Question

基本上我有一张类似的表：

time.....activities.....length  
13:00........3.............1  
13:15........2.............2  
13:00........3.............2  
13:30........1.............1  
13:45........2.............3  
13:15........5.............1  
13:45........1.............3  
13:15........3.............1  
13:45........3.............2  
13:45........1.............1  
13:15........3.............3

几点说明：

活动可以在1到5之间
长度可以在1到3之间

查询应返回：

time........count  
13:00.........2  
13:15.........2  
13:30.........0  
13:45.........1

基本上，对于每个唯一的时间，我想要计算活动值为3的行数。

那么我可以说：

At 13:00 there were X amount of activity 3s.
At 13:45 there were Y amount of activity 3s.

然后我想要计算活动1s，2s，4s和5s。所以我可以绘制每个独特时间的分布。

Answer 1

是的，您可以使用group by：

select time, activities, count(*) from table group by time, activities;

Answer 2

select time, coalesce(count(case when activities = 3 then 1 end), 0) as count
from MyTable
group by time

SQL Fiddle Example

<强>输出：

|  TIME | COUNT |
-----------------
| 13:00 |     2 |
| 13:15 |     2 |
| 13:30 |     0 |
| 13:45 |     1 |

如果要计算一个查询中的所有活动，可以执行以下操作：

select time, 
    coalesce(count(case when activities = 1 then 1 end), 0) as count1,
    coalesce(count(case when activities = 2 then 1 end), 0) as count2,
    coalesce(count(case when activities = 3 then 1 end), 0) as count3,
    coalesce(count(case when activities = 4 then 1 end), 0) as count4,
    coalesce(count(case when activities = 5 then 1 end), 0) as count5
from MyTable
group by time

这比活动分组的优点是，即使该时间段没有该类型的活动，它也会返回0。

当然，这不会返回没有任何类型活动的时间段的行。如果需要，您需要使用左连接和表，列出所有可能的时间段。

Answer 3

如果我理解你的问题，这会有用吗？（您必须替换实际的列和表名称）

SELECT time_col, COUNT(time_col) As Count
FROM time_table
GROUP BY time_col
WHERE activity_col = 3

Answer 4

您应该将查询更改为：

SELECT time_col, COUNT(time_col) As Count
FROM time_table
WHERE activity_col = 3
GROUP BY time_col

此vl正常工作。

计算每个唯一值的出现次数

4 个答案: