[英]How to write JPA query where parameter is a set?
Assuming the following class, how do you find a Person
with a particular email address? 假设下面的类,你如何找到一个
Person
与一个特定的电子邮件地址?
public class Person implements Comparable<Person> {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name="id")
private long id = 0;
@OneToMany(cascade={CascadeType.PERSIST, CascadeType.MERGE, CascadeType.REMOVE}, fetch=FetchType.LAZY)
private Set<String> email = new HashSet<String>();
}
Is it as simple as doing just this, or is there a proper way? 它就像做这个一样简单,还是有正确的方法?
select p from Person p where p.email=:email
It's not that easy. 那并没那么简单。 JPQL provides the
IN
operator for this: JPQL为此提供了
IN
运算符:
select p from Person p, IN(p.email) m where m = :email
The 'old' way (read SQL-like) would be: “旧”方式(读取类似SQL)将是:
select p from Person p join p.email m where m = :email
The SQL would look something like this: SQL看起来像这样:
WHERE email IN ('foo@yahoo.com', 'bar@gmail.com')
Unfortunately, I'm not aware of an easy way to do this. 不幸的是,我不知道有一种简单的方法可以做到这一点。 If you were doing this with raw SQL, you'd have to do it in two steps: create a bind parameter
?
如果您使用原始SQL执行此操作,则必须分两步执行此操作:创建绑定参数
?
for each value in the set, then iterate through the set and bind each value to its bind parameter. 对于集合中的每个值,然后遍历集合并将每个值绑定到其绑定参数。
I'm not aware of a way to do it cleanly in JPA, but that's what you should be looking for. 我不知道在JPA中干净利落的方法,但这就是你应该寻找的。
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