SQL，获取列的最大id

Question

我已经在MS SQL Server Management Studio中获得了此表

+------+-----------+--------
| id   | client id | bla bla blaaaa
+------+-----------+--------
|    1 |   1       | .......
|    2 |   2       | .......
|    3 |   3       | .......
|    4 |   8       | .......
|    5 |   9       | .......
|    6 |   15      | .......
|    7 |   1       | .......
|    8 |   16      | .......
|    9 |   2       | .......
|   10 |   9       | .......
|   12 |   12      | .......
+------+-----------+--------

我需要获取唯一的客户端id，最大值为id，如下所示

+------+-----------+--------
| id   | client id | bla bla blaaaa
+------+-----------+--------
|    3 |   3       | .......
|    4 |   8       | .......
|    6 |   15      | .......
|    7 |   1       | .......
|    8 |   16      | .......
|    9 |   2       | .......
|   10 |   9       | .......
|   12 |   12      | .......
+------+-----------+--------

我试过这段代码，但它不能很好地工作。有人能帮我吗？

SELECT *
FROM   table AS one 
INNER JOIN table AS two
   ON one.[client id] = two.[client id]
   WHERE one.[id] > two.[id]

Answer 1

SELECT max(id), client_id, blah_blah
FROM my_table
GROUP BY client_id, blah_blah

Answer 2

我会选择ROW_NUMBER()解决方案：

SELECT *
FROM (
    SELECT ROW_NUMBER() OVER(PARTITION BY [client id] ORDER BY [id] DESC) AS RN, *
    FROM table AS one
    ) AS T
WHERE T.RN = 1;