will do nothing.
So you need to tell Mongo to look in the document tags and then in the words using dot notation. So the correct query is.
db.example.update(
{},
{ $unset: {'tags.words':1}},
false, true
)
Just for the sake of completion, I will refer to another way of doing it, which is much worse, but this way you can change the field with any custom code (even based on another field from this document).
db.example.updateMany({},{"$unset":{"tags.words":1}})
We can also use this to update multiple documents.
db.collection.updateMany({}, {$unset: {"fieldName": ""}})
updateMany requires a matching condition for each document, since we are passing {} it is always true. And the second argument uses $unset operator to remove the required field in each document.
To remove or delete field in MongoDB
For single Record
db.getCollection('userData').update({}, {$unset: {pi: 1}})
For Multi Record
db.getCollection('userData').update({}, {$unset: {pi: 1}}, {multi: true})
Starting in Mongo 4.2, it's also possible to use a slightly different syntax:
// { name: "book", tags: { words: ["abc", "123"], lat: 33, long: 22 } }
db.collection.updateMany({}, [{ $unset: ["tags.words"] }])
// { name: "book", tags: { lat: 33, long: 22 } }
Since the update method can accept an aggregation pipeline (note the squared brackets signifying the use of an aggregation pipeline), it means the $unset operator used here is the aggregation one (as opposed to the "query" one), whose syntax takes an array of fields.
The solution for PyMongo (Python mongo):
db.example.update({}, {'$unset': {'tags.words':1}}, multi=True);
I was trying to do something similar to this but instead remove the column from an embedded document. It took me a while to find a solution and this was the first post I came across so I thought I would post this here for anyone else trying to do the same.
So lets say instead your data looks like this:
{
name: 'book',
tags: [
{
words: ['abc','123'],
lat: 33,
long: 22
}, {
words: ['def','456'],
lat: 44,
long: 33
}
]
}
To remove the column words from the embedded document, do this:
db.example.update(
{'tags': {'$exists': true}},
{ $unset: {'tags.$[].words': 1}},
{multi: true}
)
or using the updateMany
db.example.updateMany(
{'tags': {'$exists': true}},
{ $unset: {'tags.$[].words': 1}}
)
The $unset will only edit it if the value exists but it will not do a safe navigation (it wont check if tags exists first) so the exists is needed on the embedded document.
This uses the all positional operator ($[]) which was introduced in version 3.6
To Remove a field you can use this command(this will be applied to all the documents in the collection) :
db.getCollection('example').update({}, {$unset: {Words:1}}, {multi: true});
And to remove multiple fields in one command :
db.getCollection('example').update({}, {$unset: {Words:1 ,Sentences:1}}, {multi: true});
Because I kept finding this page when looking for a way to remove a field using MongoEngine, I guess it might be helpful to post the MongoEngine way here too:
Example.objects.all().update(unset__tags__words=1)
In mongoDB shell this code might be helpful:
db.collection.update({}, {$unset: {fieldname: ""}} )
By default, the update() method updates a single document. Set the Multi Parameter to update all documents that match the query criteria.
Changed in version 3.6.
Syntax :
db.collection.update(
<query>,
<update>,
{
upsert: <boolean>,
multi: <boolean>,
writeConcern: <document>,
collation: <document>,
arrayFilters: [ <filterdocument1>, ... ]
}
)
Example :
db.getCollection('products').update({},{$unset: {translate:1, qordoba_translation_version:1}}, {multi: true})
In your example :
db.getCollection('products').update({},{$unset: {'tags.words' :1}}, {multi: true})
And for mongomapper,
Document: Shutoff
Field to remove: shutoff_type
Shutoff.collection.update( {}, { '$unset' => { 'shutoff_type': 1 } }, :multi => true )
It's worth mentioning, if you are using Typegoose/Mongoose your field needs to be declared in your model otherwise $unset will silently fail.
After a refactoring of a database this caught me unaware. The redundant fields were no longer in my model and I was running an $unset but nothing was happening.
So if you are doing some refactoring make sure you $unset your redundant fields before removing them from the model.
{
name: 'book',
tags: {
words: ['abc','123'],
lat: 33,
long: 22
}
}
Ans:
db.tablename.remove({'tags.words':['abc','123']})
Checking if "words" exists and then removing from the document
db.users.update({"tags.words" :{$exists: true}},
{$unset:{"tags.words":1}},false,true);
true indicates update multiple documents if matched.
you can also do this in aggregation by using project at 3.4
{$project: {"tags.words": 0} }
To reference a package and remove various "keys",
try this
db['name1.name2.name3.Properties'].remove([
{
"key" : "name_key1"
},
{
"key" : "name_key2"
},
{
"key" : "name_key3"
}
)]
I have a mongo document that contains something like
{date: [2018, 3, 22]}
and when I try to project this into a flat JSON structure with these fields concatenated, I always get an array with 0 elements, eg. just extracting the year with
db.getCollection('blah').aggregate([
{$project: {year: "$date.0"}}
])
I get
{"year" : []}
even though matching on a similar expression works fine, eg.
db.getCollection('blah').aggregate([
{$match: {"$date.0": 2018}}
])
selects the documents I would expect just fine.
What am I doing wrong? I've searched mongo documentation and stackoverflow but could find nothing.
For $project you should use $arrayElemAt instead of dot notation which works only for queries.
db.getCollection('blah').aggregate([
{$project: {year: { $arrayElemAt: [ "$date", 0 ] }}}
])
More here
Can I somehow add custom field with static (not computed) value?
I want to prepare objects before send and I need to remove some fields with internal information and add field with some entity ID.
For example I have collection "test" with objects like this
{_id: ObjectId(...), data: {...}}
And I need to convert it to
{data: {...}, entity_id: 54}
So how can I add entity_id: 54 without looping over result in my code?
db.test.aggregate({ $project: {_id: 0, data: 1, entity_id: ? } })
Thanks
Note that $literal was implemented in Mongo 2.6.
So now you can simply write:
db.test.aggregate(
{$project: {_id: 0, data: 1, entity_id: {$literal: 54}}})
See $literal.
edit as of 2.6 the $literal expression exists so you don't have to use the workaround now.
Original answer: I know this may sound really dumb, but you can use a "no-op" expression to "compute" what you need.
Example:
db.test.aggregate( { $project : {_id:0, data:1, entity_id: {$add: [54]} } } )
There was a proposed $literal operator for exactly this use case but it hasn't been implemented yet, you can vote for it here.
{
name: 'book',
tags: {
words: ['abc','123'],
lat: 33,
long: 22
}
}
Suppose this is a document. How do I remove "words" completely from all the documents in this collection? I want all documents to be without "words":
{
name: 'book',
tags: {
lat: 33,
long: 22
}
}
Try this: If your collection was 'example'
db.example.update({}, {$unset: {words:1}}, false, true);
Refer this:
http://www.mongodb.org/display/DOCS/Updating#Updating-%24unset
UPDATE:
The above link no longer covers '$unset'ing. Be sure to add {multi: true} if you want to remove this field from all of the documents in the collection; otherwise, it will only remove it from the first document it finds that matches. See this for updated documentation:
https://docs.mongodb.com/manual/reference/operator/update/unset/
Example:
db.example.update({}, {$unset: {words:1}} , {multi: true});
In the beginning, I did not get why the question has a bounty (I thought that the question has a nice answer and there is nothing to add), but then I noticed that the answer which was accepted and upvoted 15 times was actually wrong!
Yes, you have to use $unset operator, but this unset is going to remove the words key which does not exist for a document for a collection. So basically it will do nothing.
So you need to tell Mongo to look in the document tags and then in the words using dot notation. So the correct query is.
db.example.update(
{},
{ $unset: {'tags.words':1}},
false, true
)
Just for the sake of completion, I will refer to another way of doing it, which is much worse, but this way you can change the field with any custom code (even based on another field from this document).
db.example.updateMany({},{"$unset":{"tags.words":1}})
We can also use this to update multiple documents.
db.collection.updateMany({}, {$unset: {"fieldName": ""}})
updateMany requires a matching condition for each document, since we are passing {} it is always true. And the second argument uses $unset operator to remove the required field in each document.
To remove or delete field in MongoDB
For single Record
db.getCollection('userData').update({}, {$unset: {pi: 1}})
For Multi Record
db.getCollection('userData').update({}, {$unset: {pi: 1}}, {multi: true})
Starting in Mongo 4.2, it's also possible to use a slightly different syntax:
// { name: "book", tags: { words: ["abc", "123"], lat: 33, long: 22 } }
db.collection.updateMany({}, [{ $unset: ["tags.words"] }])
// { name: "book", tags: { lat: 33, long: 22 } }
Since the update method can accept an aggregation pipeline (note the squared brackets signifying the use of an aggregation pipeline), it means the $unset operator used here is the aggregation one (as opposed to the "query" one), whose syntax takes an array of fields.
The solution for PyMongo (Python mongo):
db.example.update({}, {'$unset': {'tags.words':1}}, multi=True);
I was trying to do something similar to this but instead remove the column from an embedded document. It took me a while to find a solution and this was the first post I came across so I thought I would post this here for anyone else trying to do the same.
So lets say instead your data looks like this:
{
name: 'book',
tags: [
{
words: ['abc','123'],
lat: 33,
long: 22
}, {
words: ['def','456'],
lat: 44,
long: 33
}
]
}
To remove the column words from the embedded document, do this:
db.example.update(
{'tags': {'$exists': true}},
{ $unset: {'tags.$[].words': 1}},
{multi: true}
)
or using the updateMany
db.example.updateMany(
{'tags': {'$exists': true}},
{ $unset: {'tags.$[].words': 1}}
)
The $unset will only edit it if the value exists but it will not do a safe navigation (it wont check if tags exists first) so the exists is needed on the embedded document.
This uses the all positional operator ($[]) which was introduced in version 3.6
To Remove a field you can use this command(this will be applied to all the documents in the collection) :
db.getCollection('example').update({}, {$unset: {Words:1}}, {multi: true});
And to remove multiple fields in one command :
db.getCollection('example').update({}, {$unset: {Words:1 ,Sentences:1}}, {multi: true});
Because I kept finding this page when looking for a way to remove a field using MongoEngine, I guess it might be helpful to post the MongoEngine way here too:
Example.objects.all().update(unset__tags__words=1)
In mongoDB shell this code might be helpful:
db.collection.update({}, {$unset: {fieldname: ""}} )
By default, the update() method updates a single document. Set the Multi Parameter to update all documents that match the query criteria.
Changed in version 3.6.
Syntax :
db.collection.update(
<query>,
<update>,
{
upsert: <boolean>,
multi: <boolean>,
writeConcern: <document>,
collation: <document>,
arrayFilters: [ <filterdocument1>, ... ]
}
)
Example :
db.getCollection('products').update({},{$unset: {translate:1, qordoba_translation_version:1}}, {multi: true})
In your example :
db.getCollection('products').update({},{$unset: {'tags.words' :1}}, {multi: true})
And for mongomapper,
Document: Shutoff
Field to remove: shutoff_type
Shutoff.collection.update( {}, { '$unset' => { 'shutoff_type': 1 } }, :multi => true )
It's worth mentioning, if you are using Typegoose/Mongoose your field needs to be declared in your model otherwise $unset will silently fail.
After a refactoring of a database this caught me unaware. The redundant fields were no longer in my model and I was running an $unset but nothing was happening.
So if you are doing some refactoring make sure you $unset your redundant fields before removing them from the model.
{
name: 'book',
tags: {
words: ['abc','123'],
lat: 33,
long: 22
}
}
Ans:
db.tablename.remove({'tags.words':['abc','123']})
Checking if "words" exists and then removing from the document
db.users.update({"tags.words" :{$exists: true}},
{$unset:{"tags.words":1}},false,true);
true indicates update multiple documents if matched.
you can also do this in aggregation by using project at 3.4
{$project: {"tags.words": 0} }
To reference a package and remove various "keys",
try this
db['name1.name2.name3.Properties'].remove([
{
"key" : "name_key1"
},
{
"key" : "name_key2"
},
{
"key" : "name_key3"
}
)]