What exactly happens when I call ensureIndex(data) when typical data looks like data:{name: "A",age:"B", job : "C"} ? Will it create a compound index over these three fields or will it create only one index applicable when anything from data is requested or something altogether different ?
You can do either :
> db.collection.ensureIndex({"data.name": 1,"data.age":1, "data.job" : 1})
> db.collection.ensureIndex({"data": 1})
This is discussed in the documentation under indexes-on-embedded-fields and indexes on sub documents
The important section of the sub document section is 'When performing equality matches on subdocuments, field order matters and the subdocuments must match exactly.'
This means that the 2 indexes are the same for simple queries .
However, as the sub-document example shows, you can get some interesting results (that you might not expect) if you just index the whole sub-document as opposed to a specific field and then do a comparison operator (like $gte) - if you index a specific sub field you get a less flexible, but potentially more useful index.
It really all depends on your use case.
Anyway, once you have created the index you can check what's created with :
> db.collection.getIndexes()
[
{
"v" : 1,
"key" : {
"_id" : 1
},
"ns" : "test.collection",
"name" : "_id_"
},
{
"v" : 1,
"key" : {
"data.name" : 1,
"data.age" : 1,
"data.job" : 1
},
"ns" : "test.collection",
"name" : "data.name_1_data.age_1_data.job_1"
}
]
As you can see from the output it created a new key called data.name_1_data.age_1_data.job_1 (the _id_ index is always created).
If you want to test your new index then you can do :
> db.collection.insert({data:{name: "A",age:"B", job : "C"}})
> db.collection.insert({data:{name: "A1",age:"B", job : "C"}})
> db.collection.find({"data.name" : "A"}).explain()
{
"cursor" : "BtreeCursor data.name_1_data.age_1_data.job_1",
.... more stuff
The main thing is that you can see that your new index was used (BtreeCursor data.name_1_data.age_1_data.job_1 in the cursor field is what indicates this is the case). If you see "cursor" : "BasicCursor", then your index was not used.
For more detailed information look here.
you can try this :
db.collection.ensureIndex({"data.name": 1,"data.age":1, "data.job" : 1})
Related
My query is failing to find all matching results. If I add an additional _id parameter to a specific matching example, I am getting results
> db.reviews.count({"contentProvider":"GLORP", "responses.0": {$exists: true}})
0
> db.reviews.count({_id: "1234", "contentProvider":"GLORP", "responses.0": {$exists: true}})
1
the first query is using index:
"indexName" : "contentProvider_1_reviewDetail_1_reviewerUserName_1_providerReviewId_1",
and the query with the _id is of course using the _id_ index:
"indexName" : "_id_"
Here is the index in question:
{
"v" : 1,
"key" : {
"contentProvider" : 1,
"reviewDetail" : 1,
"reviewerUserName" : 1,
"providerReviewId" : 1
},
"name" : "contentProvider_1_reviewDetail_1_reviewerUserName_1_providerReviewId_1",
"ns" : "test.reviews",
"background" : true
}
Using mongodb version 3.2.3
Is the index corrupted? Will dropping it and readding it likely fix the problem?
It's possible and you could certainly try it, however without knowing what version of MongoDB you are using and without seeing the index definition I cannot say for certain.
There are multiple different types of indexes as well as index properties like: sparse or partial that can change behavior and may explain why the index doesn't return the results you expect.
I'd recommend checking the index first and see if the index definition has any properties that would result in the document being excluded.
If not then you can always drop the index and recreate it.
I created mongo db collection index using java code
dbCollection.createIndex("accountNumber");
When i see indices using
db.accounts.getIndexes()
I am getting the index name as "accountNumber_1"
How to get the index name also same as document field? or how to give index name?
Is naming indices important or i can ignore this?
When we create index on the document users
> db.users.createIndex({name: 1})
{
"ok" : 0,
"errmsg" : "Index with name: name_1 already exists with different option
s",
"code" : 85
}
the name: name_1 is returned, then we can get the index through getIndexes()
> db.users.getIndexes()
[
{
"v" : 1,
"key" : {
"_id" : 1
},
"name" : "_id_",
"ns" : "test.users"
},
{
"v" : 1,
"unique" : true,
"key" : {
"name" : 1
},
"name" : "name_1",
"ns" : "test.users",
"background" : true,
"safe" : null
}
]
We know, the name_1 is just the value of index name. and the key name is used to create index for document users. I think the name_1 is the value of name to meet BSON structure. We can ignore it...
You can create index with the name you wanted using the other variant of createIndex method, refer java API here.
public void createIndex(DBObject keys,
DBObject options)
Creates an index on the field specified, if that index does not already exist.
Prior to MongoDB 3.0 the dropDups option could be used with unique indexes allowing documents with duplicate values to be dropped when building the index. Later versions of MongoDB will silently ignore this setting.
Parameters:
keys - a document that contains pairs with the name of the field or fields to index and order of the index
options - a document that controls the creation of the index.
MongoDB documentation
Index Creation Tutorials
You can corresponding mongodb documentation here.
Basically the second parameter 'options' contain an option to supply index name explicitly.
I remember reading somewhere that the mongo engine was more confortable when the entire structure of a document was already in place in case of an update, so here is the question.
When dealing with "empty" data, for example when inserting an empty string, should I default it to null, "" or not insert it at all ?
{
_id: ObjectId("5192b6072fda974610000005"),
description: ""
}
or
{
_id: ObjectId("5192b6072fda974610000005"),
description: null
}
or
{
_id: ObjectId("5192b6072fda974610000005")
}
You have to remember that the description field may or may not be filled in every document (based on user input).
Introduction
If a document doesn't have a value, the DB considers its value to be null. Suppose a database with the following documents:
{ "_id" : ObjectId("5192d23b1698aa96f0690d96"), "a" : 1, "desc" : "" }
{ "_id" : ObjectId("5192d23f1698aa96f0690d97"), "a" : 1, "desc" : null }
{ "_id" : ObjectId("5192d2441698aa96f0690d98"), "a" : 1 }
If you create a query to find documents with the field desc different than null, you will get just one document:
db.test.find({desc: {$ne: null}})
// Output:
{ "_id" : ObjectId("5192d23b1698aa96f0690d96"), "a" : 1, "desc" : "" }
The database doesn't differ documents without a desc field and documents with a desc field with the value null. One more test:
db.test.find({desc: null})
// Output:
{ "_id" : ObjectId("5192d2441698aa96f0690d98"), "a" : 1 }
{ "_id" : ObjectId("5192d23f1698aa96f0690d97"), "a" : 1, "desc" : null }
But the differences are only ignored in the queries, because, as shown in the last example above, the fields are still saved on disk and you'll receive documents with the same structure of the documents that were sent to the MongoDB.
Question
When dealing with "empty" data, for example when inserting an empty string, should I default it to null, "" or not insert it at all ?
There isn't much difference from {desc: null} to {}, because most of the operators will have the same result. You should only pay special attention to these two operators:
$exists
$type
I'd save documents without the desc field, because the operators will continue to work as expected and I'd save some space.
Padding factor
If you know the documents in your database grow frequently, then MongoDB might need to move the documents during the update, because there isn't enough space in the previous document place. To prevent moving documents around, MongoDB allocates extra space for each document.
The ammount of extra space allocated by MongoDB per document is controlled by the padding factor. You cannot (and don't need to) choose the padding factor, because MongoDB will adaptively learn it, but you can help MongoDB preallocating internal space for each document by filling the possible future fields with null values. The difference is very small (depending on your application) and might be even smaller after MongoDB learn the best padding factor.
Sparse indexes
This section isn't too important to your specific problem right now, but may help you when you face similar problems.
If you create an unique index on field desc, then you wouldn't be able to save more than one document with the same value and in the previous database, we had more than one document with same value on field desc. Let's try to create an unique index in the previous presented database and see what error we get:
db.test.ensureIndex({desc: 1}, {unique: true})
// Output:
{
"err" : "E11000 duplicate key error index: test.test.$desc_1 dup key: { : null }",
"code" : 11000,
"n" : 0,
"connectionId" : 3,
"ok" : 1
}
If we want to be able to create an unique index on some field and let some documents have this field empty, we should create a sparse index. Let's try to create the unique index again:
// No errors this time:
db.test.ensureIndex({desc: 1}, {unique: true, sparse: true})
So far, so good, but why am I explaining all this? Because there is a obscure behaviour about sparse indexes. In the following query, we expect to have ALL documents sorted by desc.
