Mongodb avoid duplicate entries

问题

I am newbie to mongodb. May I know how to avoid duplicate entries. In relational tables, we use primary key to avoid it. May I know how to specify it in Mongodb using java?

回答1:

Use an index with the {unique:true} option.

// everyone's username must be unique:
db.users.createIndex({email:1},{unique:true});

You can also do this across multiple fields. See this section in the docs for more details and examples.

MongoDB indexes may optionally impose a unique key constraint, which guarantees that no documents are inserted whose values for the indexed keys match those of an existing document.

If you wish for null values to be ignored from the unique key, then you have to also make the index sparse (see here), by also adding the sparse option:

// everyone's username must be unique,
//but there can be multiple users with no email field or a null email:
db.users.createIndex({email:1},{unique:true, sparse:true});

If you want to create the index using the MongoDB Java Driver. Try:

Document keys = new Document("email", 1);
collection.createIndex(keys, new IndexOptions().unique(true));

回答2:

This can be done using "_id" field although this use is discouraged. suppose you want the names to be unique, then you can put the names in "_id" column and as you might know "_id" column is unique for each entry.

BasicDBObject bdbo = new BasicDBObject("_id","amit");

Now , no other entry can have name as "amit" in the collection.This can be one of the way you are asking for.

回答3:

As of Mongo's v3.0 Java driver, the code to create the index looks like:

public void createUniqueIndex() {
    Document index = new Document("fieldName", 1);
    MongoCollection<Document> collection = client.getDatabase("dbName").getCollection("CollectionName");
    collection.createIndex(index, new IndexOptions().unique(true));
}

// And test to verify it works as expected
@Test
public void testIndex() {
    MongoCollection<Document> collection = client.getDatabase("dbName").getCollection("CollectionName");

    Document newDoc = new Document("fieldName", "duplicateValue");
    collection.insertOne(newDoc);

    // this will throw a MongoWriteException
    try {
        collection.insertOne(newDoc);
        fail("Should have thrown a mongo write exception due to duplicate key");
    } catch (MongoWriteException e) {
        assertTrue(e.getMessage().contains("duplicate key"));
    }
}

回答4:

Theon solution didn't work for me, but this one did:

BasicDBObject query = new BasicDBObject(<fieldname>, 1);
collection.ensureIndex(query, <index_name>, true);

回答5:

I am not a Java programmer however you can probably convert this over.

MongoDB by default does have a primary key known as the _id you can use upsert() or save() on this key to prevent the document from being written twice like so:

var doc = {'name': 'sam'};
db.users.insert(doc); // doc will get an _id assigned to it
db.users.insert(doc); // Will fail since it already exists

This will stop immediately duplicates. As to multithread safe inserts under certain conditions: well, we would need to know more about your condition in that case.

I should add however that the _id index is unqiue by default.

回答6:

using pymongo it looks like:

mycol.create_index("id", unique=True)

where myCol is the collection in the DB

#!/usr/bin/env python
# -*- coding: utf-8 -*-
import pymongo

myclient = pymongo.MongoClient("mongodb://localhost:27017/")
mydb = myclient["mydatabase"]
mycol = mydb["customers"]
mycol.create_index("id", unique=True)
mydict = {"name": "xoce", "address": "Highway to hell 666", "id": 1}
x = mycol.insert_one(mydict)

来源：https://stackoverflow.com/questions/12191311/mongodb-avoid-duplicate-entries

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