Question

我想将booking_class_rcd的数量记为NoOfPax，将FareAmount的总数记为Total Fare，将commission_amount的总数记为TotalCommission特定的航班日期。我找不到正确的查询。

Booking_Class_rcd – 'N'，'B'，'C'，'D'

 Select agency_code,departure_date, booking_class_rcd, COUNT(booking_class_rcd)
       AS NoOfPax,origin_rcd, destination_rcd,SUM (fare_amount) as TotalFare,Sum(commission_amount) as TotalCommissionAmount
        from passenger_segment_mapping
    WHERE agency_code=’TEST’
    Group By booking_class_rcd
    Order by departure_date desc

我的结果看起来像

    AgencyCode TEST Departure Date Booking_Class_rcd NoOfPax, origin, destination, TotalFareAmount TotalCommissionAmount
    TEST 2018-06-03 B 10 KTM PKR 30000 3000
    TEST 201806-03 C  20 KTM PKR 20000  2000

Answer 1

您似乎正在使用允许非标准group by查询的MySQL。 SQL标准要求所有“非聚合”列均用于定义group by子句中的组。

SELECT
        /* non-aggregating columns */
    agency_code
  , departure_date
  , booking_class_rcd
  , origin_rcd
  , destination_rcd

       /* aggregating columns */
  , COUNT(booking_class_rcd) AS noofpax
  , SUM(fare_amount)         AS totalfare
  , SUM(commission_amount)   AS totalcommissionamount

FROM passenger_segment_mapping
WHERE agency_code = 'TEST'
GROUP BY
        /* repeat the non-aggregating columns here (without any column aliases) */
    agency_code
  , departure_date
  , booking_class_rcd
  , origin_rcd
  , destination_rcd
ORDER BY
    departure_date DESC

非聚合列是不使用聚合函数（例如sum（）min（）max（）count（）等）的任何列。

Answer 2

在group by中添加所有聚合和非聚集字段，我得到了解决方案。

   Select departure_date,agency_code, origin_rcd +'-'+destination_rcd as Sector,booking_class_rcd,  count(departure_date) AS NoOfPax, SUM(fare_amount) as TotalFare, SUM(commission_amount) as TotalCommission
    from passenger_segment_mapping
  WHERE agency_code='TEST'
  Group BY departure_date, agency_code, origin_rcd, passenger_check_in_status_rcd, destination_rcd,booking_class_rcd, fare_amount, commission_amount
  ORDER BY departure_date asc

按聚合函数排序SQL

2 个答案: