Mysql 的rank 函数如何实现 - 大鸟的学习乐园 - BlogJava
http://www.blogjava.net/dunkbird/archive/2011/01/28/343718.html
表特征：
mysql> select * from test;
+------+------+
| a | b |
+------+------+
| 1 | 20 |
| 1 | 21 |
| 1 | 24 |
| 2 | 20 |
| 2 | 32 |
| 2 | 14 |
+------+------+
6 rows in set (0.00 sec)
现在，我们以a分组，查询b列最大的2个值。 这条sql要怎么写了？
1.创建表
Create Table: CREATE TABLE `sam` ( `a` int(11) DEFAULT NULL, `b` int(11) DEFAULT NULL) ENGINE=MyISAM DEFAULT CHARSET=utf8
INSERT INTO `sam` VALUES (1,10),(1,15),(1,20),(1,25),(2,20),(2,22),(2,33),(2,45);
3.SQL实现
select a,b,rownum,rank from (select ff.a,ff.b,@rownum:=@rownum+1 rownum,if(@pa=ff.a,@rank:=@rank+1,@rank:=1) as rank,@pa:=ff.a FROM (select a,b from sam group by a,b order by a asc,b desc) ff,(select @rank:=0,@rownum:=0,@pa=null) tt) result having rank <br>4.结果：<br>+------+------+--------+------+<br>| a | b | rownum | rank |<br>+------+------+--------+------+<br>| 1 | 25 | 1 | 1 | <br>| 1 | 20 | 2 | 2 | <br>| 2 | 45 | 5 | 1 | <br>| 2 | 33 | 6 | 2 | <br>+------+------+--------+------+<br>4 rows in set (0.00 sec)<p>注：</p><p>@x 为一变量，<br>X:=Y 将Y值赋给X<br></p>
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