得到記錄，其中一個科拉姆具有跨越具有相同的列名

Question

與表像下面

+------+-----+------+----------+-----------+ 
| city | day | hour | car_name | car_count | 
+------+-----+------+----------+-----------+ 
| 1 | 12 | 00 | corolla |   8 | 
| 1 | 12 | 00 | city  |   9 | 
| 1 | 12 | 00 | amaze |   3 | 
| 1 | 13 | 00 | corolla |  17 | 
| 1 | 13 | 00 | city  |   2 | 
| 1 | 13 | 00 | amaze |   8 | 
| 1 | 14 | 00 | corolla |   3 | 
| 1 | 14 | 00 | amaze |   1 | 
+------+-----+------+----------+-----------+ 

需要找出city, day, hour其中car_count所有car_name記錄範圍內的值s是> = 3和< = 10

預期的結果

| city | day | hour | 
+------+-----+------+ 
| 1 | 12 | 00 | 

Answer 1

使用group by和having。

select city,day,hour 
from tablename 
group by city,day,hour 
having sum(case when car_count>=3 and car_count<=10 then 1 else 0 end) = count(*) 

Answer 2

select city, day, hour 
from t 
group by 1, 2, 3 
having bool_and(car_count >= 3) 

Answer 3

可以組由在城市，日和小時爲條件總和（你的條件）=計數（你的病情）
所以基本上我們正在創建一個標誌對於滿足條件「10> = car_count> = 3」的每一行。現在，我們正在總結所有的旗幟和他們同時計數，如果兩個計數和總和是相等的，這意味着你的條件「10> = car_count> = 3」是針對城市，日和小時

create table want as 
select city,day,hour from have 
group by city,day,hour 
having sum(car_count>=3 and car_count<=10)=count(car_count>=3 and car_count<=10); 

所有的車真

如有任何疑問，請讓我知道。