如何使SQL查询选择唯一的ID并按日期在列中排列它们？ [重复]

Question

这个问题已经在这里有了答案：

• MySQL数据透视表 8答案

大家好！ 您能帮我进行SQL查询以选择唯一的ID并将它们按日期从01.01.2018到04.01.2018排列在列中吗？ 在我的表中登录名是varchar类型。

users

id     login
-----------------
100  04.01.2018
101  01.01.2018
180  02.01.2018
103  03.01.2018
104  02.03.2018
205  03.01.2018
203  05.04.2018

Required view

01.01.2018   02.01.2018   03.01.2018   04.01.2018  
1           1            2             1          

我进行这样的查询，但它不起作用

SELECT COUNT(DISTINCT id) as '01.01.2018' FROM users
where login = '01.01.2018'
union
SELECT COUNT(DISTINCT id) as '02.01.2018' FROM users
where login = '02.01.2018';
union
SELECT COUNT(DISTINCT id) as '03.01.2018' FROM users
where login = '03.01.2018';
union
SELECT COUNT(DISTINCT id) as '06.04.2018' FROM users
where login = '04.01.2018';

我将不胜感激任何帮助和指示！

Answer 1

您可以进行条件聚合：

SELECT COUNT(DISTINCT CASE WEHN login = '01.01.2018' THEN id END) AS '01.01.2018',
       COUNT(DISTINCT CASE WEHN login = '02.01.2018' THEN id END) AS '02.01.2018',
       COUNT(DISTINCT CASE WEHN login = '03.01.2018' THEN id END) AS '01.01.2018',
       COUNT(DISTINCT CASE WEHN login = '04.01.2018' THEN id END) AS '01.01.2018'
FROM users;

但是，您需要的输出提示我：

SELECT SUM( login = '01.01.2018' ) AS '01.01.2018',
       SUM( login = '02.01.2018' ) AS '02.01.2018',
       SUM( login = '03.01.2018' ) AS '03.01.2018',
       SUM( login = '04.01.2018' ) AS '04.01.2018'
FROM users;