多连接 Sql 合并低效代码答案

【问题标题】：Multi Join Tsql Coalesce Inefficient Code多连接 Sql 合并低效代码
【发布时间】：2018-07-18 16:26:28
【问题描述】：

需要帮助以提高代码效率。

我有以下选择（它将成为更新语句），现在我必须有 8 个完全相同的数据的连接。我尝试了一些事情并失败了，但也许这里的集体智慧会对如何提高效率有所了解。

我有以下表格

#temp_data 
customer    season  id      percent
1           2016    835     35.93000    
1           2016    836     31.82000    
1           2016    837     11.88000    
1           2016    841     5.09000 
1           2016    839     15.28000    
1           2017    836     31.40000
1           2017    837     11.45000    
1           2017    839     14.04000    
1           2017    841     6.40000 
1           2017    835     36.70000

和

lt_program_data

customer season     value_835   value_836   value_837   value_838   value_839   value_840   value_841   value_842
1       2016        35.93000    31.82000    11.88000    NULL        15.28000    NULL        5.09000     NULL
1       2017        36.70000    31.40000    11.45000    NULL        14.04000    NULL        6.40000     NULL
1       2018        NULL        NULL        NULL        NULL        NULL        NULL        NULL        NULL
1       2019        NULL        NULL        NULL        NULL        NULL        NULL        NULL        NULL

发生的情况是，我们匹配了两个表的季节/客户/id 值。在#temp_data 表中不存在值的情况下，我们使用存在的最大值。我能够整理代码以使其正常工作。然而，它又大又丑，而且效率不高。

我要么必须有 8 个更新语句，其中我明确声明每个 id 值（835、836 等）。或者一个包含 8 个连接的大型语句实际上是同一件事。关于如何更新以下内容以提高效率的任何建议都会很棒。

正如您所看到的，以下内容实际上是一遍又一遍地重复完全相同的事情。虽然我知道选择部分可能需要 8 个单独的语句，但有没有办法将 8 个左外连接消除为一个。

代码：

select      a.customer, 
            fyear,
            value_835 = coalesce (( select [percent] from #temp_data d where d.season = a.fyear and d.customer = a.customer and d.id = 835),    
                                        c1.[percent] ),

            value_836 = coalesce (( select [percent] from #temp_data d where d.season = a.fyear and d.customer = a.customer and d.id = 836),    
                                        c2.[percent] ),

            value_837 = coalesce (( select [percent] from #temp_data d where d.season = a.fyear and d.customer = a.customer and d.id = 837),    
                                        c3.[percent] ),

            value_838 = coalesce (( select [percent] from #temp_data d where d.season = a.fyear and d.customer = a.customer and d.id = 838),    
                                        c4.[percent] ),

            value_839 = coalesce (( select [percent] from #temp_data d where d.season = a.fyear and d.customer = a.customer and d.id = 839),    
                                        c5.[percent] ),

            value_840 = coalesce (( select [percent] from #temp_data d where d.season = a.fyear and d.customer = a.customer and d.id = 840),    
                                        c6.[percent] ),

            value_841 = coalesce (( select [percent] from #temp_data d where d.season = a.fyear and d.customer = a.customer and d.id = 841),    
                                        c7.[percent] ),

            value_842 = coalesce (( select [percent] from #temp_data d where d.season = a.fyear and d.customer = a.customer and d.id = 842),    
                                        c8.[percent] )
from        #lt_program_data a
left outer join (   --835
                    select a.customer, a.id, a.[percent] 
                    from    #temp_data a
                    join    (   select z.customer, id, season =  max(z.season)
                                from    #temp_data z 
                                where   id = 835
                                group by z.customer, id 
                            )   b on a.customer = b.customer and a.season = b.season  and a.id = b.id 
                )  c1 on        a.customer = c1.customer 

left outer join (   --836
                    select a.customer, a.id, a.[percent]
                    from    #temp_data a
                    join    (   select z.customer, id, season =  max(z.season)
                                from    #temp_data z 
                                where   id = 836
                                group by z.customer, id 
                            )   b on a.customer = b.customer and a.season = b.season  and a.id = b.id 
                )  c2 on        a.customer = c2.customer 

left outer join (   --837
                    select a.customer, a.id, a.[percent]
                    from    #temp_data a
                    join    (   select z.customer, id, season =  max(z.season)
                                from    #temp_data z 
                                where   id = 837
                                group by z.customer, id 
                            )   b on a.customer = b.customer and a.season = b.season  and a.id = b.id 
                )  c3 on        a.customer = c3.customer 

left outer join (   --838
                    select a.customer, a.id, a.[percent]
                    from    #temp_data a
                    join    (   select z.customer, id, season =  max(z.season)
                                from    #temp_data z 
                                where   id = 838
                                group by z.customer, id 
                            )   b on a.customer = b.customer and a.season = b.season  and a.id = b.id 
                )  c4 on        a.customer = c4.customer 

