求一个 SQL 语句不知去重还是分组来实现

[feikeq]

现在学过的东西想再用时却忘干净了，就像解一元二次方程，小时候很会解，现在怎么不知怎么开始解。。。。


原表内容
+--------------------+
| id | name | fid |
+--------------------+
| 101 | aaaa | |
+--------------------+
| 102 | bbbb | 101 |
+--------------------+
| 103 | cccc | 102 |
+--------------------+
| 104 | dddd | 101 |
+--------------------+

查询结果
+--------------------+
| id | name | sum |
+--------------------+
| 101 | aaaa | 2 |
+--------------------+
| 102 | bbbb | 1 |
+--------------------+

[feikeq]

@txoooy 的方法查询要 0.07 秒，采用下面方法只需 0.02 秒

SELECT T1.`uid`,T1.`headimg`, T1.`nickname` ,T1.`referer` ,T2.`num`
FROM `user_center` AS T1 LEFT JOIN (
SELECT C.`referer` AS uid, count(`referer`) AS num
FROM `user_center` C, `user_open` O
WHERE C.`uid`= O.`uid`
AND O.`subscribe`=1
GROUP BY C.`referer`
) AS T2
ON T1.`uid`= T2.`uid`
WHERE T2.`num` IS NOT NULL
ORDER BY T2.`num` DESC

[txoooy]

这个要分情况, 有些情况中利用子查询构建中间表可以提高性能, 不过你现阶段并不需要考虑这些, sql 优化东西太多了, 多写多看, 自然知道什么时候用什么方法

[wavingclear]

地球思维的写法
SELECT id, name, num FROM test INNER JOIN ( SELECT fid, COUNT(fid) as num FROM `test` GROUP BY fid) AS temp ON ( test.id = temp.fid )

修改之后的写法，不用子查询
SELECT a.id, a.name, count(b.fid) as num FROM test as a INNER JOIN test as b ON (a.id = b.fid) GROUP BY b.fid

[wavingclear]

啊 17 楼已经写过了……