Question

首先抱歉模糊的标题不知道如何解释它。在下面的查询中，我想把事件从'truck1'参加的地方拉出来。不仅仅是自己，而是当其他车辆参加时。我确定它的东西是直接的，但无法解决。

select
i.incident_number
Vehicles,
countofvehicles

FROM

(


SELECT


i.Incident_Number,
array_agg(ir.RESOURCE) as vehicle,
count(ir.RESOURCE) as countofvehicles



FROM INCIDENT as i
JOIN RESOURCE as ir on i.Incident_Number = ir.Incident_Number 

--WHERE ir.  like '%Truck1%'

GROUP BY i.Incident_Number) i

结果

incident_Number       vehicle               countofvehicle
     1             car1,car2,bike1                 3
     2             car1,car2,truck1                3
     3             truck1                          1
     4             car1                            1

如果我想看看卡车1只参加了哪些事故，那么使用WHERE ir.RESOURCE就像'％truck1％只会带回事故3号而不是事故2，它与其他车辆一起参加。我怎么能绕过这个呢？

由于

Answer 1

您不需要子查询，只需having子句：

SELECT i.Incident_Number,
       array_agg(ir.RESOURCE) as vehicle,
       count(ir.RESOURCE) as countofvehicles
FROM INCIDENT i JOIN
     RESOURCE ir 
     ON i.Incident_Number = ir.Incident_Number 
GROUP BY i.Incident_Number
HAVING SUM(CASE WHEN ir.RESOURCE like '%truck1% THEN 1 ELSE 0 END) > 1;

事实上，您可能也不需要join，因为您从INCIDENT获取的唯一字段也位于RESOURCE中：

SELECT ir.Incident_Number,
       array_agg(ir.RESOURCE) as vehicle,
       count(ir.RESOURCE) as countofvehicles
FROM RESOURCE ir 
GROUP BY ir.Incident_Number
HAVING SUM(CASE WHEN ir.RESOURCE like '%truck1% THEN 1 ELSE 0 END) > 1;

如何根据非标准化列进行选择

1 个答案: