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如何从具有父子对关系的自引用实体中获得部分结果？

我试图获取整个实体对象，然后将其重构为部分对象。

我还尝试了 FetchGroup，选择部分列等。

但他们都没有工作。当它检索孩子时，整个对象会发生变化以带来整个实体。

@Entity
public class someClass extents Model {

    public String name;
    public String code;

    @ManyToOne
    @JsonBackReference
    public Menu parent;

    @OneToMany(mappedBy = "parent")
    @JsonManagedReference
    public Set<someClass> children;

    other columns and getters/setters.
}

这是我使用的查找器

public static List<someClass> findInTree() {
    find
            .query()
            .where()
            .isNull("parent")
            .findList();
}

我可以看到查询正在运行

[info] o.j.StatementLogger - select t0.id, t0.code, t0.name from some_entity t0 where t0.parent_id is null  and t0.is_active = Y  and  to_timestamp('2019-04-04 19:05:54.333', 'yyyy-MM-dd hh24:mi:ss.ff3') between t0.start_date and t0.end_date ;
[info] o.j.StatementLogger - select t0.parent_id, t0.id from some_entity t0 where (t0.parent_id) in (1, 2, 3, 4, 1 ) ;
[info] o.j.StatementLogger - select t0.id, t0.code, t0.name from some_entity t0 where t0.id in (5, 6, 5, 5, 5 ) ;
[info] o.j.StatementLogger - select t0.parent_id, t0.id from some_entity t0 where (t0.parent_id) in (5, 6, 5, 5, 5 ) ;
[info] o.j.StatementLogger - select t0.id, t0.code, t0.name from some_entity t0 where t0.id in (7, 8, 7, 7, 7 ) ;
[info] o.j.StatementLogger - select t0.parent_id, t0.id from some_entity t0 where (t0.parent_id) in (7, 8, 7, 7, 7 ) ;

结果是

"someClass":[  
         {  
            "id":1,
            "code":"test1",
            "name":"test1",
            "children":[  

            ],
            ... rest of columns
         },
            ... 
         {  
            "id":4,
            "code":"test1",
            "name":"test1",
            "children":[  
               {  
                  "id":5,
                  "code":"test1",
                  "name":"test1",
                  "children":[  
                     {  
                        "id":7,
                        "code":"test1",
                        "name":"test1",
                        "children":[  

                        ],
                     },
              ... rest of columns

                  ],
            ... rest of columns

               },
...
]

我担心递归查询在变得更大时会消耗太多资源。

而且，我想将这种做法用于具有不同关系的其他实体。

预期结果如下所示。

"someClass":[  
         {  
            "id":1,
            "code":"test1",
            "name":"test1",
            "children":null,
         },
            ... 
         {  
            "id":4,
            "code":"test1",
            "name":"test1",
            "children":[  
               {  
                  "id":5,
                  "code":"test1",
                  "name":"test1",
                  "children":[  
                     {  
                        "id":7,
                        "code":"test1",
                        "name":"test1",
                        "children":null,
                     },
                  ],
               },
...
]

我想从结果中删除所有不必要的数据，并希望减少检索映射查询以进行优化。

是否有我可以查找的最佳实践或示例？

此致。

4

1 回答 1

0

我这样做如下

[ModelClassName].find.select("field1, field2, ...")

.where
...
....
.findList();

发现在哪里

public static final Finder<primaryKeyType, ClassName> find = new Finder<>(ProClassNameducts.class);

请注意，字段名称就像在您的模型类中而不是在数据库中。

希望这有帮助

于 2019-07-11T13:22:02.977 回答