的Oracle SQL逗号分隔的ID

我有一个查询，像这样：的Oracle SQL逗号分隔的ID

SELECT 
    s.id, 
    CASE 
    WHEN INSTR(s.strain_value, '[') = 0 
    THEN s.strain_name 
    ELSE TRIM(SUBSTR(s.strain_name, 1, INSTR(s.strain_name, '[')-1)) 
    END AS strain_name 
FROM strain s

这给了我一个表，看起来像：

id  strain_name 
---------------------------------------------- 
100  CD-1 
101  CD-1 
102  CD-1 
200  SCID 
300  BusIp-hy 
301  BusIp-hy 
...  ...

现在由我strain_name想组结果表并获得逗号分隔id像下面这样：

id    strain_name 
---------------------------------------------- 
100, 101, 102 CD-1 
200    SCID 
300, 301  BusIp-hy 
...    ...

我尝试此查询出来，但我没有牛逼得到我想要的东西：

WITH q AS (
    SELECT 
    s.id, 
    CASE 
     WHEN INSTR(s.strain_name, '[') = 0 
     THEN s.strain_name 
     ELSE TRIM(SUBSTR(s.strain_name, 1, INSTR(s.strain_name, '[')-1)) 
    END AS strain_name 
    FROM strain s 
) 
SELECT (
    SELECT LISTAGG(id, ', ') 
    WITHIN GROUP (
    ORDER BY strain_name ASC 
) 
    FROM (SELECT DISTINCT strain_name FROM q) dc 
) AS id, strain_name 
FROM q

能否请你帮我纠正过去的查询，这样我可以得到id的每个strain_name一个逗号分隔的列表？

来源

2016-08-23 uom-pgregorio

WITH q (id, strain_name) AS (
    SELECT id, 
     CASE WHEN INSTR(strain_value, '[') = 0 
       THEN strain_name 
       ELSE TRIM(SUBSTR(strain_name, 1, INSTR(strain_name, '[')-1)) 
     END 
    FROM strain 
) 
SELECT LISTAGG(id, ', ') WITHIN GROUP (ORDER BY id) AS id, 
     strain_name 
FROM q 
GROUP BY strain_name

来源

2016-08-23 14:04:09 MT0

感谢，这对我的作品。只是想知道为什么你必须将'ORDER BY'从'strain_name'改为'id'？ –

@PatrickGregorio从您的预期结果看来，您希望以升序排列以逗号分隔的ID。 'LISTAGG'的'ORDER BY'在组内，因此所有的项目都有相同的'strain_name'（因为这是group by子句） - 使用这个顺序会导致[非确定性排序] （http://stackoverflow.com/q/35747417/1509264），但“id”值看起来都不一样，因此会导致确定性排序。 – MT0

您是否尝试过GROUP BY？

WITH q AS (
... 
) 
SELECT strain_name, listagg(id, ', ') within GROUP(ORDER BY id) 
FROM q 
GROUP BY strain_name 
ORDER BY strain_name

来源

2016-08-23 14:04:38

WITH q AS (
    SELECT 
    s.id, 
    CASE 
     WHEN INSTR(s.strain_name, '[') = 0 
     THEN s.strain_name 
     ELSE TRIM(SUBSTR(s.strain_name, 1, INSTR(s.strain_name, '[')-1)) 
    END AS strain_name 
    FROM strain s 
) 
SELECT LISTAGG(id, ', ') WITHIN GROUP (ORDER BY id ASC) as id, q.strain_name 
    FROM q 
group by q.strain_name

来源

2016-08-23 14:04:47 vercelli

的Oracle SQL逗号分隔的ID

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