java - 继续收到“java.lang.IllegalArgumentException：没有为该名称定义查询”，即使我有具有适当名称的@NamedQuery

Question

我有休眠 JPA 应用程序。我在 DAO 类的函数中不断收到“java.lang.IllegalArgumentException: No query defined for that name [Singer.findAll]”，即使我在 @NamedQuery 中正确定义了名称。这是我的实体类

package ch8.entities;

import javax.persistence.*;
import java.io.Serializable;
import java.util.Date;
import java.util.HashSet;
import java.util.Set;

@Entity
@Table(name = "singer")
@NamedQueries({
        @NamedQuery(name = Singer.FIND_BY_ID,
                query = "select distinct s from Singer s "
                        + "left join fetch s.albums a "
                        + "left join fetch s.instruments i "
                        + "where s.id = :id"),

        @NamedQuery(name = Singer.FIND_ALL_WITH_ALBUM,
                query = "select distinct s from Singer s "
                        + "left join fetch s.albums a "
                        + "left join fetch s.instruments i"),

        @NamedQuery(name = Singer.FIND_ALL,
                query = "select s from Singer s")
})
@SqlResultSetMapping(name = "singerResult",
        entities = @EntityResult(entityClass = Singer.class))
public class Singer implements Serializable {
    private Long id;
    private String firstName;
    private String lastName;
    private Date birthDate;
    private Long version;
    private Set<Album> albums = new HashSet<>();
    private Set<Instrument> instruments = new HashSet<>();
    public static final String FIND_ALL = "Singer.findAll";
    public static final String FIND_BY_ID = "Singer.findById";
    public static final String FIND_ALL_WITH_ALBUM = "Singer.findAllWithAlbum";

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "ID")
    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    @Column(name = "FIRST_NAME")
    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    @Column(name = "LAST_NAME")
    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    @Temporal(TemporalType.DATE)
    @Column(name = "BIRTH_DATE")
    public Date getBirthDate() {
        return birthDate;
    }

    public void setBirthDate(Date birthDate) {
        this.birthDate = birthDate;
    }

    @Version
    @Column(name = "VERSION")
    public Long getVersion() {
        return version;
    }

    public void setVersion(Long version) {
        this.version = version;
    }

    @OneToMany(mappedBy = "singer", cascade = CascadeType.ALL, orphanRemoval = true)
    public Set<Album> getAlbums() {
        return albums;
    }

    public void setAlbums(Set<Album> albums) {
        this.albums = albums;
    }

    @ManyToMany
    @JoinTable(name = "singer_instrument",
    joinColumns = @JoinColumn(name = "SINGER_ID"),
    inverseJoinColumns = @JoinColumn(name = "INSTRUMENT_ID"))
    public Set<Instrument> getInstruments() {
        return instruments;
    }

    public void setInstruments(Set<Instrument> instruments) {
        this.instruments = instruments;
    }

    @Override
    public String toString () {
        return "Singer - Id: " + id + ", First name: "
                + firstName + ", Last name: " + lastName
                + ", Birthday: " + birthDate;
    }

    public boolean addAlbum(Album album) {
        if (albums == null) {
            albums = new HashSet<Album>();
            albums.add(album);
            return true;
        }
        if (albums.contains(album))
            return false;
        albums.add(album);
        return true;
    }

    public boolean removeAlbum(Album album) {
        if (album == null)
            return false;
        if (!albums.contains(album))
            return false;
        albums.remove(album);
        return true;
    }
}

和 DAO 类

package ch8.dao;

import ch8.entities.Singer;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
import org.springframework.stereotype.Repository;
import org.springframework.transaction.annotation.Transactional;

import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import java.util.Collection;

@Transactional
@Repository("singerDao")
public class SingerDao implements Dao<Singer> {
    @PersistenceContext
    private EntityManager entityManager;

...some other functions...

    @Override
    @Transactional(readOnly = true)
    public Collection<Singer> findAll() {
        return entityManager.createNamedQuery(Singer.FIND_ALL, Singer.class).getResultList();
    }

...some other functions...

}

我建议它可能与类映射有关，因为常规 createQuery(String, Class) 也不起作用。

标签： javahibernatejpa

Answer 1

感谢 JLazar0 的建议，我找到了解决问题的方法。与往常一样，问题不是因为错误的逻辑、错误配置或持久性。尽管一切似乎都很好，但我一直收到“没有为该名称定义查询”的原因是因为我在 LocalContainerEntityManagerFactoryBean.setPackagesToScan() 中的包名称错误。在方法参数的小修正后，我的代码通过了所有测试。无论如何，感谢所有试图提供帮助的人