Question

在小组项目期间，我们最近发送了一份关于我们正在建设的网站的调查。我已将数据放入mysql数据库，我试图弄清楚如何计算每个类别中给定的分数的次数表格看起来像这样

+-----------------+--------------+-------------------+
| Design          | Ease of use  | Responsiveness    |
+-----------------+--------------+-------------------+
| 5               | 5            | 5
| 4               | 4            | 4
| 3               | 3            | 3
| 2               | 2            | 2
| 1               | 1            | 1
| 5               | 4            | 2
| 5               | 4            | 4
| 3               | 3            | 3
| 1               | 2            | 2
| 1               | 2            | 2

我找到了一个适用于一个列的查询

SELECT Design, COUNT(*) AS num FROM table GROUP BY Design

我会得到

Design | num
------------- 
 5     | 3
 4     | 1
 3     | 2
 2     | 1  
 1     | 3

如果我要尝试

SELECT Design, COUNT(*) AS num1, Ease of use, COUNT(*) as num2 FROM table 
GROUP BY Design, Ease of use

桌子完全搞砸了。

我想要的是

Design | num1 | Ease of use  | num2 | Responsiveness | num3 
------------- --------------------------------------------------
 5     | 3    | 5            | 1    | 5              | 1
 4     | 1    | 4            | 3    | 4              | 2
 3     | 2    | 3            | 2    | 3              | 2
 2     | 1    | 2            | 3    | 2              | 4
 1     | 3    | 1            | 1    | 1              | 1

非常感谢任何帮助

Answer 1

您可以取消忽略值，然后进行汇总。在MySQL中，通常使用union all：

select val, count(*)
from ((select design as val from table) union all
      (select ease_of_use from table) union all
      (select responsiveness from table
     ) der
group by val
order by val desc;

对于你想要的，你可以这样做：

select val, sum(design) as design, sum(ease_of_use) as ease_of_use,
       sum(responsiveness) as responsiveness
from ((select design as val, 1 as design, 0 as ease_of_use, 0 as responsiveness from table) union all
      (select ease_of_use, 0, 1, 0 from table) union all
      (select responsiveness, 0, 0, 1 from table
     ) der
group by val
order by val desc;

我认为没有理由重复这个值三次。

Answer 2

使用具有不同值的合成表，并将其与获得每个分数计数的子查询联系起来。

SELECT nums.num AS Design, t1.count AS num1,
       nums.num AS `Ease of Use`, t2.count AS num2,
       nums.num AS Responsiveness, t3.count AS num3
FROM (SELECT 1 AS num UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5) AS nums
LEFT JOIN (
    SELECT Design, COUNT(*) AS count
    FROM yourTable
    GROUP BY Design) AS t1 ON t1.Design = nums.num
LEFT JOIN (
    SELECT `Ease of Use`, COUNT(*) AS count
    FROM yourTable
    GROUP BY `Ease of Use`) AS t2 ON t2.`Ease of Use` = nums.num
LEFT JOIN (
    SELECT Responsiveness, COUNT(*) AS count
    FROM yourTable
    GROUP BY Responsiveness) AS t3 ON t3.Responsiveness = nums.num

DEMO

Answer 3

以下是三种方式：

has_many :answers_projects
has_many :projects, through: :answers_projects

http://sqlfiddle.com/#!9/002303/2

select s.score,
  (select count(*) from tbl where `Design`         = s.score) as `Design`,
  (select count(*) from tbl where `Ease of use`    = s.score) as `Ease of use`,
  (select count(*) from tbl where `Responsiveness` = s.score) as `Responsiveness`
from (
  select Design as score from tbl
  union select `Ease of use` from tbl
  union select Responsiveness from tbl
) s
order by score desc

http://sqlfiddle.com/#!9/002303/4

select s.score,
  (select count(*) from tbl where `Design`         = s.score) as `Design`,
  (select count(*) from tbl where `Ease of use`    = s.score) as `Ease of use`,
  (select count(*) from tbl where `Responsiveness` = s.score) as `Responsiveness`
from (select 1 as score union select 2 union select 3 union select 4 union select 5) s
order by score desc

http://sqlfiddle.com/#!9/002303/5

他们都返回相同的结果：

select s.score,
  sum(`Design`         = score) as `Design`,
  sum(`Ease of use`    = score) as `Ease of use`,
  sum(`Responsiveness` = score) as `Responsiveness`
from (select 1 as score union select 2 union select 3 union select 4 union select 5) s
cross join tbl t
group by s.score
order by s.score desc

@futureweb在评论中写道，我没有理由重复三次得分。虽然你可以使用别名。

如果您有数百万行;-)而且没有索引，您只希望通过一次表扫描获得结果。这可以通过以下方式实现：

| score | Design | Ease of use | Responsiveness |
|-------|--------|-------------|----------------|
|     5 |      3 |           1 |              1 |
|     4 |      1 |           3 |              2 |
|     3 |      2 |           2 |              2 |
|     2 |      1 |           3 |              4 |
|     1 |      3 |           1 |              1 |

这将返回您需要的数据，但不会返回您想要的数据：

select
  sum(`Design` = 1) as d1,
  sum(`Design` = 2) as d2,
  sum(`Design` = 3) as d3,
  sum(`Design` = 4) as d4,
  sum(`Design` = 5) as d5,
  sum(`Ease of use` = 1) as e1,
  sum(`Ease of use` = 2) as e2,
  sum(`Ease of use` = 3) as e3,
  sum(`Ease of use` = 4) as e4,
  sum(`Ease of use` = 5) as e5,
  sum(`Responsiveness` = 1) as r1,
  sum(`Responsiveness` = 2) as r2,
  sum(`Responsiveness` = 3) as r3,
  sum(`Responsiveness` = 4) as r4,
  sum(`Responsiveness` = 5) as r5
from tbl

所以你需要发布它。

mysql计算在多个列中显示相同值的次数

3 个答案: