I'm building a simple REST api with spring-boot-starter-data-mongodb and I always get a E11000 duplicate key error
when attempting to insert my 2nd row.
Spring's getting started guide has a pretty simple configuration that I followed, but I must be missing something.
I've dropped the collection, and started fresh, the 1st document saves fine, but the second one tries to save as id=0 as well. How do I get Spring/Mongo to increment properly?
Here's the error I'm getting:
org.springframework.dao.DuplicateKeyException: { "serverUsed" : "localhost:27017" , "ok" : 1 , "n" : 0 , "err" : "E11000 duplicate key error index: test.game.$_id_ dup key: { : 0 }" , "code" : 11000}; nested exception is com.mongodb.MongoException$DuplicateKey: { "serverUsed" : "localhost:27017" , "ok" : 1 , "n" : 0 , "err" : "E11000 duplicate key error index: test.game.$_id_ dup key: { : 0 }" , "code" : 11000}
Game
package com.recursivechaos.boredgames.domain;
import org.springframework.data.annotation.Id;
import org.springframework.data.mongodb.core.mapping.Document;
@Document
public class Game {
@Id
private long id;
private String title;
private String description;
public Game() {
}
public long getId() {
return id;
}
public String getTitle() {
return title;
}
public void setTitle(String title) {
this.title = title;
}
public String getDescription() {
return description;
}
public void setDescription(String description) {
this.description = description;
}
}
Game Repository
package com.recursivechaos.boredgames.repository;
import com.recursivechaos.boredgames.domain.Game;
import org.springframework.data.mongodb.repository.MongoRepository;
import org.springframework.data.repository.query.Param;
import java.util.List;
public interface GameRepository extends MongoRepository<Game, Long> {
List<Game> findByTitle(@Param("title") String title);
}
AppConfig
package com.recursivechaos.boredgames.configuration;
import com.mongodb.Mongo;
import org.springframework.context.annotation.Bean;
import org.springframework.data.authentication.UserCredentials;
import org.springframework.data.mongodb.MongoDbFactory;
import org.springframework.data.mongodb.core.MongoTemplate;
import org.springframework.data.mongodb.core.SimpleMongoDbFactory;
public class AppConfig {
public
@Bean
MongoDbFactory mongoDbFactory() throws Exception {
UserCredentials userCredentials = new UserCredentials("username", "password");
SimpleMongoDbFactory boredgamesdb = new SimpleMongoDbFactory(new Mongo(), "boredgamesdb", userCredentials);
return boredgamesdb;
}
public
@Bean
MongoTemplate mongoTemplate() throws Exception {
return new MongoTemplate(mongoDbFactory());
}
}
Thanks for looking!
You'r using a primitive long
which has an implicit, pre-assigned value. Hence, that value is passed to MongoDB and it persists it as is as it assumes you'd like to define identifiers manually.
The trick is to just use a wrapper type, as this can be null
, MongoDB detects the absence of the value and will auto-populate an ObjectID
for you. However, Long
is not going to work as ObjectID
s do not fit into the number space of a Long
. The id types supported for auto-generation are ObjectId
, String
and BigInteger
.
All of this is documented in the reference documentation.