嘿,大家,我有一个关系太多的类,我正在使用springboot,当我想通过Hql从@Query中提取数据时,我得到了数组的结果,当我使用NativeQuery时,我不能选择一个元素,它总是选择*
这是我的课
import java.util.HashSet;
import java.util.Set;
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.JoinTable;
import javax.persistence.ManyToMany;
import javax.persistence.ManyToOne;
import javax.persistence.Table;
@Entity
@Table(name="\"DEV\"")
public class Dev {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
@Column(name="\"IdDev\"")
private int id ;
@Column(name="\"NomDev\"")
private String nomdev;
@Column(name="\"NomDLL\"")
private String dll ;
@ManyToOne(fetch=FetchType.LAZY, cascade = CascadeType.ALL)
@JoinColumn(name="\"IdEtatDev\"")
private EtatDev etatdev ;
@ManyToOne(fetch=FetchType.LAZY, cascade = CascadeType.ALL)
@JoinColumn(name="\"IdEcu\"")
private Ecu ecu ;
@ManyToMany(fetch=FetchType.LAZY)
@JoinTable(name="\"VEH_BY_DEV\""
,joinColumns={@JoinColumn(name="\"IdDev\"")},
inverseJoinColumns={@JoinColumn(name="\"GRPMOD\"")})
private Set<Vehid> vehid = new HashSet<>();
@ManyToOne(fetch=FetchType.LAZY)
private Maj maj ;
public Set<Vehid> getVehid() {
return vehid;
}
public void setVehid(Set<Vehid> vehid) {
this.vehid = vehid;
}
public Ecu getEcu() {
return ecu;
}
public void setEcu(Ecu ecu) {
this.ecu = ecu;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getNomdev() {
return nomdev;
}
public void setNomdev(String nomdev) {
this.nomdev = nomdev;
}
public String getDll() {
return dll;
}
public void setDll(String dll) {
this.dll = dll;
}
public EtatDev getEtatdev() {
return etatdev;
}
public void setEtatdev(EtatDev etatdev) {
this.etatdev = etatdev;
}
public Dev() {
super();
}
public Dev(int id, String nomdev, String dll, EtatDev etatdev, Ecu ecu, Set<Vehid> vehid) {
super();
this.id = id;
this.nomdev = nomdev;
this.dll = dll;
this.etatdev = etatdev;
this.ecu = ecu;
this.vehid = vehid;
}
}这是我的存储库类
import java.util.List;
import javax.transaction.Transactional;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.jpa.repository.Query;
@Transactional
public interface DevRepository extends JpaRepository<Dev, Integer> {
@Query(value="SELECT \"IdDev\" "\NomDev\" FROM \"DEV\" ",nativeQuery=true)
List<Dev> getDevwithEtat() ;
}执行此操作时,未找到saye ERROR SQL "NomDLL“列。
好像只允许选择all或not,可能是结果列表,所以它不能返回一个字段,所以如果有人想要更改结果类型,那么它就可以得到两个或三个字段的结果,而不是全部考虑。请随意回答( ps )是逃避,当我添加"\NomDev\“时,我得到了一个错误
找不到错误的SQL "IdEtatDev“)
如能提供任何帮助,将不胜感激。
发布于 2016-04-08 13:21:16
首先用这个List<Dev> getDevwithEtat();改变List<Object[]> getDevwithEtat();
在转换json之后,您的devRepository.getDevwithEtat()结果如下
Map<Integer, String> map = new HashMap<Integer, String>();
for (Object[] result : devRepository.getDevwithEtat())
map.put((Integer)result[0], (String)result[1]);与Google Gson转换为json
String json = new Gson().toJson(map);https://stackoverflow.com/questions/36498538
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