MySQL - 获取行号

Question

如果项目已排序，我可以运行select语句并获取行号吗？

我有这样的一个表：

SELECT itemID, COUNT(*) as ordercount 
FROM orders 
GROUP BY itemID ORDER BY ordercount DESC; 

这让我在每个itemID的计数：

mysql> describe orders; 
+-------------+---------------------+------+-----+---------+----------------+ 
| Field  | Type    | Null | Key | Default | Extra   | 
+-------------+---------------------+------+-----+---------+----------------+ 
| orderID  | bigint(20) unsigned | NO | PRI | NULL | auto_increment | 
| itemID  | bigint(20) unsigned | NO |  | NULL |    | 
+-------------+---------------------+------+-----+---------+----------------+ 

然后我就可以运行此查询通过ID来获得的订单数这样的表：

+--------+------------+ 
| itemID | ordercount | 
+--------+------------+ 
| 388 |   3 | 
| 234 |   2 | 
| 3432 |   1 | 
| 693 |   1 | 
| 3459 |   1 | 
+--------+------------+ 

我想要得到行号，所以我可以告诉itemID=388是第一行，234是第二等，（基本上是订单的排名，而不仅仅是一个原始计数）。我知道当我得到结果集时，我可以用Java来做到这一点，但我想知道是否有办法纯粹用SQL来处理它。

更新

设置排名将其添加到结果集，但不能正常下令：

mysql> SET @rank=0; 
Query OK, 0 rows affected (0.00 sec) 

mysql> SELECT @rank:=@rank+1 AS rank, itemID, COUNT(*) as ordercount 
    -> FROM orders 
    -> GROUP BY itemID ORDER BY rank DESC; 
+------+--------+------------+ 
| rank | itemID | ordercount | 
+------+--------+------------+ 
| 5 | 3459 |   1 | 
| 4 | 234 |   2 | 
| 3 | 693 |   1 | 
| 2 | 3432 |   1 | 
| 1 | 388 |   3 | 
+------+--------+------------+ 
5 rows in set (0.00 sec) 

Answer 1

看看this。

查询更改为：

SET @rank=0; 
SELECT @rank:=@rank+1 AS rank, itemID, COUNT(*) as ordercount 
    FROM orders 
    GROUP BY itemID 
    ORDER BY ordercount DESC; 

Answer 2

您可以使用MySQL变量来做到这一点。像这样的东西应该可以工作（但是，这是两个查询）。

SELECT 0 INTO @x; 

SELECT itemID, COUNT(*) as ordercount, (@x:=@x+1) as rownumber FROM orders GROUP BY itemID ORDER BY ordercount DESC; 

Answer 3

SELECT @rn:=@rn+1 AS rank, itemID, ordercount 
FROM (
    SELECT itemID, COUNT(*) AS ordercount 
    FROM orders 
    GROUP BY itemID 
    ORDER BY ordercount DESC 
) t1, (SELECT @rn:=0) t2; 

Answer 4

Swamibebop的解决方案工作，但通过采取table.*语法的优势，我们可以avoid repeating内select的列名，并得到一个简单的/更短的结果：

select @r := @r+1 , z.* from(

    /* your original select statement goes in here */ 

)z, (select @r:=0)y; 

所以那会给你：

select @r := @r+1 , z.* from(

    select itemID, count(*) as ordercount 
    from orders 
    group by itemID 
    order by ordercount desc 

)z, (select @r:=0)y;