假设您具有以下一对多关系:School->Student->ScientificWork。现在,您要选择所有学生名为“ John”且其科学工作称为“Black Holes”的学校。
School->Student->ScientificWork
我这样做如下,但是由于某种原因,它使我无法接受所有可能的学校。
public static Specification<School> spec() { return (root, query, cb) -> { final SetJoin<School, Student> studs = root.joinSet("students", JoinType.LEFT); final SetJoin<Student, ScientificWork> works = root.joinSet("works", JoinType.LEFT); return cb.and( cb.equal(studs.get(Student_.name), 'John'), cb.equal(nodes.get(ScientificWork_.name), 'Black Holes') ); }; }
找到此答案后,我尝试了以下操作,但结果相同(它使我返回所有学校而不是所有学校):
public static Specification<School> spec() { return (root, query, cb) -> { final SetJoin<School, Student> studs = root.joinSet("students", JoinType.LEFT); studs.on(cb.equal(studs.get(Student_.name), 'John')); final SetJoin<Student, ScientificWork> works = root.joinSet("works", JoinType.LEFT); return cb.equal(nodes.get(ScientificWork_.name), 'Black Holes'); }; }
public static Specification spec() { return (root, query, cb) -> { final Join studs = root.join(“students”, JoinType.LEFT); studs.on(cb.equal(studs.get(Student_.name), “John”)); final Join works = studs.join(“works”, JoinType.LEFT); return cb.equal(works.get(ScientificWork_.name), “Black Holes”); };
}
我用 join 代替 joinSet, 然后用put**works**.get(ScientificWork_.name)代替**nodes**.get(ScientificWork_.name)
**works**.get(ScientificWork_.name)
**nodes**.get(ScientificWork_.name)