Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
537 views
in Technique[技术] by (71.8m points)

hibernate - Dynamic Queries in Spring Data JPA

I am looking for a solution to dynamically build queries using Spring Data JPA. I have a GameController which has a RESTful service endpoint /games which takes 4 optional parameters: genre, platform, year, title. The API may be passed none of those, all 4, and every combination in between. If any parameter is not passed it defaults to null. I need a method in the Repository that will build the appropriate query and ideally also still allow Spring Data JPA Paging, although I'm not sure if that is possible.

I found this article but this doesn't seem to be what I need unless I am misunderstanding. http://spring.io/blog/2011/04/26/advanced-spring-data-jpa-specifications-and-querydsl/

I know JPA has a Query Criteria API but really have no idea how to implement this.

I realize I could create a method for each possible scenario but that seems like really bad practice and a lot of unnecessary code.

GameRepository:

package net.jkratz.igdb.repository;

import net.jkratz.igdb.model.Game;
import org.springframework.data.domain.Page;
import org.springframework.data.domain.Pageable;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.jpa.repository.Query;
import org.springframework.data.repository.query.Param;

public interface GameRepository extends JpaRepository<Game, Long> {

    @Query("select g from Game g, GamePlatformMap gpm, Platform p where g = gpm.game and gpm.platform = p and p.id = :platform")
    Page<Game> getGamesByPlatform(@Param("platform") Long platformId, Pageable pageable);

    @Query("select g from Game g where g.title like :title")
    Page<Game> getGamesByTitle(@Param("title") String title, Pageable pageable);

    @Query("select g from Game g, GameGenreMap ggm, Genre ge where g = ggm.game and ggm.genre = ge and ge.id = :genreId")
    Page<Game> getGamesByGenre(@Param("genre") Long genreId, Pageable pageable);
}
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I would say that using QueryDSL is one way of doing what you want.

For example I have a repository defined as below:

public interface UserRepository extends PagingAndSortingRepository<User, Long>, QueryDslPredicateExecutor<User> {

    public Page<User> findAll(Predicate predicate, Pageable p);
}

I can call this method with any combination of parameters, like below:

public class UserRepositoryTest{

    @Autowired
    private UserRepository userRepository;

    @Test
    public void testFindByGender() {
        List<User> users = userRepository.findAll(QUser.user.gender.eq(Gender.M));
        Assert.assertEquals(4, users.size());

        users = userRepository.findAll(QUser.user.gender.eq(Gender.F));
        Assert.assertEquals(2, users.size());
    }

    @Test
    public void testFindByCity() {

        List<User> users = userRepository.findAll(QUser.user.address.town.eq("Edinburgh"));
        Assert.assertEquals(2, users.size());

        users = userRepository.findAll(QUser.user.address.town.eq("Stirling"));
        Assert.assertEquals(1, users.size());
    }

    @Test
    public void testFindByGenderAndCity() {
        List<User> users = userRepository.findAll(QUser.user.address.town.eq("Glasgow").and(QUser.user.gender.eq(Gender.M)));
        Assert.assertEquals(2, users.size());

        users = userRepository.findAll(QUser.user.address.town.eq("Glasgow").and(QUser.user.gender.eq(Gender.F)));
        Assert.assertEquals(1, users.size());
    }
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...