SELECT name,
  SUM(IF(`month` = 'July',1,0)) AS July,
  SUM(IF(`month` = 'August',1,0)) AS August,
  SUM(IF(`month` = 'September',1,0)) AS September,
  COUNT(*)  FROM test
GROUP BY name;

参考文献：https://stackoverflow.com/a/13075582/1688441

COUNT 计数非空值，请重试 SUM ```
SELECT Name,
SUM(IF(Date = 'July',1,0)) AS July,
SUM(IF(Date = 'August',1,0)) AS August,
SUM(IF(Date = 'September',1,0)) AS September,
COUNT(*) AS All FROM table
GROUP BY Name

发生这种情况的原因很简单： COUNT() 只计算值的发生次数。（=数据集中有多少个值？）
这样看， 0 只是另一个值 1 也是。
您可以：

SELECT Name, 
COUNT(IF(`Date` = 'July', 1, NULL)) AS July,
COUNT(IF(`Date` = 'August', 1, NULL)) AS August,
COUNT(IF(`Date` = 'September', 1, NULL)) AS September,
COUNT(*) AS All FROM table
GROUP BY Name

这是因为聚合函数 COUNT() 忽略 NULL 价值观。
或者，就像其他人回答的那样，你可以 SUM() 所有的 1 价值观：

SELECT Name, 
SUM(`Date` = 'July') AS July,
SUM(`Date` = 'August') AS August,
SUM(`Date` = 'September') AS September,
COUNT(*) AS All FROM table
GROUP BY Name

这是写得更短，因为评估 field = value 已经回来了 1 如果匹配，则返回0。不需要 Package IF() 周围的人也做同样的事。

select date, count(*) 
from table1
group by date

应该给你一些

|DATE       |Count(*)|
----------------------
|August     |    5   |
|July       |    3   |
|September  |    2   |

http://sqlfiddle.com/#!9/c0a483/2号机组

最简单的方法就是试试这个。带bool（0或1）的条件聚合函数

SELECT Name, 
    SUM(`Date` = 'July') AS July,
    SUM(`Date` = 'August') AS August,
    SUM(`Date` = 'September') AS September,
    COUNT(*) AS All 
FROM table
GROUP BY Name

count 计算不存在的值 null -包括零。带着那些 if s、 你可以使用 sum 模仿 count -您将有效地计算 1 学生：

SELECT Name, 
SUM(IF(`Date` = 'July',1,0)) AS July,
SUM(IF(`Date` = 'August',1,0)) AS August,
SUM(IF(`Date` = 'September',1,0)) AS September,
SUM(*) AS All FROM table
GROUP BY Name

尝试 SELECT Name, Date, Count(*) as Count FROM table GROUP BY Name,Date 它应该返回如下内容：

| Name  | Date      | Count |
-----------------------------
| Carl  | August    |   3   |
| Carl  | September |   1   |
| John  | July      |   3   |
| John  | August    |   1   |
| Robert| September |   1   |
| Rober | August    |   1   |

看看这把小提琴。