我有一个要总结的简单mysql表。这是我的桌子
表名称:hauling_trip
列:
ID, CONTRACTOR, HAULING_SHEET_NO, FROM_LOCATION, TO_LOCATION, LOAD_WEIGHT_KG, PASS_NO
数据:
1 PM A00001 Stumping A Pangkalan 940.00 112233
2 PM A00002 Stumping B Pangkalan 530.00 112233
3 PM A00003 Stumping B Pangkalan 970.00 112244
4 PM A00004 Stumping A Pangkalan 450.00 112244
5 TKF A00005 Stumping A Pangkalan 850.00 112255
6 TKF A00006 Stumping B Pangkalan 780.00 112255
7 TKF A00007 Stumping A Pangkalan 903.00 112266
我想做的是,我想通过显示每个PASS_NO的总行程数和负载重量(KG)来汇总此表。这是预期的结果
PASS_NO CONTRACTOR WEIGHT_A TRIP_A WEIGHT_B TRIP_B
112233 PM 940.00 1 530.00 0
112244 PM 450.00 0 970.00 1
112255 TKF 850.00 1 780.00 0
112266 TKF 903.00 1 0.00 0
相同的PASS_NO表示,牵引单被假定为1次行程。例如,牵引工作表A00001和A00002具有相同的PASS_NO,即112233。因此,这两个牵引工作表都属于1次行程,并且此行程属于FROM_LOCATION装卸站A,因为重量大于装卸站B。
这是我目前正在处理的查询。但是我不知道如何正确地旅行。
SELECT
a.PASS_NO,
a.CONTRACTOR,
( SELECT sum( LOAD_WEIGHT_KG ) FROM hauling_trip WHERE PASS_NO = a.PASS_NO AND FROM_LOCATION = 'Stumping A' ) AS WEIGTH_A,
( SELECT count( PASS_NO ) FROM hauling_trip WHERE PASS_NO = a.PASS_NO AND FROM_LOCATION = 'Stumping A' ) AS TRIP_A,
( SELECT sum( LOAD_WEIGHT_KG ) FROM hauling_trip WHERE PASS_NO = a.PASS_NO AND FROM_LOCATION = 'Stumping B' ) AS WEIGHT_B,
( SELECT count( PASS_NO ) FROM hauling_trip WHERE PASS_NO = a.PASS_NO AND FROM_LOCATION = 'Stumping B' ) AS TRIP_B,
SUM(a.LOAD_WEIGHT_KG) AS TOTAL_WEIGHT
FROM
hauling_trip AS a
GROUP BY
a.PASS_NO, a.CONTRACTOR
答案 0 :(得分:0)
使用条件聚合:
SELECT ht.PASS_NO, ht.CONTRACTOR,
sum(case when ht.from_location = 'Stumping A' then ht.load_weight_kg end) as weight_a,
sum(ht.from_location = 'Stumping A') as trip_a,
sum(case when ht.from_location = 'Stumping B' then ht.load_weight_kg end) as weight_b,
sum(ht.from_location = 'Stumping B') as trip_b
sum(ht.LOAD_WEIGHT_KG) AS TOTAL_WEIGHT
FROM hauling_trip ht
GROUP BY ht.PASS_NO, ht.CONTRACTOR