获取“Count(*)”占“GROUP BY”中所有项目数的百分比

Question

假设我需要“某个类别的可用项目数”与“所有项目数”的比率。请考虑这样的 MySQL 表：

/*

mysql> select * from Item;
+----+------------+----------+
| ID | Department | Category |
+----+------------+----------+
|  1 | Popular    | Rock     |
|  2 | Classical  | Opera    |
|  3 | Popular    | Jazz     |
|  4 | Classical  | Dance    |
|  5 | Classical  | General  |
|  6 | Classical  | Vocal    |
|  7 | Popular    | Blues    |
|  8 | Popular    | Jazz     |
|  9 | Popular    | Country  |
| 10 | Popular    | New Age  |
| 11 | Popular    | New Age  |
| 12 | Classical  | General  |
| 13 | Classical  | Dance    |
| 14 | Classical  | Opera    |
| 15 | Popular    | Blues    |
| 16 | Popular    | Blues    |
+----+------------+----------+
16 rows in set (0.03 sec)

mysql> SELECT Category, COUNT(*) AS Total
    -> FROM Item
    -> WHERE Department='Popular'
    -> GROUP BY Category;
+----------+-------+
| Category | Total |
+----------+-------+
| Blues    |     3 |
| Country  |     1 |
| Jazz     |     2 |
| New Age  |     2 |
| Rock     |     1 |
+----------+-------+
5 rows in set (0.02 sec)

*/

我需要的基本上是一个类似于这个的结果集：

/*
+----------+-------+-----------------------------+
| Category | Total | percentage to the all items | (Note that number of all available items is "9")
+----------+-------+-----------------------------+
| Blues    |     3 |                          33 | (3/9)*100
| Country  |     1 |                          11 | (1/9)*100
| Jazz     |     2 |                          22 | (2/9)*100
| New Age  |     2 |                          22 | (2/9)*100
| Rock     |     1 |                          11 | (1/9)*100
+----------+-------+-----------------------------+
5 rows in set (0.02 sec)

*/

如何在单个查询中实现这样的结果集？

提前致谢。

Answer 1

SELECT Category, COUNT(*) AS Total , (COUNT(*) / (SELECT COUNT(*) FROM Item WHERE Department='Popular')) * 100 AS 'Percentage to all items', 
FROM Item
WHERE Department='Popular'
GROUP BY Category;

我不确定 MySql 语法，但您可以使用如图所示的子查询。

Answer 2

应该这样做：

SELECT I.category AS category, COUNT(*) AS items, COUNT(*) / T.total * 100 AS percent
FROM Item as I,
     (SELECT COUNT(*) AS total FROM Item WHERE Department='Popular') AS T
WHERE Department='Popular'
GROUP BY category;

Answer 3

SET @total=0;

SELECT Category, count(*) as Count, count(*) / @total * 100 AS Percent FROM (
    SELECT Category, @total := @total + 1
    FROM Item
    WHERE Department='Popular') temp
GROUP BY Category;

这样做的一个好处是您不必复制WHERE 条件，这是下次有人来更新条件时的定时炸弹，但没有意识到它在两个不同的地方.

避免重复的WHERE 条件也提高了可读性，尤其是如果您的WHERE 更复杂（具有多个连接等）。