198

如果项目已排序，我可以运行 select 语句并获取行号吗？

我有一张这样的桌子：

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

然后我可以运行此查询以按 ID 获取订单数：

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

这给了我表中每个的计数，itemID如下所示：

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

我也想得到行号，所以我可以说这itemID=388是第一行，234是第二行，等等（本质上是订单的排名，而不仅仅是原始计数）。我知道当我得到结果集时我可以在 Java 中执行此操作，但我想知道是否有一种方法可以纯粹在 SQL 中处理它。

更新

设置排名会将其添加到结果集中，但排序不正确：

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)
4

6 回答 6

195

看看这个。

将您的查询更改为：

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

最后一个选择是你的计数。

于 2010-03-26T00:15:00.210 回答
182 
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;
于 2010-12-17T20:10:33.453 回答
36

Swamibebop 的解决方案有效，但通过利用table.*语法，我们可以避免重复内部的列名select并获得更简单/更短的结果：

SELECT @r := @r+1 , 
       z.* 
FROM(/* your original select statement goes in here */)z, 
(SELECT @r:=0)y;

所以这会给你：

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;
于 2015-04-24T11:33:11.640 回答
12

您可以使用 MySQL 变量来执行此操作。像这样的东西应该可以工作（尽管它由两个查询组成）。

SELECT 0 INTO @x;

SELECT itemID, 
       COUNT(*) AS ordercount, 
       (@x:=@x+1) AS rownumber 
FROM orders 
GROUP BY itemID 
ORDER BY ordercount DESC;
于 2010-03-26T00:19:12.317 回答
7

它现在内置在 MySQL 8.0 和 MariaDB 10.2 中：

SELECT
  itemID, COUNT(*) as ordercount,
  ROW_NUMBER OVER (PARTITION BY itemID ORDER BY rank DESC) as rank
FROM orders
GROUP BY itemID ORDER BY rank DESC
于 2019-12-02T08:22:34.950 回答
0 
SELECT RANK() OVER(ORDER BY Employee.ID) rank, forename, surname, Department.Name, Occupation.Name  
FROM Employee  
JOIN Occupation ON Occupation.ID = Employee.OccupationID  
JOIN Department ON Department.ID = Employee.DepartmentID 
WHERE DepartmentID = 2;
于 2022-02-17T16:01:47.207 回答