SQL多重左加入子查询？

有关如何解决此查询依赖性问题的任何想法？子查询会有帮助吗？我使用的数据库是SQL Server 2012的SQL多重左加入子查询？

FROM [Scheduling].[studentsection] AS [table027] 

Left JOIN [Grading].[StudentGradeBucket] AS [table028] 
ON ([table028].[StudentSectionID] = [table027].[StudentSectionID]) 
    And (@0 = [table002].[label]) 

Left JOIN [Grading].[GradingPeriodGradeBucket] AS [table029] 
ON [table028].[GradingPeriodGradeBucketID] = [table029].[GradingPeriodGradeBucketID] 

Left JOIN [Grading].[GradeBucket] AS [table002] 
ON [table029].[GradeBucketID] = [table002].[GradeBucketID] 

Left JOIN [Grading].[GradeBucketType] AS [table001] 
ON [table002].[GradeBucketTypeID] = [table001].[GradeBucketTypeID] 

Left JOIN [Grading].[GradeMark] AS [table022] 
ON [table028].[GradeMarkID] = [table022].[GradeMarkID]

的依赖问题，我已经是这个：

@0 = [table002].[label] //@0 is a string variable

像加入尚未创建，但我需要用它来创建联接的关系[评分] [StudentGradeBucket]或[table028]

来源

2013-10-23 user2529249

您需要对您的问题进行更具体的描述......您尝试解决哪些依赖性问题？（侧面说明...这是最古怪的表格别名我已经看到编码一段时间） – Twelfth

我更新了我的帖子，更详细。让我知道你是否需要更多。 – user2529249

什么是@ 0 - 你是指其中一个表中的一列，还是它是一个变量/参数？你使用的是哪个数据库？ – JohnLBevan

试试这个：

declare @0 nvarchar(max) = 'label value' 

select * 

from [Scheduling].[studentsection] as ss 

left join [Grading].[StudentGradeBucket] as sgb 
    on sgb.[StudentSectionID] = ss.[StudentSectionID] 

inner join [Grading].[GradingPeriodGradeBucket] as gpgb 
    on gpgb.[GradingPeriodGradeBucketID] = sgb.[GradingPeriodGradeBucketID] 

inner join [Grading].[GradeBucket] as gb 
    on gb.[GradeBucketID] = gpgb.[GradeBucketID] 
    and gb.[label] = @0 

left join [Grading].[GradeBucketType] as gbt 
    on gbt.[GradeBucketTypeID] = gb.[GradeBucketTypeID] 

left join [Grading].[GradeMark] as gm 
    on gm.[GradeMarkID] = sgb.[GradeMarkID]

或者这一点，如果你真的想左外连接：

declare @0 nvarchar(max) = 'label value' 

select * 

from [Scheduling].[studentsection] as ss 

left join [Grading].[StudentGradeBucket] as sgb 
    on sgb.[StudentSectionID] = ss.[StudentSectionID] 

left join [Grading].[GradingPeriodGradeBucket] as gpgb 
    on gpgb.[GradingPeriodGradeBucketID] = sgb.[GradingPeriodGradeBucketID] 

left join [Grading].[GradeBucket] as gb 
    on gb.[GradeBucketID] = gpgb.[GradeBucketID] 
    and gb.[label] = @0 

left join [Grading].[GradeBucketType] as gbt 
    on gbt.[GradeBucketTypeID] = gb.[GradeBucketTypeID] 

left join [Grading].[GradeMark] as gm 
    on gm.[GradeMarkID] = sgb.[GradeMarkID]

或者，如果您需要时，有一个匹配GradeBucket记录StudentGradeBucket只返回结果的逻辑;这样的：

declare @0 nvarchar(max) = 'label value' 

select * 

from [Scheduling].[studentsection] as ss 

left join [Grading].[StudentGradeBucket] as sgb 
    on sgb.[StudentSectionID] = ss.[StudentSectionID] 

left join [Grading].[GradingPeriodGradeBucket] as gpgb 
    on gpgb.[GradingPeriodGradeBucketID] = sgb.[GradingPeriodGradeBucketID] 

left join [Grading].[GradeBucket] as gb 
    on gb.[GradeBucketID] = gpgb.[GradeBucketID] 

left join [Grading].[GradeBucketType] as gbt 
    on gbt.[GradeBucketTypeID] = gb.[GradeBucketTypeID] 

left join [Grading].[GradeMark] as gm 
    on gm.[GradeMarkID] = sgb.[GradeMarkID] 

where 
(--filter on the gb label only if there's a result from the sgb table; 
    sgb.[StudentSectionID] is null 
    or and gb.[label] = @0 
)

来源

2013-10-23 18:46:44 JohnLBevan

这是奇怪的圆形连接使用的是逻辑和你选择是使它更加混乱，这些表的别名的。别名应该简化，不要混淆。这么说，我觉得它可以只被移动到WHERE子句：

... 
Left JOIN [Grading].[GradeMark] AS [table022] ON [table028].[GradeMarkID] = [table022].[GradeMarkID] 
where @0 = [table002].[label]

如果失败，则可能需要重新定义你的逻辑......似乎有点圆了我。

来源

2013-10-23 18:47:29 Twelfth

这是不是会返回更多的记录，然后必要的性能也会受到打击？ – user2529249

使用此方法进行表格别名的原因是由于标识符的最大长度为128个字符。我们有很深的关系，当它超过128个字符时，它就无法工作。所以这是解决这个问题的方法。基本上这是通过数据层（ORM）生成的，并且在生成sql时，实际的表名称作为注释包含在顶部。 – user2529249

这不是完整的查询，如果你有兴趣，我可以给你发送完整的查询。我想我们将尝试使用通用表格表达式（CTE）http://technet.microsoft.com/en-us/library/ms190766(v=sql.105).aspx – user2529249

SQL多重左加入子查询？

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