SpringJpa:如何获取具有唯一字段或第一个字段的对象列表

问题描述

我有一个名为 Message 的实体类:

I have an Entity class titled Message:

    @Entity
@Table(name = "Message")
public class Message extends DefaultEntity {


    @Valid
    @ManyToOne(targetEntity = User.class)
    @JoinColumn(name="sender")
    private User sender;

    @Valid
    @ManyToOne(targetEntity = User.class)
    @JoinColumn(name="receiver")
    private User receiver;


    @NotNull
    private String message;

    @Enumerated(EnumType.STRING)
    private MessageStatus messageStatus;

    private Boolean isRead;
}

顺便说一下，我删除了 getter 和 setter，所以它不会那么长.

by the way i removed the getters and setters, so it wouldnt be so long.

我想检索 MessageStatus 为READ"且发件人为唯一/最后/最新的聊天列表(因为两个特定用户(发件人和收件人)之间会有很多记录)，我想要根据创建日期进入数据库的最新记录.

I want to retrieve a list of chats where the MessageStatus is "READ" and the Sender is Unique/Last/Latest (because there would be a lot of records between two specific users(sender and receiver)), I want the latest record to enter the DB based on date created.

我试过了，但我被卡住了，因为我不知道如何让应用程序根据创建的日期获取单个发件人；

I tried this, but i am stuck as i dont know how to make the app get a single sender based on the Date created;

List<Message> findMessagesByMessageStatusAndReceiver(Enum.MessageStatus messageStatus, User Receiver);

所以现在我有一些消息实体像这样保存到我的数据库中；

so now i have some message enitity persisted into my database like this;

{date, sender, receiver} , {2/5/7, 1, 2}, {2/6/7, 1, 2}, {2/8/7, 1, 2}, {2/9/7, 1, 2} , {2/5/7, 4, 2}, {2/5/7, 5, 2} , {2/5/7, 10, 2}, {1/5/7, 10, 2}, {2/3/7, 10, 2}

................我想要一种情况，当同一发件人有多个记录时，我可以获取所有消息，但具有唯一记录，因此在此我们有多个具有相同发件人 ID 的记录的情况:{2/5/7, 1, 2}, {2/6/7, 1, 2}, {2/8/7, 1, 2},{2/9/7, 1, 2} 仅根据最近的日期选择一个并添加到列表中，因此选择 {2/9/7, 1, 2} 以及此 {2/5/7, 10, 2}, {1/5/7, 10, 2}, {2/3/7, 10, 2} 这个 {2/5/7, 10, 2} 被选中.所以在一天结束时，我的列表是 {2/9/7, 1, 2}, {2/5/7, 4, 2}, {2/5/7, 5, 2}, {2/5/7, 10, 2}

................i want a situation where i can get all the messages but with a unique record when there are more than one record with the same sender, so in this situation where we have more that one record with the same sender ID : {2/5/7, 1, 2}, {2/6/7, 1, 2}, {2/8/7, 1, 2}, {2/9/7, 1, 2} only one gets picked based on the recent date and gets added to the list, so {2/9/7, 1, 2} gets picked and also for this {2/5/7, 10, 2}, {1/5/7, 10, 2}, {2/3/7, 10, 2} this {2/5/7, 10, 2} is picked. so at the end of the day my list is {2/9/7, 1, 2},{2/5/7, 4, 2}, {2/5/7, 5, 2}, {2/5/7, 10, 2}

推荐答案

您可以使用两种查询方法，findFirstBy.. 和 findTopBy...这两个关键字用于限制查询的结果.

There are two query methods that you can use for this, the findFirstBy.. and findTopBy... These two keywords are used to limit the results of a query.

您可以查看参考 示例.

编辑

做一个这样的查询:

消息 findFirstBySenderOrderByDate(Sender sender)

通过它您可以找到发件人的最新记录.然后，您可以从外部循环所有发件人，并为每个发件人调用查询方法.

with which you find most recent record for a sender. Then, from outside you can loop all senders, and call the query method for each one.

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