Question

我有这张桌子：

╔════════════════╤═══════════════════╤═══════════╤═══════════╗
║ question1      │ question2         │ question3 │ question4 ║
╠════════════════╪═══════════════════╪═══════════╪═══════════╣
║ Agree          │ Disagree          │ Agree     │ Disagree  ║
╟────────────────┼───────────────────┼───────────┼───────────╢
║ Strongly Agree │ Strongly Disagree │ Agree     │ Disagree  ║
╚════════════════╧═══════════════════╧═══════════╧═══════════╝

我正在尝试使用COUNT（）编写一个查询，它显示每个问题的回复数量，如下所示：

╔══════════╤════════════════╤═══════╤══════════╤═══════════════════╗
║ Question │ Strongly Agree │ Agree │ Disagree │ Strongly Disagree ║
╠══════════╪════════════════╪═══════╪══════════╪═══════════════════╣
║ Q1       │ 1              │ 1     │ 0        │ 0                 ║
╟──────────┼────────────────┼───────┼──────────┼───────────────────╢
║ Q2       │ 0              │ 0     │ 1        │ 1                 ║
╟──────────┼────────────────┼───────┼──────────┼───────────────────╢
║ Q3       │ 0              │ 2     │ 0        │ 0                 ║
╟──────────┼────────────────┼───────┼──────────┼───────────────────╢
║ Q4       │ 0              │ 0     │ 2        │ 0                 ║
╚══════════╧════════════════╧═══════╧══════════╧═══════════════════╝

我尝试了几次查询，但总是给我错误的结果。任何帮助，将不胜感激。感谢。

Answer 1

不确定为什么选择按照显示的方式构建表格，但如果您可以灵活地更改它，我建议您这样做。使用现在提供的结构，您不仅在使用所需结果获得正确查询时遇到问题，而且还有一个结构无法在没有DB模式更新的情况下添加新问题。

如果您无法修改表格结构 的 SQL DEMO

SELECT 'Q1' as Question , Count(CASE WHEN Question1 = 'Strongly Agree' THEN 1 END) AS 'Strongly Agree', Count(CASE WHEN Question1 = 'Agree' THEN 1 END) AS 'Agree', Count(CASE WHEN Question1 = 'Disagree' THEN 1 END) AS 'Disagree', Count(CASE WHEN Question1 = 'Strongly Disagree' THEN 1 END) AS 'Strongly Disagree' FROM QandR UNION ALL SELECT 'Q2' as Question , Count(CASE WHEN Question2 = 'Strongly Agree' THEN 1 END) AS 'Strongly Agree', Count(CASE WHEN Question2 = 'Agree' THEN 1 END) AS 'Agree', Count(CASE WHEN Question2 = 'Disagree' THEN 1 END) AS 'Disagree', Count(CASE WHEN Question2 = 'Strongly Disagree' THEN 1 END) AS 'Strongly Disagree' FROM QandR UNION ALL SELECT 'Q3' as Question , Count(CASE WHEN Question3 = 'Strongly Agree' THEN 1 END) AS 'Strongly Agree', Count(CASE WHEN Question3 = 'Agree' THEN 1 END) AS 'Agree', Count(CASE WHEN Question3 = 'Disagree' THEN 1 END) AS 'Disagree', Count(CASE WHEN Question3 = 'Strongly Disagree' THEN 1 END) AS 'Strongly Disagree' FROM QandR UNION ALL SELECT 'Q4' as Question , Count(CASE WHEN Question4 = 'Strongly Agree' THEN 1 END) AS 'Strongly Agree', Count(CASE WHEN Question4 = 'Agree' THEN 1 END) AS 'Agree', Count(CASE WHEN Question4 = 'Disagree' THEN 1 END) AS 'Disagree', Count(CASE WHEN Question4 = 'Strongly Disagree' THEN 1 END) AS 'Strongly Disagree' FROM QandR

如果您可以更改结构

以下是我的建议：

2表：问题＆amp; QuestionResponse

问题有2列

id（int; autoincrement）

问题（varchar）

QuestionRresponse有3列

id（int; autoincrement）

QuestionId（int; FK to Question：id）

响应（varchar）

然后，您可以使用此查询和输出获取您正在查找的数据：

SELECT q.Question, qr.Response, Count(qr.Response) as Count FROM `Question` q LEFT JOIN QuestionResponse qr ON q.id = qr.QuestionId GROUP BY q.Question,qr.Response

Answer 2

首先，您需要两个表{{1}}和Questions，以便执行Answers并填写LEFT JOIN

NULL's

然后你需要打开你的桌子。 MySQL - turn table into different table

CREATE TABLE `Questions` (`id` int, `Question` varchar(17));  
INSERT INTO `Questions`  (`id`, `Question`)
VALUES
    (1, 'question1'),(2, 'question2'),(3, 'question3'),(4, 'question4');

CREATE TABLE `Answers`  (`id` int, `choice` varchar(17));
INSERT INTO `Answers`   (`id`, `choice`)
VALUES
    (1, 'Strongly Agree'),(2, 'Agree'),(3, 'Disagree'),(4, 'Strongly Disagree');

然后，您将两个结果连接在一起并执行数据透视MySQL pivot table

<强> SQL Fiddle Demo

select c.col,
       case c.col
            when 'question1' then question1
            when 'question2' then question2
            when 'question3' then question3
            when 'question4' then question4
       end as `data` 
from yourTable t     
cross join
       (
        select 'question1' as col
        union all select 'question2'
        union all select 'question3'
        union all select 'question4'
       ) c

<强>输出

计算多列中的不同值

2 个答案: