[英]How to use projection and where clause in JPA criteria?
我有實體人
@Entity(name = "Person")
public class Person {
@Id
@GeneratedValue
private Long id;
private String name;
@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy = "person")
private Set<Phone> phones=new HashSet<Phone>();
public Person() {
}
public Person(String name) {
this.name = name;
}
廣告實體電話:
@Entity(name = "Phone")
public class Phone {
@Id
@GeneratedValue
private Long id;
@Column(name = "`number`")
private String number;
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "person_id", nullable = false)
private Person person;
public Phone() {
}
他們有一對多的關系。 現在,我想在jpa標准中建立這樣的查詢:
select p.phones from person p join phone ph where p.name = :name;
因此,我想從人員名稱為參數的人員實體中提取Set<Phone> phones
。
我已經編寫了這個jpa條件查詢:
CriteriaBuilder builder = session.getCriteriaBuilder();
CriteriaQuery<Person> query = builder.createQuery(Person.class);
Root<Person> root = query.from(Person.class);
CriteriaQuery<Person> where = query.where(builder.equal(root.get("name"), "Mary Dick"));
CompoundSelection<Set> projection = builder.construct(Set.class, root.get("phones"));
where.select(projection); //compile error: The method select(Selection<? extends Person>) in the type CriteriaQuery<Person> is not applicable for the arguments (CompoundSelection<Set>)
}
但是它會產生編譯錯誤:
The method select(Selection<? extends Person>) in the type CriteriaQuery<Person> is not applicable for the arguments (CompoundSelection<Set>)
正確嗎? 我需要元模型類嗎?
CompoundSelection<Y> construct(Class<Y> result, Selection<?>... terms)
僅當查詢涉及某些未完全由單個實體類封裝的投影時,此方法才有用。 在這種情況下,第一個參數將是自定義POJO類(具有適當的構造函數),該類具有與查詢的select子句相對應的字段。
在這種情況下,選擇已經是實體類的一部分。 因此,您只需選擇所需的字段即可。
CriteriaQuery<Person> query = builder.createQuery(Person.class);
Root<Person> root = query.from(Person.class);
query.where(builder.equal(root.get("name"), "Mary Dick"));
query.select(root.get("phones"));
上面的查詢將返回人員列表。 但是,如果您僅查找可迭代的電話列表,請嘗試使用稍有不同的查詢。
select ph from phone ph join ph.person p where p.name = :name;
及其等效的CriteriaQuery:
CriteriaBuilder builder = session.getCriteriaBuilder();
CriteriaQuery<Phone> query = builder.createQuery(Phone.class);
Root<Phone> root = query.from(Phone.class);
Join<Phone, Person> join = root.join(root.get("person"))
query.where(builder.equal(join.get("name"), "Mary Dick"));
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