基于最新交易日期和先前交易日期的价格增长百分比

Question

，

我想根据最新交易日期和上一交易日期，通过国家和城市获得价格上涨百分比。

我如何查询？我没有得到它。这是我尝试过的：

SELECT Country,City, Price
From tbl
Group by Country,City 

增长百分比= [（最新价格 - 前一个价格）/上一个价格] * 100

预计出局： 独特的国家和城市名称+价格增长百分比 国家|城市|百分比

Answer 1

使用lag()：

select t.*,
       100 * ((price / lag(price) over (partition by country, city order by t_date) - 1) as increase
from t;

Answer 2

这可能过于复杂。

设置一些随机数据：

IF OBJECT_ID('tempdb..#Cities') IS NOT NULL
BEGIN
    DROP TABLE #Cities;
END;

CREATE TABLE #Cities
(
    Country VARCHAR(20)
   , City    VARCHAR(20)
);

IF OBJECT_ID('tempdb..#Data') IS NOT NULL
BEGIN
    DROP TABLE #Data;
END;

CREATE TABLE #Data
(
    Country VARCHAR(20)
   , City    VARCHAR(20)
   , Price   DECIMAL(13, 4)
   , Date    DATETIME
);

INSERT INTO #Cities
VALUES      ('Country 1', 'City 1'), ('Country 1', 'City 2'), ('Country 1', 'City 3'), ('Country 2', 'City 4'), ('Country 2', 'City 5');


INSERT INTO #Data
SELECT Country
    , City
    , ROUND(RAND(CHECKSUM(NEWID())) * 100, 4) AS Price
    , DATEADD(DAY, ROUND(RAND(CHECKSUM(NEWID())) * 10, 0), GETDATE()) AS Date
FROM     #Cities
UNION
SELECT Country
    , City
    , ROUND(RAND(CHECKSUM(NEWID())) * 100, 4)
    , DATEADD(DAY, ROUND(RAND(CHECKSUM(NEWID())) * 10, 0), GETDATE())
FROM     #Cities;

--Delete duplicate dates

WITH data3 AS
(
SELECT *,ROW_NUMBER() OVER (PARTITION BY Country,City,Date ORDER BY Country,City,Date) AS RN
FROM #Data
)
DELETE FROM data3 WHERE RN<>1

查询数据以获取最新价格，日期和百分比增加：

SELECT        Dates.*
          , Latest.Price AS Latestprice
          , Previous.Price AS Previousprice
          , ((Latest.Price - Previous.Price) / Previous.Price) * 100 AS Percentageincrease
FROM
(
    SELECT    C.*
          , Latestdate.Latestdate
          , Previousdate.Previousdate
    FROM      #Cities AS C
            LEFT JOIN
    (
       --Latest Date for each county, city
       SELECT Country
           , City
           , MAX(Date) AS Latestdate
       FROM   #Data
       GROUP BY Country
            , City
    ) AS Latestdate ON Latestdate.Country = C.Country
                   AND Latestdate.City = C.City
            LEFT JOIN
    (
       --Previous Date for each county, city
       SELECT Country
           , City
           , Date AS Previousdate
       FROM
       (
          SELECT Country
              , City
              , Date
              , RANK() OVER(PARTITION BY Country
                                  , City ORDER BY Date DESC) AS Rank
          FROM   #Data
       ) AS A
       WHERE  Rank = 2
    ) AS Previousdate ON Previousdate.Country = C.Country
                    AND Previousdate.City = C.City
) AS Dates
JOIN #Data AS Latest ON Latest.Country = Dates.Country
                    AND Latest.City = Dates.City
                    AND Latest.Date = Dates.Latestdate
JOIN #Data AS Previous ON Previous.Country = Dates.Country
                     AND Previous.City = Dates.City
                     AND Previous.Date = Dates.Previousdate

为了进行比较，使用lag（）来获得每个日期的百分比增长。与戈登的回答类似：

SELECT D.Country
    , D.City
    , D.Date
    , Lag(Date) OVER(PARTITION BY Country
                           , City ORDER BY Date) AS Previousdate
    , D.Price
    , Lag(Price) OVER(PARTITION BY Country
                            , City ORDER BY Date) AS Previousprice
    , 100 * (Price / Lag(Price) OVER(PARTITION BY Country
                                        , City ORDER BY Date) - 1) AS PercentageIncrease
FROM   #Data AS D;

使用滞后来获得与我的第一个查询相同的结果（每个城市的最新信息）：

SELECT *
FROM
(
    SELECT D.Country
        , D.City
        , D.Date
        , Lag(Date) OVER(PARTITION BY Country
                              , City ORDER BY Date) AS Previousdate
        , D.Price
        , Lag(Price) OVER(PARTITION BY Country
                               , City ORDER BY Date) AS Previousprice
        , 100 * (Price / Lag(Price) OVER(PARTITION BY Country
                                           , City ORDER BY Date) - 1) AS Percentageincrease
        , ROW_NUMBER() OVER(PARTITION BY Country
                                , City ORDER BY Date DESC) AS Rn
    FROM   #Data AS D
) AS A
WHERE  Rn = 1
ORDER BY Country
      , City;