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？