我的数据如下:
MEDIA_ID | CHANNEL_NAME
EH/123A CH-1
EH/123A CH-4
EH/132A CH-5
ES/133B CH-1
ES/133B CH-2
ES/133B CH-5
我想要的是:
EH/123A | CH-1,CH-4,CH-5
ES/123B | CH-1,CH-2,CH-5
我在Oracle中使用这个SQL:
SELECT DISTINCT
PR.MEDIA_ID
, LISTAGG(PR.CHANNEL_NAME, ', ') WITHIN GROUP (ORDER BY CHANNEL_NAME) AS PREM_CHAN
FROM PREM_REPORT PR
GROUP BY PR.MEDIA_ITEM, PR.CHANNEL_NAME;
我得到的是:
MEDIA_ID | CHANNEL_NAME
EH/123A CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1
EH/123A CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4,CH-4
EH/132A CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5
ES/133B CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1,CH-1
ES/133B CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2,CH-2
ES/133B CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5,CH-5
想法?
感谢。 本
答案 0 :(得分:1)
我认为您想要的查询是:
SELECT PR.MEDIA_ID,
LISTAGG(PR.CHANNEL_NAME, ', ') WITHIN GROUP (ORDER BY CHANNEL_NAME) AS PREM_CHAN
FROM PREM_REPORT PR
GROUP BY PR.MEDIA_ITEM;
也就是说,从查询中删除PR.CHANNEL_NAME
。我不确定您为什么会通过您提供的查询获得结果。也许select distinct
和group by
之间存在一些奇怪的互动。您几乎从未将select distinct
与group by
一起使用。
编辑:
要在LIST_AGG()
中返回不同的值,您需要使用子查询。在这种情况下,一种简单的方法是:
SELECT PR.MEDIA_ID,
LISTAGG(PR.CHANNEL_NAME, ', ') WITHIN GROUP (ORDER BY CHANNEL_NAME) AS PREM_CHAN
FROM (SELECT DISTINCT MEDIA_ID, CHANNEL_NAME
FROM PREM_REPORT PR
) PR
GROUP BY PR.MEDIA_ITEM;
答案 1 :(得分:1)
您可以删除GROUP BY
,只需添加PARTITION BY
:
SELECT DISTINCT PR.MEDIA_ID
,LISTAGG(PR.CHANNEL_NAME, ', ')
WITHIN GROUP (ORDER BY CHANNEL_NAME) OVER (PARTITION BY PR.MEDIA_ID) AS PREM_CHAN
FROM PREM_REPORT PR;
的 SqlFiddleDemo
强>
输出:
╔═══════════╦══════════════════╗
║ MEDIA_ID ║ PREM_CHAN ║
╠═══════════╬══════════════════╣
║ ES/133B ║ CH-1, CH-2, CH-5 ║
║ EH/123A ║ CH-1, CH-4, CH-5 ║
╚═══════════╩══════════════════╝
答案 2 :(得分:0)
我认为你的GROUP BY错了。这对我有用。
WITH prem_report AS (
SELECT 'EH/123A' media_id, 'CH-1' channel_name FROM DUAL
UNION
SELECT 'EH/123A' media_id, 'CH-4' channel_name FROM DUAL
UNION
SELECT 'EH/132A' media_id, 'CH-5' channel_name FROM DUAL
UNION
SELECT 'ES/133B' media_id, 'CH-1' channel_name FROM DUAL
UNION
SELECT 'ES/133B' media_id, 'CH-2' channel_name FROM DUAL
UNION
SELECT 'ES/133B' media_id, 'CH-5' channel_name FROM DUAL
)
SELECT DISTINCT pr.media_id, LISTAGG(pr.channel_name, ', ') WITHIN GROUP (ORDER BY channel_name) AS prem_chan
FROM prem_report pr
GROUP BY pr.media_id