多对多和 HQL
我上了一堂课：
@Entity
@Table(name="restaurants")
public class Restaurant {
@Id
@GeneratedValue
private int id;
private String name;
private String street;
(...)
@ManyToMany
@JoinTable(name="user_restaurant_owner",
joinColumns={@JoinColumn(name="restaurant_id")},
inverseJoinColumns={@JoinColumn(name="username")})
private List<User> owner;
如果我知道用户（“所有者”）用户名，如何获取餐厅？
Query q = session.createQuery("from Restaurant as r where r.owner = :username").setString("username", username);
它不起作用
I got class:
@Entity
@Table(name="restaurants")
public class Restaurant {
@Id
@GeneratedValue
private int id;
private String name;
private String street;
(...)
@ManyToMany
@JoinTable(name="user_restaurant_owner",
joinColumns={@JoinColumn(name="restaurant_id")},
inverseJoinColumns={@JoinColumn(name="username")})
private List<User> owner;
How to get Restaurant if I know User ("owner") username?
Query q = session.createQuery("from Restaurant as r where r.owner = :username").setString("username", username);
it doesn;t work
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评论（3）
如果要选择集合中元素的属性，则需要加入 HQL 中的集合。
You need to join to a collection in HQL if you are selecting on properties of the element within the collection.
from Restaurant as r应该是from Restaurant r而不是as（不确定 as 是否是可选的）。另外，您执行
r.owner = ...这应该是r.owner.username = :usernamefrom Restaurant as rshould befrom Restaurant rnoas(not sure if as is optional).Also, you do
r.owner = ...this should ber.owner.username = :usernamer.owner引用 Owners 表的 PRIMARY KEY。你想要的是一个子选择（为了简单起见，这里用 SQL 完成 - 我对 HQL 不太有信心）：
这样，你可以将外键与其引用的字段而不是名称字段相匹配。
r.ownerreferences the PRIMARY KEY of the Owners table.what you want is a subselect (done in SQL here for simplicity - I'm not too confident with HQL):
this way, you match the foreign key with the field it references, rather than the name field.