Dense_rank首先将Oracle转换为Postgresql [英] Dense_rank first Oracle to Postgresql convert
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问题描述
我正在尝试将以下Oracle查询转换为Postgresql.我可以转换其余的块,问题是我不知道如何转换此块:
I'm trying to convert the following Oracle query into Postgresql. I could convert the rest of the blocks, the problem is I don't know how to convert this block:
SELECT ai.uid
,max(ai.OWNER) KEEP (
dense_rank first ORDER BY ai.AGENT_ID DESC
) AS OWNER
,max(ai.EMPLOYEE_KEY) KEEP (
dense_rank first ORDER BY ai.AGENT_ID DESC
) AS EMPLOYEE_KEY
,max(ai.MANAGER_LOGIN) KEEP (
dense_rank first ORDER BY ai.AGENT_ID DESC
) AS MANAGER
,max(ai.CALL_CENTER_NAME) KEEP (
dense_rank first ORDER BY ai.AGENT_ID DESC
) AS CALL_CENTER_NAME
,max(ai.CITY) KEEP (
dense_rank first ORDER BY ai.AGENT_ID DESC
) AS CITY
FROM agent_info ai
WHERE translate(ai.UID, 'X0123456789', 'X') IS NULL
GROUP BY ai.UID
推荐答案
我认为您可以将其转换为包含单个DENSE_RANK()
并在第一条记录中对所有内容进行MAX
.
I think you can convert it to contain a single DENSE_RANK()
and do a MAX
of all on the first record.
SELECT UID,
MAX(OWNER),
MAX(EMPLOYEE_KEY),
MAX(MANAGER),
MAX(CALL_CENTER_NAME),
MAX(CITY)
FROM (SELECT ai.UID,
ai.OWNER AS OWNER,
ai.EMPLOYEE_KEY AS EMPLOYEE_KEY,
ai.MANAGER_LOGIN AS MANAGER,
ai.CALL_CENTER_NAME AS CALL_CENTER_NAME,
ai.CITY AS CITY,
DENSE_RANK () OVER (PARTITION BY ai.UID ORDER BY ai.AGENT_ID DESC) rnk
FROM agent_info ai
WHERE TRANSLATE (ai.UID, 'X0123456789', 'X') IS NULL)
WHERE rnk = 1
GROUP BY UID;
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