从Java中的Json字符串生成聚合的Json数据

如何解决从Java中的Json字符串生成聚合的Json数据

我已经从db2中获取了JSON字符串，需要对数据进行分组。

SELECT JSON_ARRAY(JSON_OBJECT('id' VALUE id,'items' VALUE JSON_OBJECT('item' value item,'itemScore' value itemScore,'stage' value stage,'reco' VALUE JSON_OBJECT('product' value product,'url' value url,'score' value score
                                                                         format json)format json absent on null)
                   format json absent on null)format json absent on null)
FROM Product
GROUP BY id,item,itemScore,stage,url,product,score

如果Java或DB2查询中有解决方案，这将有所帮助。

需要使用id对JSON字符串执行分组。

JSON字符串：

[{
    "id": 2078012,"items": {
        "item": "cos","itemScore": -0.1334,"stage": 0.5,"reco": {
            "product": "Language","url": "https://","score": 0.019
        }
    }
},{
    "id": 2078012,"items": {
        "item": "build","itemScore": -0.2858,"reco": {
            "product": "Notifications","score": 0.0169
        }
    }
}]

预期输出： 需要使用ID对项目进行分组

我已经修改了预期的JSON，请帮助我创建更新的JSON。

[{
    "id": 2078012,"reco": [{
            "product": "Language","score": 0.019
        },{
            "product": "Notifications","score": 0.0169
        }]
    },"score": 0.0169
        }]
    }
}]

解决方法

您可能需要将其分解成几个步骤。

首先，一个与您的基表相似的基表。

db2 => SELECT * FROM PROD_T

ID          ITEM                 ITEMSCORE STAGE PRODUCT              URL                  SCORE
----------- -------------------- --------- ----- -------------------- -------------------- -------
    2078012 cos                    -0,1334   0,5 Language             https://              0,0190
    2078012 build                  -0,2585   0,5 Notification         https://              0,1690
    2100000 alpha                  -0,5 Letter               https://              0,0190
    2100000 beta                   -0,5 anouncement          https://              0,1690

步骤1：创建基本级别的json对象

select id,item,JSON_OBJECT('item' value item,'itemScore' value itemScore,'stage' value stage,'reco' VALUE JSON_OBJECT('product' value product,'url' value url,'score' value score 
                                        FORMAT JSON ) 
                FORMAT JSON ABSENT ON NULL  
               RETURNING VARCHAR(200) FORMAT JSON) ITEM_JSON  
FROM PROD_T 

ID          ITEM                 ITEM_JSON
----------- -------------------- --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
    2078012 cos                  {"item":"cos","itemScore":-0.1334,"stage":0.5,"reco":{"product":"Language","url":"https:\/\/","score":0.0190}}
    2078012 build                {"item":"build","itemScore":-0.2585,"reco":{"product":"Notification","score":0.1690}}
    2100000 alpha                {"item":"alpha","reco":{"product":"Letter","score":0.0190}}
    2100000 beta                 {"item":"beta","reco":{"product":"anouncement","score":0.1690}}

步骤2：按ID分组，然后将对象汇总到一个数组中。

WITH BASE AS ( 
select id,'score' value score 
                                        FORMAT JSON ) 
                FORMAT JSON ABSENT ON NULL  
               RETURNING VARCHAR(200) FORMAT JSON) ITEM_JSON  
FROM PROD_T  ) 
 SELECT ID,JSON_ARRAY ( LISTAGG( ITEM_JSON,',') WITHIN GROUP (ORDER BY ITEM) FORMAT JSON ) ITEMS_ARRAY
 FROM BASE GROUP BY ID 


ID          ITEMS_ARRAY
----------- --------------------------------------------------
    2078012 [{"item":"build","score":0.1690}},{"item":"cos","score":0.0190}}] 
    2100000 [{"item":"alpha","score":0.0190}},{"item":"beta","score":0.1690}}] 

第3步：为每个id行创建整个json_object

WITH BASE AS ( 
select id,'score' value score 
                                        FORMAT JSON ) 
                FORMAT JSON ABSENT ON NULL  
               RETURNING VARCHAR(200) FORMAT JSON) ITEM_JSON  
FROM PROD_T  ) 
 SELECT JSON_OBJECT ( KEY 'id' VALUE ID,KEY 'itens' VALUE 
                   JSON_ARRAY ( LISTAGG( ITEM_JSON,') WITHIN GROUP (ORDER BY ITEM) FORMAT JSON ) 
            FORMAT JSON ) json_objects 
 FROM BASE GROUP BY ID 


json_objects 
-----------------------------------------------------------------------------------------------------------------------------------
{"id":2078012,"itens":[{"item":"build","score":0.0190}}]} 
{"id":2100000,"itens":[{"item":"alpha","score":0.1690}}]}

第4步：完整的ID的json文档数组

下面的查询应该可以工作

WITH BASE AS ( 
select id,'score' value score 
                                        FORMAT JSON ) 
                FORMAT JSON ABSENT ON NULL  
               RETURNING VARCHAR(200) FORMAT JSON) ITEM_JSON  
FROM PROD_T  ),PROD_OBJS AS ( 
     SELECT  JSON_OBJECT ( KEY 'id' VALUE ID,') WITHIN GROUP (ORDER BY ITEM) FORMAT JSON ) 
            FORMAT JSON ) json_objects 
     FROM BASE GROUP BY ID ) 
SELECT JSON_ARRAY (select json_objects FROM PROD_OBJS format json) FROM SYSIBM.SYSDUMMY1

但是，在我的系统（DB2 v11.5.0.1077）中，它由于内部编译错误而失败：

SQL0901N由于数据库，SQL语句或命令失败 系统错误。 （原因为“意外的agg opparm”。）SQLSTATE = 58004

但是，如果您遇到相同的情况，则可以在临时表中实现来自step3的输出，并从中执行最终数组。

DECLARE GLOBAL TEMPORARY TABLE prod_json_objcts AS ( 
WITH BASE AS ( 
select id,') WITHIN GROUP (ORDER BY ITEM) FORMAT JSON ) 
            FORMAT JSON ) json_objects 
 FROM BASE GROUP BY ID 
) WITH DATA 

SELECT JSON_ARRAY (select json_objects FROM session.prod_json_objcts format json) FROM SYSIBM.SYSDUMMY1 

产生最终的json数组：

[{"id":2078012,"itens":[
      {"item":"build","score":0.0190}}
    ]
 },{"id":2100000,"itens":[
      {"item":"alpha","score":0.1690}}
    ]
 }
]