Oracle SQL：计算具有特定值的行数
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Oracle SQL: Count number of rows with specific value
提问by George Newton
I have a table like this:
我有一张这样的表：
ID | AGE
--------
1 | 27
2 | 24
1 | 25
4 | 25
1 | 25
5 | 25
5 | 25
...|...
As you can see, ID and AGE are not unique (there is a second ID that is unique, but that's irrelevant here).
如您所见，ID 和 AGE 不是唯一的（有第二个 ID 是唯一的，但这在这里无关紧要）。
I want to return all the distinct IDs with more than one specific value: for example, counting all the IDs with more than one AGE = 25; in the table above, this should yield [1, 5], (ID = 4 is not counted because 25 occurs only once; ID = 1 and ID = 5 IS included because 25 occurs more than once in AGE).
我想返回具有多个特定值的所有不同 ID：例如，计算具有多个 AGE = 25 的所有 ID；在上表中，这应该产生 [1, 5]，（ID = 4 不计算在内，因为 25 只出现一次；ID = 1 和 ID = 5 被包括在内，因为 25 在 AGE 中出现不止一次）。
Basically what I want in invalid SQL:
基本上我想要的无效SQL：
SELECT DISTINCT id FROM table WHERE count(AGE = 25) > 1
Unfortunately, count(AGE = 25) is invalid. How would I write this out in SQL?
不幸的是，count(AGE = 25) 无效。我将如何在 SQL 中写出它？
回答by Mithrandir
Do you want something like this?
你想要这样的东西吗？
SELECT
ID, COUNT(ID)
FROM table
WHERE Age = 25
GROUP BY ID
HAVING COUNT(ID) > 1
