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我正在尝试使用 JPA Criteria API 编写以下 SQL 查询

SELECT * FROM roles WHERE roles.name IN (SELECT users.role FROM users where name="somename");

这对我来说有点过分（我刚刚开始学习 Criteria API）。我得到了这样的东西：

    CriteriaBuilder criteriaBuilder = manager.getCriteriaBuilder();
    CriteriaQuery<RoleEntity> criteriaQuery = criteriaBuilder.createQuery(RoleEntity.class);
    Root<RoleEntity> root = criteriaQuery.from(RoleEntity.class);

    Subquery<UserEntity> subquery = criteriaQuery.subquery(UserEntity.class);
    Root<UserEntity> subqueryRoot = subquery.from(UserEntity.class);
    subquery.where(criteriaBuilder.equal(subqueryRoot.get(UserEntity_.username), username));
    subquery.select(subqueryRoot);

我不知道如何把它们放在一起。

最好的问候， 巴特克

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1 回答 1

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这里是 JPA 学习者。这是我设置它的尝试：

// Get the criteria builder from the entity manager
CriteriaBuilder cb = manager.getCriteriaBuilder();

// Create a new criteria instance for the main query, the generic type indicates overall query results
CriteriaQuery<RoleEntity> c = cb.createQuery(RoleEntity.class);
// Root is the first from entity in the main query
Root<RoleEntity> role = criteriaQuery.from(RoleEntity.class);

// Now setup the subquery (type here is RETURN type of subquery, should match the users.role)
Subquery<RoleEntity> sq = cb.subquery(RoleEntity.class);
// Subquery selects from users
Root<UserEntity> userSQ = sq.from(UserEntity.class);
// Subquery selects users.role path, NOT the root, which is users
sq.select(userSQ.get(UserEntity_.role))
  .where(cb.equal(userSQ.get(UserEntity_.username), username)); // test for name="somename"

// Now set the select list on the criteria, and add the in condition for the non-correlated subquery
c.select(role)
  .where(cb.in(role).value(sq));  // can compare entities directly, this compares primary key identities automatically

希望这会有所帮助！

于 2012-09-25T00:30:23.150 回答