[英]How to format an aggregate function result in SQL select query of Oracle Database?
[英]How to aggregate from result data Oracle SQL?
我有桌子:
+------+-------+-----------------+
| id | name | code | desc |
+------+-------+-----------------+
| 1 | aa | 032016 | grape |
| 1 | aa | 012016 | apple |
| 1 | aa | 032016 | grape |
| 1 | aa | 022016 | orange |
| 1 | aa | 012016 | apple |
| 1 | aa | 032016 | grape |
+------+-------+-----------------+
我嘗試查詢:
SELECT id, name, code, desc, COUNT(code) as view
FROM mytable
GROUP BY id, name, code, desc
結果是:
+------+-------+------------------------+
| id | name | code | desc | view |
+------+-------+------------------------+
| 1 | aa | 012016 | apple | 2 |
| 1 | aa | 022016 | orange | 1 |
| 1 | aa | 032016 | grape | 3 |
+------+-------+------------------------+
我所期望的是這樣的:
+------+-------+----------------------------------------------------+
| id | name | code | desc | view |
+------+-------+----------------------------------------------------+
| 1 | aa | 012016,022016,032016 | apple,orange,grape | 2,1,3 |
+------+-------+----------------------------------------------------+
誰能幫我匯總結果? 提前致謝
您的餐桌設計讓我有些擔心。 一個水果在表中總是具有相同的代碼是否是偶然的? 那為什么要多余地存放呢? 應該有一個水果表,每個水果及其代碼只能容納一次。 您知道為什么將其稱為關系數據庫系統,不是嗎?
但是,通過查詢,您幾乎可以到達。 您具有每個id,名稱,代碼和desc的計數。 現在,您想進一步匯總。 因此,在下一步中,按ID和名稱分組,因為您似乎希望每個ID和名稱都包含一個結果行。 使用LISTAGG
連接組中的字符串:
SELECT
id,
name,
listagg(code, ',') within group(order by code) as codes,
listagg(desc, ',') within group(order by code) as descs,
listagg(view, ',') within group(order by code) as views
FROM
(
SELECT id, name, code, desc, COUNT(*) as view
FROM mytable
GROUP BY id, name, code, desc
)
GROUP BY id, name
ORDER BY id, name;
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