SQL：在同一字段中使用不同where子句的2个不同计数

Question

如何将以下2个查询合并到一个查询中？

select count(distinct user_id) as action_1_count, date 
from table  
where action = "action_1" 
group by date; 

select count(distinct user_id) as action_2_count, date 
from table 
where action = "action_2" 
group by date; 

Answer 1

SELECT COUNT(user_id), date 
FROM table 
WHERE action IN ('action_1', 'action_2) 
GROUP BY action, date 

Answer 2

您的查询，可就是这样

SELECT COUNT(user_id), date FROM table WHERE action IN ('action_1', 'action_2') GROUP BY date 

下面是使用SUM()功能本应该做的伎俩

SELECT 
    SUM(action ='action_1') count_1, 
    SUM(action ='action_2') count_2, 
    SUM(action ='action_1' OR action ='action_2') `full_count`, 
    date 
FROM 
table 
GROUP BY date 

与表达式中使用SUM()将导致另一条路在布尔值中，如果action ='action_1'那么结果将是1如果没有的话0，这一切都归结生产基于表达式的结果计数

Answer 3

做这样的合并上面的查询：

SELECT date, 
     CASE WHEN action = "action_1" THEN COUNT(user_id) END AS Action1Count, 
     CASE WHEN action = "action_2" THEN COUNT(user_id) END AS Action2Count 
    FROM table  
    WHERE action IN ("action_1", "action_2") 
    GROUP BY date 

Answer 4

以上答案会给你ACTION_2多行，而看起来你只需要一行action_2。

你可以使用一个联盟：

select '1', count(distinct user_id), date 
from table 
where action = "action_1" 
group by 3 
union 
select '2', count(distinct user_id), max(date) 
from table 
where action = "action_2" 
group by 3;