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java - JPA/JPQL JOIN 子选择/子查询

转载 作者:搜寻专家 更新时间:2023-11-01 03:20:41 24 4
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我无法将一些简单的 SQL 语句转换为 JPQL,因为使用了 JPQL 不支持的子查询。

谁能给我一个提示,如何使用 JPQL 或 JPA2 Criteria API 获得相同的结果?

给定(简化的假数据来证明问题):

CREATE TABLE person (id integer, name text);
CREATE TABLE phone (id integer, person_id integer, type text, number text);

INSERT INTO person VALUES (1, "John");
INSERT INTO person VALUES (2, "Mike");
INSERT INTO person VALUES (3, "Paul");
INSERT INTO person VALUES (4, "Walter");

INSERT INTO phone VALUES (1, 1, "MOBILE", "+49-123-11111");
INSERT INTO phone VALUES (2, 1, "HOME" , "+49-123-22222");
INSERT INTO phone VALUES (3, 2, "WORK" , "+49-123-33333");
INSERT INTO phone VALUES (4, 4, "MOBILE", "+49-123-44444");

-- Select all from person and their mobile number if possible
-- This query has to be translated to JPQL

SELECT person.name, mobile.number FROM person LEFT JOIN (
SELECT * FROM phone WHERE type = "MOBILE"
) AS mobile ON person.id = mobile.person_id;

预期结果:

| name   | number        |
|--------|---------------|
| John | +49-123-11111 |
| Mike | |
| Paul | |
| Walter | +49-123-44444 |

Java:

class Person {
String name;
List<Phone> phones;
}

class Phone {
String type;
String number;
}

JPQL(没有按预期工作 :-( ):

SELECT person.name, phone.number FROM Person person
LEFT JOIN person.phones AS phone
WHERE phone.type = "MOBILE"

最佳答案

SELECT person.name, phone.number 
FROM Person AS person LEFT JOIN person.phones AS phone
ON phone.type = 'MOBILE'

您还可以将 ON 关键字替换为 hibernate 特定的 WITH:

SELECT person.name, phone.number 
FROM Person AS person LEFT JOIN person.phones AS phone
WITH phone.type = 'MOBILE'

关于java - JPA/JPQL JOIN 子选择/子查询,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31393246/

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