SQL中实例的乘以数

Question

我有一个数据库，我需要查询来统计和计算篮球比赛中球员传球产生的积分数。例如，如果玩家传递给队友并且以2分结果，我的数据库当前将其作为一个实例存储在某个表中。如果通过结果为3分，还有另一个表格存储数据。我想查询一下，所有得到2分的球员传球实例都会被计数，然后乘以2，并且所有得到3分的球员传球实例都会被计数，然后乘以3。

这里是我的相关表和SELECT语句：

CREATE TABLE Passer(
PasserID int identity PRIMARY KEY not null 
, Forename char(30) not null 
, Surname char (30) not null) 

CREATE TABLE Teammate(
TeammateID int identity PRIMARY KEY not null 
, Forename char(30) not null 
, Surname char(30) not null 
, PasserID int FOREIGN KEY REFERENCES Passer(PasserID) not null) 

CREATE TABLE TwoPointsFromShot(
    TwoPointsFromShotID int identity PRIMARY KEY not null 
    , PasserID int FOREIGN KEY REFERENCES Passer(PasserID) not null 
    , TeammateID int FOREIGN KEY REFERENCES Teammate(TeammateID) not null) 

CREATE TABLE ThreePointsFromShot(
    ThreePointsFromShotID int identity PRIMARY KEY not null 
    , PasserID int FOREIGN KEY REFERENCES Passer(PasserID) not null 
    , TeammateID int FOREIGN KEY REFERENCES Teammate(TeammateID) not null 

--First and Last Name of Passer from TwoPointsFromShot-- 
SELECT Forename, Surname 
FROM Passer 
JOIN TwoPointsFromShot ON TwoPointsFromShot.PasserID = Passer.PasserID 

--First And Last name of Passer from ThreePointsFromShot-- 
SELECT Forename, Surname 
FROM Passer 
JOIN ThreePointsFromShot ON ThreePointsFromShot.PasserID = Passer.PasserID 

当我查询从TwoPointsFromShot表PasserID，我收到一个表，看起来像：

    | PasserID 
------------------- | -------- 
       1  | 1 
       2  | 3 
       3  | 3 
       4  | 2 

我收到了类似的表当从ThreeP查询PasserID时ointsFromShot表。

     | PasserID 
--------------------- | -------- 
       1  | 3 
       2  | 1 
       3  | 3 
       4  | 4 

我想返回计数从TwoPointsFromShot的PasserID的实例的数量，并乘以通过2，计算从ThreePointsFromShot和倍数PasserID的实例数量由3，总结这两个查询值，并用播放器的名称替换PasserID。因此，它看起来像这样（如果朱利叶斯·兰德尔是PlayerID 1，达里奥·萨里奇是PlayerID 2，TJ·麦康奈尔是PlayerID 3，和布兰登英格拉姆是PlayerID 4）：

| PasserName | PointsFromTwo | PointsFromThree | PassToPoints 
--- | ---------  | --------------- | ------------------ | ------------ 
    1 | Julius Randle |  2   |  3   |  5 
    2 | Dario Saric |  2   |  0   |  2 
    3 | TJ McConnell |  4   |  6   |  10 
    4 | Brandon Ingram|  0   |  3   |  3 

任何帮助将不胜感激！

Answer 1

我能够通过使用两个common table expressions获得想要的结果。

--CTE to get number of 2 pointers 
WITH twoPointers AS 
(
    SELECT PasserId,COUNT(*) '2ptCount' 
    FROM TwoPointsFromShot 
    GROUP BY PasserID 
), 
--CTE to get number of 3 pointers 
threePointers As 
(
    SELECT PasserId,COUNT(*) '3ptCount' 
    FROM ThreePointsFromShot 
    GROUP BY PasserID 
) 
--Join the Passer table with 2 CTE's and 
--calculate results 
SELECT RTRIM(Forename) + ' ' + RTRIM(Surname) AS 'PasserName', 
     ISNULL(two.[2ptCount] * 2, 0.00) AS 'PointsFromTwo', 
     ISNULL(three.[3ptCount] * 3, 0.00) AS 'PointsFromThree', 
     ISNULL(two.[2ptCount] * 2, 0.00) + ISNULL(three.[3ptCount] * 3, 0.00) AS 'PassToPoints' 
FROM Passer p 
    LEFT JOIN twoPointers two ON p.PasserID = two.PasserID 
    LEFT JOIN threePointers three ON p.PasserID = three.PasserID