Java 将jpa(Hibernate)中的本机(join)查询转换为json
我正在从事一个SpringBoot项目,并使用jpa进行数据持久化。现在我有两个相互关联的表,用户和项目。一个用户可以拥有任意数量的项目,而一个项目只能由一个用户拥有。Java 将jpa(Hibernate)中的本机(join)查询转换为json,java,json,hibernate,spring-mvc,jpa,Java,Json,Hibernate,Spring Mvc,Jpa,我正在从事一个SpringBoot项目,并使用jpa进行数据持久化。现在我有两个相互关联的表,用户和项目。一个用户可以拥有任意数量的项目,而一个项目只能由一个用户拥有。 这是我的POJO: 用户 @Entity @Table(name="users", uniqueConstraints = { @UniqueConstraint(columnNames = { "email" }) }) @EntityLi
这是我的POJO:
用户
@Entity
@Table(name="users", uniqueConstraints = {
@UniqueConstraint(columnNames = {
"email"
})
})
@EntityListeners(AuditingEntityListener.class)
@JsonIgnoreProperties(value = {"createdAt", "updatedAt"}, allowGetters = true)
public class User {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@NotBlank
@NaturalId
@Column(unique = true)
private String email;
@NotBlank
@JsonIgnore
private String password;
@NotBlank
private String first_name;
@NotBlank
private String last_name;
@OneToMany(cascade = CascadeType.REMOVE, orphanRemoval = true)
@JsonIgnore
private Set<Item> items;
@ManyToMany(fetch = FetchType.LAZY)
@JoinTable(name = "user_roles",
joinColumns = @JoinColumn(name = "user_id"),
inverseJoinColumns = @JoinColumn(name = "role_id"))
private Set<Role> roles = new HashSet<>();
@Column(nullable = false, updatable = false)
@Temporal(TemporalType.TIMESTAMP)
@CreatedDate
private Date createdAt;
@Column(nullable = false)
@Temporal(TemporalType.TIMESTAMP)
@LastModifiedDate
private Date updatedAt;
@Entity
@Table(name="items")
@EntityListeners(AuditingEntityListener.class)
@JsonIgnoreProperties(value = {"createdAt", "updatedAt"}, allowGetters = true)
public class Item {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@NotBlank
@NotNull
private String name;
private String description;
@NotNull
private Date purchase_date;
private double price;
@ManyToOne(fetch = FetchType.LAZY, optional = true)
@JoinColumn(name = "owner", nullable = true)
private User owner;
@Column(nullable = false, updatable = false)
@Temporal(TemporalType.TIMESTAMP)
@CreatedDate
private Date createdAt;
@Column(nullable = false)
@Temporal(TemporalType.TIMESTAMP)
@LastModifiedDate
private Date updatedAt;
现在我想将所有项目都作为restfulljson获取。我需要使用项目加入用户,以便获得以下JSON:
{
"item_id":1,
"name":"item1",
"price":120.00,
etc .....
"owner_id":1,
"owner_name":"Jon"
etc ....
}
因此,我使用的是一个自定义本机查询
SELECT i.id, i.name, i.description ...... u.id, u.name ..... from items i , users u where i.owner = u.id
然后我返回query.getResultList()
,但是这会返回字符串的数组,而不是像这样的json
[
[ 1 , "item1" , 120.00 , ..... , 1 , "Jon" , ....]
[ 2 , "item2" , 420.00 .... ]
etc...
]
如何将返回的对象直接转换为JSON,或转换为将列名映射到值的映射列表,然后将其转换为JSON 可以使用构造函数表达式创建包含所需数据的DTO(数据传输对象)
package dto;
public class ItemDTO {
private final Long item_id;
private final String name;
private final Long owner_id;
private final String owner_name;
public ItemDTO(Long item_id, String name, Long owner_id, String owner_name) {
// set the fields
}
// getters
}
然后在构造函数表达式查询中使用此DTO(重要说明:这仅适用于JPQL查询,而不是本机查询)
此DTO可用于序列化为JSON
请在此处阅读有关Constructor表达式的更多信息:
您可以使用此功能,无需在每次请求时都使用新的DTO
@SuppressWarnings("unchecked")
public static List<ObjectNode> getQueryResult(EntityManager entityManager, String nativeQuery, Object... parameters) {
Query localQuery = entityManager.createNativeQuery(nativeQuery,Tuple.class);
for(int i = 0; i < parameters.length; i++)
localQuery.setParameter(i+1, parameters[i]);
return toJson(localQuery.getResultList());
}
private static List<ObjectNode> toJson(List<Tuple> results) {
List<ObjectNode> json = new ArrayList<>();
ObjectMapper mapper = new ObjectMapper();
for (Tuple tuple : results)
{
List<TupleElement<?>> cols = tuple.getElements();
ObjectNode node = mapper.createObjectNode();
for (TupleElement col : cols)
node.put(col.getAlias(), tuple.get(col.getAlias()).toString());
json.add(node);
}
return json;
}
@SuppressWarnings(“未选中”)
公共静态列表getQueryResult(EntityManager EntityManager、字符串nativeQuery、对象…参数){
Query localQuery=entityManager.createNativeQuery(nativeQuery,Tuple.class);
对于(int i=0;i 列表他还必须删除“nativeQuery=true”因为您给出的示例是一个JPQL表达式。另外,我会检索整个对象,然后使用MapStruct…@dbl之类的映射器来构造DTO。从性能角度来看,构造函数表达式比读取实体和映射到DTO要好得多,因为它只会执行一条SQL语句,而您不会有n+1炸弹。这是真的,特别是对于最终甚至需要分页的大型列表…但是如果列表像几个简单对象一样计数,并且请求不太频繁,那么我看不到任何好处。但是这是否意味着我需要为我所做的每个查询编写一个新的DTO类?
@SuppressWarnings("unchecked")
public static List<ObjectNode> getQueryResult(EntityManager entityManager, String nativeQuery, Object... parameters) {
Query localQuery = entityManager.createNativeQuery(nativeQuery,Tuple.class);
for(int i = 0; i < parameters.length; i++)
localQuery.setParameter(i+1, parameters[i]);
return toJson(localQuery.getResultList());
}
private static List<ObjectNode> toJson(List<Tuple> results) {
List<ObjectNode> json = new ArrayList<>();
ObjectMapper mapper = new ObjectMapper();
for (Tuple tuple : results)
{
List<TupleElement<?>> cols = tuple.getElements();
ObjectNode node = mapper.createObjectNode();
for (TupleElement col : cols)
node.put(col.getAlias(), tuple.get(col.getAlias()).toString());
json.add(node);
}
return json;
}