[英]Convert native (join) queries in jpa (Hibernate) to json
我正在開發一個Spring Boot項目,並且正在使用jpa進行數據持久化。 現在,我有兩個相互關聯的表:用戶和項目。 一個用戶可以擁有任意數量的項目,而一個項目只能由一個用戶擁有。
這是我的pojos:
用戶
@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;
項目
@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;
現在,我想將所有項目都獲取為RESTfull JSON。 我需要為用戶添加項目,以便獲得以下JSON:
{
"item_id":1,
"name":"item1",
"price":120.00,
etc .....
"owner_id":1,
"owner_name":"Jon"
etc ....
}
所以我正在使用自定義的本機查詢
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()
但是這返回的是字符串數組,而不是像這樣的json
[
[ 1 , "item1" , 120.00 , ..... , 1 , "Jon" , ....]
[ 2 , "item2" , 420.00 .... ]
etc...
]
如何將返回的對象直接轉換為JSON或映射到將列名映射到值的映射列表,然后將其轉換為JSON?
您可以使用構造函數表達式創建一個包含所需數據的DTO(數據傳輸對象)。
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
}
然后在構造函數表達式查詢中使用此DTO(重要說明:這僅適用於JPQL查詢而不適用於本機查詢)
SELECT new dto.ItemDTO(i.id, i.name, i.owner.id, i.owner.name) from Item i where i.owner.id = u.id
此DTO可用於序列化為JSON。
在此處閱讀有關構造函數表達式的更多信息: https : //vladmihalcea.com/the-best-way-to-map-a-projection-query-to-a-dto-with-jpa-and-hibernate/
您可以使用它,不需要在每個請求上都使用新的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;
}
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