db.test.find().sort({desc: 1})
// Output:
{ "_id" : ObjectId("5192d23f1698aa96f0690d97"), "a" : 1, "desc" : null }
{ "_id" : ObjectId("5192d23b1698aa96f0690d96"), "a" : 1, "desc" : "" }
The result seems weird. What happened to the missing document? Let's try the query without sorting it:
{ "_id" : ObjectId("5192d23b1698aa96f0690d96"), "a" : 1, "desc" : "" }
{ "_id" : ObjectId("5192d23f1698aa96f0690d97"), "a" : 1, "desc" : null }
{ "_id" : ObjectId("5192d2441698aa96f0690d98"), "a" : 1 }
All documents were returned this time. What's happening? It's simple, but not so obvious. When we sort the result by desc, we use the sparse index created previously and there is no entries for the documents that haven't the desc field. The following query show us the use of the index to sort the result:
db.test.find().sort({desc: 1}).explain().cursor
// Output:
"BtreeCursor desc_1"
We can skip the index using a hint:
db.test.find().sort({desc: 1}).hint({$natural: 1})
// Output:
{ "_id" : ObjectId("5192d23f1698aa96f0690d97"), "a" : 1, "desc" : null }
{ "_id" : ObjectId("5192d2441698aa96f0690d98"), "a" : 1 }
{ "_id" : ObjectId("5192d23b1698aa96f0690d96"), "a" : 1, "desc" : "" }
Summary
Sparse unique indexes don't work if you include {desc: null}
Sparse unique indexes don't work if you include {desc: ""}
Sparse indexes might change the result of a query
There is little difference between the null value field and a document without the field. The main difference is that the former consumes a little disk space, while the latter does not consume at all. They can be distinguished by using $exists operator.
The field with an empty string is quite different from them. Though it depends on purpose I don't recommend to use it as a replacement for null. To be precise, they should be used to mean different things. For instance, think about voting. A person who cast a blank ballot is different from a person who wasn't permitted to vote. The former vote is an empty String, while the latter vote is null.
There is already a similar question here.
I can't remove the _id index, why?
When I try running the dropIndexes command, it removes all indexes but not the _id index.
Doing 'db.runCommand' doesn't work either:
> db.runCommand({dropIndexes:'fs_files',index:{_id:1}})
{ "nIndexesWas" : 2, "errmsg" : "may not delete _id index", "ok" : 0 }
not ok.
Can i use a field including _id in a composite index?
I couldn't find anything online, the ensureindex command can't do it.
db.fs_files.ensureIndex({'_id':1, 'created':1});
the above command just created a new composite index. i haven't found some similar 'create Index' command.
the default _id index is a unique index?
the getIndexes returns it's not a unique index.
{
"v" : 1,
"key" : {
"_id" : 1
},
"ns" : "gridfs.fs_files",
"name" : "_id_"
},
{
"v" : 1,
"key" : {
"created" : 1
},
"unique" : true,
"ns" : "gridfs.fs_files",
"name" : "created_1"
}
There is a createIndex command in addition to ensureIndex also.
E.g.
db.<coll>.createIndex({foo:1})
You cannot delete the index on "_id" in mongodb.
Please see the documentation here
When I call ensureIndex from the mongo shell on a collection for a compound index an _id field of type ObjectId is auto-generated in the index object.
> db.system.indexes.find();
{ "name" : "_id_", "ns" : "database.coll", "key" : { "_id" : 1 } }
{ "_id" : ObjectId("4ea78d66413e9b6a64c3e941"), "ns" : "database.coll", "key" : { "a.b" : 1, "a.c" : 1 }, "name" : "a.b_1_a.c_1" }
This makes intuitive sense as all documents in a collection need an _id field (even system.indexes, right?), but when I check the indexes generated by morphia's ensureIndex call for the same collection *there is no _id property*.
Looking at morphia's source code, it's clear that it's calling the same code that the shell uses, but for some reason (whether it's the fact that I'm creating a compound index or indexing an Embedded document or both) they produce different results. Can anyone explain this behavior to me?
Not exactly sure how you managed to get an _id field in the indexes collection but both shell and Morphia originated ensureIndex calls for compound indexes do not put an _id field in the index object :
> db.test.ensureIndex({'a.b':1, 'a.c':1})
> db.system.indexes.find({})
{ "v" : 1, "key" : { "_id" : 1 }, "ns" : "test.test", "name" : "_id_" }
{ "v" : 1, "key" : { "a.b" : 1, "a.c" : 1 }, "ns" : "test.test", "name" : "a.b_1_a.c_1" }
>
Upgrade to 2.x if you're running an older version to avoid running into now resolved issues. And judging from your output you are running 1.8 or earlier.