mysql 从 where 子句中获取总值和总数

Question

我正在尝试编写一个 mysql 查询，它将 return 列中值的总数以及基于同一列中 where 子句的值总数。

我有一个 table 这样的：

+------------------------+-------+
| color                  | code  |
+------------------------+-------+
| red                    |   200 |
| red                    |   202 |
| blue                   |   105 |
| yellow                 |   136 |
| green                  |   561 |
| red                    |   198 |
| blue                   |   414 |
| green                  |   11  |
| yellow                 |   600 |
| green                  |   155 |
| red                    |   865 |
| blue                   |   601 |
| green                  |   311 |
+------------------------+-------+

如果我运行这个查询：

select 
    color, 
    count(*) as count 
from colors 
where code > 0 && 
    code <= 500 
group by color 
order by count(*) desc;

我得到了这个很棒的结果，因为它几乎就是我想要的：

+------------------------+-------+
| color                  | count |
+------------------------+-------+
| red                    |     3 |
| green                  |     3 |
| blue                   |     2 |
| yellow                 |     1 |
+------------------------+-------+

我还需要 returned 是列中值的总数，所以结果 table 看起来像这样。

+------------------------+--------------+-------+
| color                  | total        | count |
+------------------------+--------------+-------+
| red                    |            4 |     3 |
| green                  |            4 |     3 |
| blue                   |            3 |     2 |
| yellow                 |            2 |     1 |
+------------------------+--------------+-------+

所以total就是color列中每个值的个数，count就是匹配where子句的总数。

谢谢:)

这是 SQLFiddle 的 link。

http://sqlfiddle.com/#!9/777f93/2

Answer 1

您需要使用条件聚合来处理计数并让引擎处理总数。

SELECT color
     , count(*) as Total
     , sum(case when code > 0 and code <= 500  then 1 else 0 end) as cnt 
FROM  colors 
GROUP BY color 
ORDER BY cnt desc;

Answer 2

您可以 JOIN 您的查询与另一个查询，例如：

select color, count(*) as count , a.total
from colors JOIN (
    SELECT color, count(*) as `total` FROM colors GROUP BY color 
 ) a ON colors.code = a.color
where code > 0 && code <= 500 
group by color 
order by count(*) desc;

Answer 3

您的案例使用 case when 语句来仅计算满足您条件的项目。像这样：

Select 
color
, count(*) as total
, SUM(CASE WHEN code > 0 && code <= 500 THEN 1 ELSE 0 END ) as Count
group by color order by count(*) desc;

Answer 4

您可以使用连接查询和 table 别名

    select t1.color,t1.total,t2.count 
FROM 
(select color, count(*) as total from colors group by color) t1,
(select color, count(*) as `count` from colors where `code` > 0 && `code` <= 500 group by color) t2
WHERE
t1.color=t2.color order by `count` desc;

这是 SQLFiddle 的 link。

http://sqlfiddle.com/#!9/777f93/2

Answer 5

您可以使用条件聚合：

select color,
    count(*) as Total,
    sum(code > 0 and code <= 500) as count_0_to_500
from colors
group by color
order by count_0_to_500 desc

它利用了 MySQL 中 true 的计算结果为 1 而 false 的计算结果为 0 的事实。

这基本上就是 xQbert 回答的内容，只是没有 case 表达式。

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