Question

我想计算具有完全相同结构的两个表的行之间的增量，但不一定是相同的数据。

表1（预测）

Id |名称| GP | G |甲

表1（实际数据）

Id |名称| GP | G |甲

结果（与Id匹配）：

Id | （GP1-GP2）AS DeltaGP | （G1-G2）AS DeltaG | （A1-A2）AS DeltaA |

抓住我的漂移？这是为了解决SSAS中预测模型的错误

这是我的代码：

    SELECT P.[Player_id]
          ,P.[PlayerName]
          ,sum(P.[Games Played])-sum(S.[GamesPlayed]) AS GP
          ,sum(P.[Total Goals])-sum(s.[TotalGoals]) AS Goals
          ,sum(P.[Total Assists])-sum(s.[TotalAssists]) AS Assists   
FROM [PredictionsPlayersHistory] as P   
JOIN [V_StatsPlayers] AS S on p.pLAYER_id = s.Player_id    
where S.LastPlayedDate >= '2010-02-01' and P.Player_id
    = 8471675   group by P.[Player_id],P.[PlayerName]   
order by Goals desc, Assists desc, GP asc

问题是结果不对：

预测（SUM）

PlayerName GP目标助攻

Sidney Crosby 56 28 34

实际数据（SUM）

PlayerName GP目标助攻

Sidney Crosby 26 17 24

结果

Sidney Crosby 28 -42 -98

Answer 1

你可以通过ID加入桌子吗？如果我做对了，那么：

SELECT table1.id, table1.name,
       (table1.gp - table2.gp) as DeltaGP,
       (table1.g - table2.g) as DeltaG,
       (table1.a - table2.a) as DeltaA
FROM table1
JOIN table2
ON table1.id = table2.id

Answer 2

您需要加入id列上的两个表：

SELECT a.Id,a.Name, a.GP - p.GP AS DeltaGP, a.G - p.G AS DeltaG,  a.A - p.A AS DeltaA
FROM dbo.table1 AS a --actual values
JOIN dbo.table2 AS p --predictions
ON a.Id = p.Id;

有关SQL Server联接的介绍，请查看我的帖子系列：http://sqlity.net/en/1146/a-join-a-day-introduction/

编辑：如果每个表中每个玩家有多行，则需要在加入之前聚合它们：

SELECT  a.Id,
        a.Name,
        a.GP - p.GP AS DeltaGP,
        a.G - p.G AS DeltaG,
        a.A - p.A AS DeltaA
FROM    (
          SELECT  Id,
                  Name,
                  SUM(GP) AS GP,
                  SUM(G) AS G,
                  SUM(A) AS A
          FROM    dbo.table1
          GROUP BY Id,
                  Name
        ) AS a --actual values
JOIN    (
          SELECT  Id,
                  SUM(GP) AS GP,
                  SUM(G) AS G,
                  SUM(A) AS A
          FROM    dbo.table2
          GROUP BY Id
        ) AS p
        --predictions
  ON a.Id = p.Id;

根据您的要求，您可能希望使用AVG而不是SUM。

比较SQL中两个表之间的差异（错误）

2 个答案: