Recursive query in DB2 to get all items in the chain

流过昼夜 提交于 2019-12-01 00:25:34

Here is the answer to your question using recursive CTE query:

WITH links AS
( SELECT 
    loan_id, 
    client_id as c1, 
    client_id as c2, 0 as distance 
  FROM 
    myTable 
  -- recursion 
  UNION ALL
  SELECT 
     t.loan_id, 
     l.c1 as c1, 
     tt.client_id as c2, 
     distance = distance + 1 
  FROM 
    links l INNER JOIN
    myTable t ON l.c2 = t.client_id
    AND l.loan_id != t.loan_id INNER JOIN
    myTable tt  ON t.loan_id = tt.loan_id 
     AND t.client_id != tt.client_id
 )
SELECT * FROM myTable t
WHERE EXISTS 
    (SELECT * FROM links 
     WHERE c2 = t.client_id and c1 = 7);

http://sqlfiddle.com/#!3/8394d/16

I have left distance inside the query to make it easier to understand.

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!