mysql - 相当于MySQL $ i++

我正在使用的查询：

SELECT COUNT(*),
    SUM(amount) AS amount,
    FROM_UNIXTIME(added, '%W (%e/%m)') AS daily
FROM affiliates_earnings
WHERE added >= '1319074836'
    AND added <= '1319679636'
GROUP BY daily

输出：

+----------+--------+-------------------+
| COUNT(*) | amount | daily             |
+----------+--------+-------------------+
|        1 |    195 | Tuesday (25/10)   |
|        4 |    470 | Wednesday (26/10) |
+----------+--------+-------------------+

我想显示的是：

+-------+----------+--------+-------------------+
| i     |COUNT(*)  | amount | daily             |
+-------+----------+--------+-------------------+
| 1     |        1 |    195 | Tuesday (25/10)   |
| 2     |        4 |    470 | Wednesday (26/10) |
+-------+----------+--------+-------------------+

本质上i是序列号（即1,2,3,4,5,6 ...）。

我将如何完成？

最佳答案

如果没有ROW_NUMBER（），则可以per this answer或this one对其进行伪造：

  SELECT @i := @i + 1 AS i, COUNT(*), ...
    FROM affiliates_earnings
    JOIN (SELECT @i := 0) dummy
   WHERE ...
GROUP BY ...;

关于mysql - 相当于MySQL $ i++，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/7910989/

于Mysql

mysql - 相当于MySQL $ i++