的Oracle SQL逗號分隔的ID

我有一個查詢，像這樣：的Oracle SQL逗號分隔的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

這給了我一個表，看起來像：

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

現在由我strain_name想組結果表並獲得逗號分隔id像下面這樣：

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

我嘗試此查詢出來，但我沒有牛逼得到我想要的東西：

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

能否請你幫我糾正過去的查詢，這樣我可以得到id的每個strain_name一個逗號分隔的列表？

來源

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

來源

2016-08-23 14:04:09 MT0

感謝，這對我的作品。只是想知道爲什麼你必須將'ORDER BY'從'strain_name'改爲'id'？ –

@PatrickGregorio從您的預期結果看來，您希望以升序排列以逗號分隔的ID。 'LISTAGG'的'ORDER BY'在組內，因此所有的項目都有相同的'strain_name'（因爲這是group by子句） - 使用這個順序會導致[非確定性排序] （http://stackoverflow.com/q/35747417/1509264），但「id」值看起來都不一樣，因此會導致確定性排序。 – MT0

您是否嘗試過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

來源

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

來源

2016-08-23 14:04:47 vercelli

的Oracle SQL逗號分隔的ID

回答

相關問題