Question

鉴于下表（请参见下文），我需要创建一个视图，以显示表MedDRARelations中的每个代码（SOC，HLGT，HLT，PT和LLT）及其各自的值（Soc_Value，HLGT_value，HLT_Value，表MedDRANames中的PT_Value和LLT_Value。

我尝试过多次联接MedDRANames表并创建一个标量函数，该函数返回给定代码（键）的值。

表定义：

CREATE TABLE dbo.MedDRANames 
(
    [Key]   INT NOT NULL,
    [Value] VARCHAR(100) NULL,
    CONSTRAINT [PK_MedDRANames] PRIMARY KEY CLUSTERED ([Key])
);

CREATE TABLE dbo.MedDRARelations 
(
    [SOC]  INT NOT NULL,
    [HLGT] INT NOT NULL,
    [HLT]  INT NOT NULL,
    [PT]   INT NOT NULL,
    [LLT]  INT NOT NULL,    
    CONSTRAINT [PK_MedDRARelations] 
        PRIMARY KEY CLUSTERED ([SOC] ASC, [HLGT] ASC, [HLT] ASC, [PT] ASC, [LLT] ASC),
    CONSTRAINT [FK_MedDRANameSOC] 
        FOREIGN KEY ([SOC]) REFERENCES dbo.MedDRANAmes([Key]),
    CONSTRAINT [FK_MedDRANameHLGT] 
        FOREIGN KEY ([HLGT]) REFERENCES dbo.MedDRANAmes([Key]),
    CONSTRAINT [FK_MedDRANameHLT] 
        FOREIGN KEY ([HLT]) REFERENCES dbo.MedDRANAmes([Key]),
    CONSTRAINT [FK_MedDRANamePT] 
        FOREIGN KEY ([PT]) REFERENCES dbo.MedDRANAmes([Key]),
    CONSTRAINT [FK_MedDRANameLLT] 
        FOREIGN KEY ([LLT]) REFERENCES dbo.MedDRANAmes([Key])
);

观看次数：

此视图在22.1 s（SD 1.25）内执行，影响156867行

SELECT 
    [X].[SOC] AS [Código SOC],
    [SOC].[Value] AS [Término SOC],
    [X].[HLGT] AS [Código HLGT],
    [HLGT].[Value] AS [Término HLGT],
    [X].[HLT] AS [Código HLT],
    [HLT].[Value] AS [Término HLT],
    [X].[PT] AS [Código PT],
    [PT].[Value] AS [Término PT],
    [X].[LLT] AS [Código LLT],
    [LLT].[Value] AS [Término LLT]
FROM 
    dbo.MedDRARelations AS [X]
INNER JOIN 
    dbo.MedDRANames AS [SOC] ON [X].[SOC] = [SOC].[Key]
INNER JOIN 
    dbo.MedDRANames AS [HLGT] ON [X].[HLGT] = [HLGT].[Key]
INNER JOIN 
    dbo.MedDRANames AS [HLT] ON [X].[HLT] = [HLT].[Key]
INNER JOIN 
    dbo.MedDRANames AS [PT] ON [X].[PT] = [PT].[Key]
INNER JOIN 
    dbo.MedDRANames AS [LLT] ON [X].[LLT] = [LLT].[Key]

此视图在35.3 s（SD 2.1）中执行，影响156867行

SELECT 
    [X].[SOC] AS [Código SOC],
    dbo.FindMedDRA(x.SOC) AS [Término SOC],
    [X].[HLGT] AS [Código HLGT],
    dbo.FindMedDRA(x.HLGT) AS [Término HLGT],
    [X].[HLT] AS [Código HLT],
    dbo.FindMedDRA(x.HLT) AS [Término HLT],
    [X].[PT] AS [Código PT],
    dbo.FindMedDRA(x.PT) AS [Término PT],
    [X].[LLT] AS [Código LLT],
    dbo.FindMedDRA(x.LLT) AS [Término LLT]
FROM 
    dbo.MedDRARelations AS [X]

标量函数：

CREATE FUNCTION [dbo].[FindMedDRA]
    (@code INT)
RETURNS VARCHAR(100)
AS
BEGIN
    DECLARE @returning VARCHAR(100)

    SELECT @returning = dbo.MedDRANames.Value 
    FROM dbo.MedDRANames 
    WHERE dbo.MedDRANames.[Key] = @code

    RETURN @returning
END

我想知道什么是实现此目的的正确方法，因为创建联接表耗时12.2 s（SD 0.1），但成本是原始空间的5倍。

谢谢。

更新：执行计划

使用功能：https://www.brentozar.com/pastetheplan/?id=rkso0aKnV
在创建IX之前：https://www.brentozar.com/pastetheplan/?id=HkFlyRth4
在Ixs创建之后（如Laughing Vergil的建议）：https://www.brentozar.com/pastetheplan/?id=B13ZW0t3V

Answer 1

第一种方法是正确的方法。单列整数主键上的INNER JOIN很快，通常通过MERGE JOIN运算符执行。结果肯定可以证明，肯定比创建函数要快。

Answer 2

这里最大的问题是MedRARelations中的值没有索引。注意：一个常见的误解是创建外键会创建索引。不会。

因此，请尝试运行此命令：

CREATE INDEX Idx_MedDRARelations_SOC ON MedDRARelations (SOC);
CREATE INDEX Idx_MedDRARelations_HGLT ON MedDRARelations (HGLT);
CREATE INDEX Idx_MedDRARelations_HLT ON MedDRARelations (HLT);
CREATE INDEX Idx_MedDRARelations_PT ON MedDRARelations (PT);
CREATE INDEX Idx_MedDRARelations_LLT ON MedDRARelations (LLT);

然后再次尝试您的视图和其他方法。

SQL Server如何在依赖表上正确创建视图？

2 个答案: