Question

问题

如何通过SQL查询获得每个酒店的总费用和每8天转移一次？

详情

8天= 7晚

意思是8天=每个酒店住宿7天

因为最后一天他乘坐的航班没有留在酒店。

结果我需要得到它

我使用以下查询：

var myApp = angular.module("myApp", []);

// Main controller
myApp.controller("appCtrl", ["$scope", "$http", "$q", function ($scope, $http, $q) {

    // Get stations data  
    $http.get("https://rata.digitraffic.fi/api/v1/metadata/stations.json").then(function (response) {
        // Load the stations data
        $scope.stations = response.data;

        // We need to map the following requests into an array of promises so we can resolve them concurrently.
        var stationPromises = $scope.stations.map(function (station) {
            // Use the "angular way" to define query parameters.
            return $http.get("https://rata.digitraffic.fi/api/v1/live-trains", {
                params: { station: station.stationShortCode }
            }).then(function (response) {

                // Assign this set of trains to the station so we don't overwrite other staions trains...
                station.trains = response.data;

                // Another set of promises to get all of the departure information for each train.
                var trainPromises = station.trains.map(function (train) {
                    // Lets use the "angular way" to define parameters again.
                    return $http.get("https://rata.digitraffic.fi/api/v1/compositions/" + train.trainNumber, {
                        params: { departure_date: train.departureDate }
                    }).then(function (response) {
                        // Assign this set of compositions to the train so we don't overwrite other compositions for other trains...
                        train.compositions = response.data;
                    });
                });

                return $q.all(trainPromises);
            });
        });

        // Resolve all of the promises.
        return $q.all(stationPromises);
    }).catch(function (err) {
        // Something bad happended...
        console.error(err);
    });
}]);

我在这个小提琴中找到的样本数据

http://dbfiddle.uk/?rdbms=sqlserver_2016&fiddle=0096a903948a93c1269e931328648be2

更新原始帖子

在下面查询答案1 不能正确计算转移价值它给我1300转移成本但是正确的是700

的sampleData

;with cte_HotelPrice
as
(
select 
T6.HotelPrice,
T4.HotelID,
T5.HotelName,
T3.DetailsDurationID from package T 
inner join StartPackage T1 on T.PackageId=T1.PackageId
inner join packageduration T2 on T.PackageId=T2.PackageId
inner join (SELECT *, RN = ROW_NUMBER() OVER (PARTITION BY PackageDurationsId ORDER BY Days) 
    FROM DurationDetails) T3 on T2.PackageDurationsId=T3.PackageDurationsID
inner join DayDetails T4 on T3.DetailsDurationID=T4.DetailsDurationID
left join Hotel T5 on T4.HotelID=T5.HotelID
cross apply (select HotelPrice from HotelPrice where HotelID=T4.HotelID and FromDate<=DATEADD(day, T3.RN - 1, T1.StartDate) and ToDate>=DATEADD(day, T3.RN - 1, T1.StartDate)) T6
)
,TransferPrice as
(
select 
ttd.Price,
dds.DetailsDurationID
from package p 
inner join StartPackage s on p.PackageId=s.PackageId
inner join packageduration pd on p.PackageId=pd.PackageId
inner join (SELECT *, RN = ROW_NUMBER() OVER (PARTITION BY PackageDurationsId ORDER BY Days) 
    FROM DurationDetails) dd on pd.PackageDurationsId=dd.PackageDurationsID
inner join DayDetails dds on dd.DetailsDurationID=dds.DetailsDurationID
left join TransferType tt on dds.TransferTypeID=tt.TransferID
cross apply (select Price from TransferPeriod where TransferTypeID=dds.TransferTypeID and FromDate<=DATEADD(day, dd.RN - 1, s.StartDate) and Todate>=DATEADD(day, dd.RN - 1, s.StartDate)) ttd
)
select 
S4.HotelID,S4.HotelName, S.PackageName, S1.StartDate, S1.EndDate, 
sum(S4.HotelPrice) AS cost,
sum(S5.Price) as transfercost 
from package S 
inner join StartPackage S1 on S.PackageId=S1.PackageId
inner join packageduration S2 on S.PackageId=S2.PackageId
inner join DurationDetails S3 on S2.PackageDurationsId=S3.PackageDurationsID
left join  cte_HotelPrice S4 on S3.DetailsDurationID=S4.DetailsDurationID
left join  TransferPrice S5 on S3.DetailsDurationID=S5.DetailsDurationID
GROUP BY S4.HotelID, S4.HotelName,S.PackageName, S1.StartDate, S1.EndDate

Answer 1

这将输出所需的结果，但不确定这是否已优化或符合您的要求。我刚编辑了你的选择陈述，没有费心去检查ctes。

SELECT  S4.HotelID ,
        S4.HotelName ,
        S.PackageName ,
        S1.StartDate ,
        S1.EndDate ,
        SUM(S4.HotelPrice) AS cost ,
        TransferCost
FROM    package S
        INNER JOIN StartPackage S1 ON S.PackageId = S1.PackageId
        INNER JOIN packageduration S2 ON S.PackageId = S2.PackageId
        INNER JOIN DurationDetails S3 ON S2.PackageDurationsId = S3.PackageDurationsID
        LEFT JOIN cte_HotelPrice S4 ON S3.DetailsDurationID = S4.DetailsDurationID
        OUTER APPLY ( SELECT    SUM(Price) AS TransferCost ,
                                HotelId
                      FROM      TransferPrice
                      WHERE     S4.HotelID = HotelID
                      GROUP BY  HotelID
                    ) tp
WHERE   S4.HotelId IS NOT NULL
GROUP BY S4.HotelID ,
        S4.HotelName ,
        S.PackageName ,
        S1.StartDate ,
        S1.EndDate ,
        TransferCost;

编辑 - 对于您的新样本数据，您的CTE - TransferPrice返回4条记录，总计为1000条。有没有理由说它不应该是1000？我不熟悉业务需求。

Price  DetailsDurationID
200    DD01
200    DD01
300    DD08
300    DD08

更新 - 请参阅新查询;还将HotelID添加到您的CTE，现在我们有您喜欢的转移成本。由于转账是在第8天向Hotel1收费，因此费用列已经更改，这是我想要实现的目标。

如何通过SQL查询获得每个酒店的总费用和每8天转移一次？

1 个答案: