db.entities.aggregate([$match : {"Company.Id" : {$gt : 11}}])
above code works in mongo shell but below does not work why ??
db.entities.aggregate([{$match : {Company : {Id : {$gt : 11}}}}])
it displays nothing.
Here is my mongodocument
"_id" : ObjectId("552ca154993cfc98fef1e13c"),
"Name" : "R",
"Address" : "RAdd",
"Company" : {
"Name" : "something",
"Id" : 14
}
From mongodb docs
When the field holds an embedded document, a query can either specify an exact match on the embedded document or specify a match by individual fields in the embedded document using the dot notation.
So this will work
db.entities.aggregate([{
'$match':{
"Company" : {
"Name" : "something",
"Id" : 14
}
}
}
])
Related
I have a collection with some documents like below:
{
"_id" : ObjectId("1"),
"className" : "model.MyClass",
"createdOn" : ISODate("2018-10-23T11:00:00.000+01:00"),
"status" : "A"
}
{
"_id" : ObjectId("2"),
"className" : "model.MyClass",
"createdOn" : ISODate("2018-10-23T11:01:00.000+01:00"),
"status" : "B"
}
{
"_id" : ObjectId("3"),
"className" : "model.MyClass",
"createdOn" : ISODate("2018-10-23T11:02:00.000+01:00"),
"status" : "C"
}
{
"_id" : ObjectId("4"),
"className" : "model.MyClass",
"createdOn" : ISODate("2018-10-23T11:03:00.000+01:00"),
"status" : "D"
}
Given a specific ID, how can I get the previous document that whose status not equals a specific status.
For example, I give the ID 4 and like to get the last document that status not is B neither C. So, I get the Object with Id 1.
How to create this query?
you could try this:
db.yourcollection.find( {"status":{"$nin":["B","C"]}}
).sort({_id:-1}).limit(1);
so use not in operator i.e. $nin, then sort the data in descending order and limit the records to 1
see below documentations for details.
$nin operator
mongo sort
I am new to MongoDB and trying to make MongoDB throw an error when I insert another document with the same index. According to this answer MongoDB should throw an error.
The steps I did are:
1.) Add an index to Name field. I verified that it is added:
> db.room.getIndexes()
[
{
"v" : 1,
"key" : {
"_id" : 1
},
"name" : "_id_",
"ns" : "test.room"
},
{
"v" : 1,
"key" : {
"Name" : 1
},
"name" : "Name_1",
"ns" : "test.room"
}
]
2.) I tried to add document with the same name and was able to add it:
> db.room.find().pretty()
{
"_id" : 1,
"ModifiedDate" : ISODate("2017-02-12T10:59:35.394Z"),
"CreatedDate" : ISODate("2017-02-12T10:59:35.394Z"),
"Name" : "Sample"
}
{
"_id" : 2,
"ModifiedDate" : ISODate("2017-02-12T10:59:39.474Z"),
"CreatedDate" : ISODate("2017-02-12T10:59:39.474Z"),
"Name" : "Sample"
}
I am using C# MongoDB Driver 2.4.
You have to specify that the index you are creating is unique, otherwise MongoDB will not enforce it. You can do that with the C# driver using the CreateIndexOptions class.
roomCollection.Indexes
.CreateOne(
Builders<Room>.IndexKeys.Ascending(r => r.Name),
new CreateIndexOptions() { Unique = true });
Note that index creation will fail if there are currently duplicate names in the collection.
I have a mongodb database named "eagle" and am trying to replace all email records of "blue#domain.com" with "pink#domain.com" (within collection "all collections".
db.eagle.find({}).forEach(function(e,i) {
e.email=e.email.replace("//blue#domain.com","//pink#domain.com");
db.eagle.save(e);
});
I am very new to mongodb...so I'm not even sure that I'm in "eagle" or "falcon"....I just run mongo. This query doesn't do anything.
Here is what my Communications object looks like:
{
"_id" : ObjectId("redacted"),
"timestamp" : ISODate("2016-08-03T15:08:07.000Z"),
"thread_index" : "",
"updated_at" : ISODate("2016-09-01T17:49:31.401Z"),
"from" : {
"username" : "None",
"name" : "Pinky Jones",
"email" : "blue#domain.com"
},
"to" : {
"username" : "redude",
"name" : "Red Baron",
"email" : "red#domain.com"
},
"created_at" : ISODate("2016-09-01T17:49:31.401Z"),
}
Use findAndModify instead find
Documentación findAndModify
I have a collection which have many duplicates due to the routines that populated it in the first place. How to dedupe these?
e.g.
{ "_id" : ObjectId("531a5fe448757e00244096fa"), "code" : "ap", "name" : "[Almost Perfect]", "value" : "[u'*']" }
{ "_id" : ObjectId("531a731148757e17587a6e04"), "code" : "ap", "name" : "[Almost Perfect]", "value" : "[u'*']" }
{ "_id" : ObjectId("531a7bb848757e1f7c0ca702"), "code" : "ap", "name" : "[Almost Perfect]", "value" : "[u'*']" }
I want it to be just one (don't care which objectID gets picked)
{ "_id" : ObjectId("531a5fe448757e00244096fa"), "code" : "ap", "name" : "[Almost Perfect]", "value" : "[u'*']" }
You should use an Index over you code field:
db.<collection>.ensureIndex({'code' : 1}, {unique : true, dropDups : true})
unique will ensure you will not have duplicates anymore.
dropDups will delete all your duplicate documents when the ensureIndex operation is run
I'm kind of stuck with the following problem. I have a MongoDB filled with documents, of these documents (I have a list with Id's) I need to insert a field.
I have this document:
{
"id" : 3639,
"type" : "P",
"createdate" : "2011-10-19T11:45:14+0200",
"name_creator" : "",
"latitude" : "50.887",
"longitude" : "9.14999",
"themes" : [{
"name" : "Fun",
"id" : "4"
}, {
"name" : "Kids",
"id" : "5"
}]
}
I need a query the can insert the themes field into the document, the current themes field does not have to be updates, just 1 new one. I have over 300 Id's where this has to be done.
The document should then look like this:
(all the other fields in themes should be removed, just one new one 'Outside')
{
"id" : 3639,
"type" : "P",
"createdate" : "2011-10-19T11:45:14+0200",
"name_creator" : "",
"latitude" : "50.887",
"longitude" : "9.14999",
"themes" : [{
"name" : "Outside",
"id" : "6"
}]
}
I would normally write a bit of Java code that would loop over the documents and change them, but I believe (hope) this could be done in a query.
Anyone got an idea on how I could do this?
Thanks for any help!
All you need to do is
db.collection.update(
{id : {$in : [your list of ids]}},
{$set : {
theme : [{
"name" : "Outside",
"id" : "6"
}]
}},
{multi : true}
)