Question

我使用spring mvc。这是我的旧代码，它正确完美。

public List<Details> filter(String search) {
    TypedQuery<Details> query = em.createQuery("SELECT DISTINCT um FROM Details um INNER JOIN um.admin ad WHERE um.name like :search and ad.status =:status", Details.class );
    String searchkey = search;
    String status = "True";
    query.setParameter("search", searchkey + '%');
    query.setParameter("status", status);
    return query.getResultList();
}

在Details类中包含名称为字符串。

public class Details implements Serializable {
@Column(name="name")
private String name;
@OneToMany(cascade={CascadeType.ALL})
@JoinColumn(name ="id")
private List<Email> email= new ArrayList<Email>();
 }

就像我想用搜索作为电子邮件一样，而不是名字我想使用电子邮件。现在我尝试下面的代码，但它显示错误。

public List<Details> filterEmail(String search) {
    System.out.println("Email filtering query..");
    TypedQuery<Details> query = em.createQuery("SELECT DISTINCT um FROM Details um INNER JOIN um.admin ad WHERE elements(um.email) like :search and ad.status =:status", Details.class );
    String status = "True";
    query.setParameter("search",'%' +search+ '%');
    query.setParameter("status", status);
    return query.getResultList();
}

我该如何使用？提前谢谢。

Answer 1

为什么不像这样搜索您的电子邮件对象的邮件属性

SELECT DISTINCT um FROM Details um 
INNER JOIN um.admin ad 
WHERE ad.mailproperty like :search and ad.status =:status

在列表值中如何在hql中使用类似查询？

1 个答案: