我设置了一个这样的NamedQuery:
@Entity
@Table(name = "channel")
@NamedQuery(name = "Channel.getPrivateChannel", query = "SELECT pvt from Channel pvt WHERE pvt.state = 3 AND pvt.channelOwnerWorkspace = :channelOwnerWorkspace AND pvt.channelSubscribers = :channelSubscribers")
public class Channel implements Serializable {
...
@Column(nullable = false)
private int type;
@ManyToOne
@JoinColumn(name = "workspaceId")
private Workspace channelOwnerWorkspace;
@ManyToMany(mappedBy = "subscribedChannels", fetch = FetchType.EAGER)
private Set<User> channelSubscribers;所有的变量都不为空并且存在;并且,当调用所述查询时,我会得到一个如下所示的IllegalArgumentException:
Parameter value [pt.project.entity.User@4125ce40] did not match expected type [java.util.Set (n/a)]异常在这里抛出:
public Channel findPrivateChannel1(String workspace, Set<User> channelSubscribers) {
///
Workspace selectedWorkspace = workspaceDAO.findByTitle(workspace);
try {
Channel pvtChannel = em.createNamedQuery("Channel.getPrivateChannel", Channel.class)
.setParameter("channelOwnerWorkspace", selectedWorkspace)
.setParameter("channelSubscribers", channelSubscribers).getSingleResult();我做错了什么?是否可以将HashSet设置为参数?
提前谢谢。
发布于 2020-04-23 05:54:26
根据JSR-338的4.6.7项,等于比较运算符不能与整个集合一起使用:
顺便说一句,你使用两个完整列表进行比较的目的是什么还不清楚。更具体地说,您可以使用一些非常有用的集合表达式来选择您的实体,例如IN、MEMBER OF或IS EMPTY,甚至一些子选择(ANY、some或ALL)。
https://stackoverflow.com/questions/61372818
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