jpa标准按子字段排序

yftpprvb  于 2021-07-13  发布在  Java
关注(0)|答案(1)|浏览(579)

我有三个实体。客户、流程和文件。
一个客户有很多流程,一个流程有很多文档。
我想按文档的更新日期对客户进行排序。
我的实体如下;
顾客;

@Entity
public class Customer {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    private String name;

    @OneToMany(mappedBy = "customer", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
    private List<Process> processes = new ArrayList<>();

    // getter, setter etc.

}

过程;

@Entity
public class Process {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    private String type;

    @ManyToOne(fetch = FetchType.LAZY)
    private Customer customer;

    @OneToMany(mappedBy = "process", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
    private List<Document> documents = new ArrayList<>();

    //getter, setter etc.

}

文件;

@Entity
public class Document {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    private String note;

    private LocalDateTime updateDate;

    @ManyToOne(fetch = FetchType.LAZY)
    private Process process;

}

我试过那种规格;

public static Specification<Customer> orderByDocumentUploadDate() {
        return (root, query, criteriaBuilder) -> {
            ListJoin<Customer, Process> processJoin = root.join(Customer_.processes);
            ListJoin<Process, Document> documentJoin = processJoin.join(Process_.documents);

            query.orderBy(criteriaBuilder.desc(documentJoin.get(Document_.updateDate)));
            query.distinct(true);
            return null;
        };
    }

它给出了错误;
错误:对于select distinct,order by表达式必须出现在选择列表中
生成sql;

select distinct customer0_.id   as id1_0_,
                customer0_.name as name2_0_
from customer customer0_
         inner join
     process processes1_ on customer0_.id = processes1_.customer_id
         inner join
     document documents2_ on processes1_.id = documents2_.process_id
order by documents2_.update_date desc
limit ?

我也尝试过分组。如下图所示;

public static Specification<Customer> orderByDocumentUploadDate() {
    return (root, query, criteriaBuilder) -> {
        ListJoin<Customer, Process> processJoin = root.join(Customer_.processes);
        ListJoin<Process, Document> documentJoin = processJoin.join(Process_.documents);

        query.orderBy(criteriaBuilder.desc(documentJoin.get(Document_.updateDate)));
        query.groupBy(root.get(Customer_.id));
        return null;
    };
}

比它给的错误;
错误:列“documents2\u0.update\u date”必须出现在group by子句中或在聚合函数中使用
并生成sql;

select
    customer0_.id as id1_0_,
    customer0_.name as name2_0_ 
from
    customer customer0_ 
inner join
    process processes1_ 
        on customer0_.id=processes1_.customer_id 
inner join
    document documents2_ 
        on processes1_.id=documents2_.process_id 
group by
    customer0_.id 
order by
    documents2_.update_date desc limit ?

我可以用那个sql来做;max()在sql中解决了这个问题

select  customer.* from customer
inner join process p on customer.id = p.customer_id
inner join document d on p.id = d.process_id
group by customer.id
order by max(d.update_date);

但是我不能通过criteriaapi来模拟它。
我怎样才能解决那个问题?你有什么建议吗?

ql3eal8s

ql3eal8s1#

这是一个概念上的误解。
首先,你必须了解内在是如何与工作者结合的。在这种情况下这部分是可以的[加入 process 带的表格 document 表基于 document.process_id = process.id ]
其次,您需要根据文档的更新日期对客户进行排序。
不幸的是,你用 group by 在这里。 GROUP BY 只返回它所在的列 grouped by “在这种情况下,它只会返回 customer id .
此外,还可以使用聚合函数,如count()、sum()。。。在分组数据上。
所以,当你试图进入 update_date ,它将通过下面的错误。

ERROR: column "documents2_.update_date" must appear in the GROUP BY clause or be used in an aggregate function

现在我们怎样才能骑上它:
因此,首先我们需要加入以获取客户id。在获取客户id之后,我们应该根据客户id对数据进行分组,然后使用max()获取每个组的max\u日期(如果需要,则使用minimum)

SELECT 
   customer_id,
   max(date) AS max_date
FROM    
   document 
   JOIN process ON process.id = document.process_id
GROUP BY customer_id

它将返回一个临时表,如下所示:
客户编号:12020-10-2422021-03-1532020-09-2442020-03-15
使用临时表,您现在可以按日期对customer\u id进行排序

SELECT
    customer_id,
    max_date
FROM    
    (SELECT 
        customer_id,
        max(date) AS max_date
    FROM    
        document 
        JOIN process ON process.id = document.process_id
    GROUP BY customer_id) AS pd
ORDER BY max_date DESC

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