left outer join (   --839
                    select a.customer, a.id, a.[percent]
                    from    #temp_data a
                    join    (   select z.customer, id, season =  max(z.season)
                                from    #temp_data z 
                                where   id = 839
                                group by z.customer, id 
                            )   b on a.customer = b.customer and a.season = b.season  and a.id = b.id 
                )  c5 on        a.customer = c5.customer 

left outer join (   --840
                    select a.customer, a.id, a.[percent]
                    from    #temp_data a
                    join    (   select z.customer, id, season =  max(z.season)
                                from    #temp_data z 
                                where   id = 840
                                group by z.customer, id 
                            )   b on a.customer = b.customer and a.season = b.season  and a.id = b.id 
                )  c6 on        a.customer = c6.customer 


left outer join (   --841
                    select a.customer, a.id, a.[percent]
                    from    #temp_data a
                    join    (   select z.customer, id, season =  max(z.season)
                                from    #temp_data z 
                                where   id = 841
                                group by z.customer, id 
                            )   b on a.customer = b.customer and a.season = b.season  and a.id = b.id 
                )  c7 on        a.customer = c7.customer 


left outer join (   --842
                    select a.customer, a.id, a.[percent]
                    from    #temp_data a
                    join    (   select z.customer, id, season =  max(z.season)
                                from    #temp_data z 
                                where   id = 842
                                group by z.customer, id 
                            )   b on a.customer = b.customer and a.season = b.season  and a.id = b.id 
                )  c8 on        a.customer = c8.customer

这是最后的示例输出。

customer    fyear   value_835   value_836   value_837   value_838   value_839   value_840   value_841   value_842
1           2016    35.93000    31.82000    11.88000    NULL        15.28000    NULL        5.09000     NULL
1           2017    36.70000    31.40000    11.45000    NULL        14.04000    NULL        6.40000     NULL
1           2018    36.70000    31.40000    11.45000    NULL        14.04000    NULL        6.40000     NULL
1           2019    36.70000    31.40000    11.45000    NULL        14.04000    NULL        6.40000     NULL

【问题讨论】：

看起来您可以使用UNPIVOT 来规范您的第二个表并使查询更简单。但不确定它会更有效。
@YossiVainshtein 我不太关心UNPIVOT 我需要做更多的挖掘工作。我的意思是代码运行得非常快（对于一个相当大的表来说不到 2 秒），但它看起来很难看，我认为它可以更好地工作。所以我愿意接受所有/任何建议。
等等，我不明白，这两个表中的数据不是完全一样吗？
不管怎样，你可以看看这个sqlfiddle，它可能对你有帮助sqlfiddle.com/#!18/7b35f/2
@YossiVainshtein 我正在查看共享的代码。示例是相同的，但 lt_ 表中有更多数据，这些值存储在 #temp 中，但在 lt_data 表中是必需的

标签： sql-server tsql join pivot coalesce

【解决方案1】：

我发现将我的逻辑查询块包装在 CTE 中有助于我更清楚地思考它。这是pivots #temp_data 而不是unpivot-ing lt_program_data。如果我对您的理解正确，您应该能够 ~~left outer join~~ cross apply 与 LatestData 获得您需要的东西。

;with LatestData as (
    select Customer, Season, [835],[836],[837],[838],[839],[840],[841],[842] from 
    (select Customer, Season, [percent], id from temp_data td) as TempSource
    pivot 
    ( sum([percent]) for id in ([835],[836],[837],[838],[839],[840],[841],[842])) as TempPivoted
    )

 select a.customer,
    a.season,
    value_835 = coalesce(a.value_835, latest.[835]),
    value_836 = coalesce(a.value_836, latest.[836]),
    value_837 = coalesce(a.value_837, latest.[837]),
    value_838 = coalesce(a.value_838, latest.[838]),
    value_839 = coalesce(a.value_839, latest.[839]),
    value_840 = coalesce(a.value_840, latest.[840]),
    value_841 = coalesce(a.value_841, latest.[841]),
    value_842 = coalesce(a.value_842, latest.[842])
from lt_program_data a
cross apply (select top 1 * from LatestData where LatestData.customer = a.customer and latestdata.season <= a.season order by season desc) as latest

【讨论】：

哦，还有一件事：如果您熟悉 CTE，请忽略我的详细说明，但如果您不熟悉它们，请注意它们只是“即时的、一次性的”，因为CTE（在这种情况下是最新数据）只能由紧跟其声明后的命令引用。简而言之，select * from LatestData 必须替换为加入它的最终选择。
我对 CTE 很熟悉，谢谢你，我会去看看。
如何将其转换为我需要的...我得到了数据的透视，但我不知道接下来会发生什么？
我已经根据自己的想法更新了答案中的示例。
@YelizavetaYR 这给了你想要的东西吗？