MySQL选择查询：计算计数和获取总计

Question

我需要将所有Start_Stop_ID和End_Stop_ID组合到一个ID列中，但我需要将每个Stop_ID的Start和End数量计算到单独的列中。

Start_Stop_ID的每一行都有Current_Fare。我需要计算与Start_Stop_ID关联的Current_Fare的总数，而不是End_Stop_ID。

The columns would be:
ID | Start | END | TOTAL

Example:
There would be rows such as
E1      | 1 | 0 | 3.50
BUSDOME | 3 | 0 | 18.50 (BUSDOME has 3 Start_Stop_ID, 10.50, 4.00 and 4.00)
N4      | 1 | 3 | 1.50
E5      | 2 | 0 | 6.00  (E5 has 2 Start_Stop_ID, 3.00 and 3.00)
E7      | 0 | 1 | 0.00

旅行表

Answer 1

这是我的解决方案：

SELECT i.ID, SUM(i.Start) Start, SUM(i.End) End, SUM(i.Total) Total FROM (
    SELECT Start_Stop_ID ID, COUNT(Start_Stop_ID) AS Start, 0 AS End, SUM(Current_Fare) AS Total 
    FROM trip_table GROUP BY ID
    UNION ALL 
    SELECT End_Stop_ID ID, 0 AS Start, COUNT(End_Stop_ID) AS End, 0 AS Total 
    FROM trip_table WHERE End_Stop_ID IS NOT NULL GROUP BY ID
    ORDER BY ID ASC) i 
GROUP BY i.ID

此查询的作用是基于Start_Stop_ID创建结果，另一个结果集基于End_Stop_ID，然后使用UNION ALL将它们组合在一起，然后对总计进行求和。

创建第一个结果集时，End将为0，因为我们只计算起始停止ID。创建第二个结果集时，Start将为0，Total（票价）也为0，因为票价仅与开始结果集相关联。

*修正了查询中的拼写错误

Answer 2

这将计算票价总额并执行计数，但另一列|End|的逻辑不明确。

select Start_Stop_ID, count(*) num_stops, sum(Current_Fare) AS total
from trip_tableclear
group by Start_Stop_ID

Answer 3

解决方案查询：

SELECT Start_Stop_ID as 'ID', count(num_stops) as 'Count', SUM(Current_Fare) as 'Total Fare' FROM trip_tableclear GROUP BY Start_Stop_ID;