Question

所以我有一张名为＆＃39; User＆＃39;有字段

用户id
的userName
supervisorId

我想获取特定用户的主管的userName。

List<BitmapImage> Images = new List<BitmapImage>
{
    new BitmapImage(new Uri(@"/Images/Car.bmp", UriKind.Relative)),
};

现在我还有许多其他条件，除了这个supervisorId。所以我必须使用相同的条件查询。

感谢。

Answer 1

Criteria API适用于您需要build queries dynamically时，您应该始终记住它基于遍历实体。

你需要的是一个像这样的简单SQL查询：

select sp.userName
from users u
join users sp on sp.userId = u.supervisorId
where userId = ?

仅仅因为你使用了JPA和Hibernate，it does not mean that you should not use SQL queries。

如果map the supervisor as a @ManyToOne association：

，可以非常轻松地编写Criteria API查询

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name="supervisorId", referencedColumnName="userId")
private User supervisor;

然后，联接变为：

Join<User, User> selfJoin = rootUser.join("supervisor", JoinType.LEFT);

Answer 2

我实现并为我工作的示例。这不是确切的代码，但此代码应该可以工作。我尚未在同一实体中创建引用变量。

SQL

select right.* from MyTable left, MyTable right where left.subjectid = 7 and left.studentid = right.studentid
and left.subjectid != right.subjectid;

条件代码

CriteriaBuilder builder = entityManager.getCriteriaBuilder();
CriteriaQuery<MyTable]> criteria = builder.createQuery(MyTable.class);
Root<MyTable> MyTableRootLeft = criteria.from(MyTable.class);
Root<MyTable> MyTableRootRight = criteria.from(MyTable.class);
Predicate selfJoinPredicate = 
builder.and(
            builder.equal(MyTableRootLeft.get(subjectid), subjectid),
            builder.equal(MyTableRootLeft.get(studentid),MyTableRootRight.get(studentid)),
              builder.notEqual(MyTableRootLeft.get(subjectid),
MyTableRootRight.get(subjectid)));

自我加入条件查询

2 个答案:

SQL

条件代码