将列转换为 json 数组的最佳方法 (Oracle)

Question

将列的数据联合到 json 数组的最佳方法是什么？

我有 table

Foreign_ID | Primary_ID | Position | Name

143         2344        Postman     Ann
143         2345        Postman     Tim
144         2346        Postman     Bob
144         2347        Postman     Dan
143         2348        Student     Jim
143         2349        Student     Corey
146         2350        Student     Justin
146         2351        Student     Keith

而且我需要将名称列中的数据联合到每个外国 ID 的 json 数组，其中行在位置列中具有相同的数据。然后，应将此 json 数组插入值合并的行之一的名称列中。结果 table 应如下所示：

  Foreign_ID | Primary_ID | Position  | Name

     143         2344       Postman     ["Ann","Tim"]
     144         2346       Postman     ["Bob","Dan"]
     143         2348       Student     ["Jim","Corey"]
     146         2350       Student     ["Justin","Keith"]

谢谢。

Answer 1

看来我找到了解决方案，查询是

MERGE INTO TESTSCHEDULER TS
USING (SELECT Foreign_ID,
  Position,
  MIN(Primary_ID) AS MINID,
  COUNT(*),
  JSON_ARRAYAGG(Name) AS JSONNAME
FROM TESTSCHEDULER
GROUP BY Position,
  Foreign_ID
HAVING COUNT(*) > 1
) JSONTABLE ON (TS.Position = JSONTABLE.Position AND TS.Foreign_ID = JSONTABLE.Foreign_ID)
WHEN MATCHED THEN
  UPDATE SET TS.NAME = JSONTABLE.JSONNAME WHERE TS.Primary_ID = JSONTABLE.MINID;

DELETE FROM TESTSCHEDULER WHERE Primary_ID IN (   
SELECT Primary_ID FROM TESTSCHEDULER
INNER JOIN  (SELECT Foreign_ID,
  Position,
  MIN(Primary_ID) AS MINID,
  COUNT(*)
FROM TESTSCHEDULER
GROUP BY Position,  Foreign_ID
HAVING COUNT(*) > 1) JSONTABLE
ON TESTSCHEDULER.FOREIGN_ID = JSONTABLE.Foreign_ID 
AND TESTSCHEDULER.Position = JSONTABLE.Position 
AND TESTSCHEDULER.PRIMARY_ID > JSONTABLE.MINID);

其中 JSON_ARRAYAGG(col) 函数适用于数据库版本 12+ ...对于低于 12.2 的版本，替代 JSON_ARRAYAGG 是：

'[' || replace(rtrim (xmlagg (xmlelement (e, '"' || Name || '",')).extract ('//text()'), ','), '&quot;','"') || ']' AS JSONNAME

这是用于 clob 列，对于 varchar 使用 LISTAGG

将列转换为 json 数组的最佳方法 (Oracle)

best way to convert column to json array (Oracle)

sql

oracle

oracle12c