Sql server 如何在SQL中将列设为行

如何使列QuotedReal并作为“ActualStatus”列值的一部分进行评级

select
        EffectiveDate,
        ActualStatus,
        SUM(case when ActualStatus = 'Bound' then 1 
                 when ActualStatus = 'Declined' then 1
                 when ActualStatus = 'Quoted' then 1 
                 when ActualStatus = 'Not Taken Up' then 1
                 when ActualStatus = 'Indication' then 1
                 when ActualStatus = 'Submitted' then 1
                 when ActualStatus = 'Lost' then 1
                 when ActualStatus = 'Indicated' then 1             
        else 0 end) as CountActual,
        SUM(Quoted) as QuotedReal,
        SUM(Case when QuotedPremium is not null then 1 else 0 end) as Rated
from #Olegggg
group by EffectiveDate,
        ActualStatus

这就是我现在拥有的：

但我需要这样的东西：

---

您可能需要使用联合将3个记录集连接在一起

select      EffectiveDate,
            ActualStatus,
            COUNT(*) as CountActual
from        #Olegggg
where       ActualStatus IN ('Bound','Declined',...'Indicated')
group by    EffectiveDate,
            ActualStatus

union all 
select      EffectiveDate,
            'Quoted Real',
            SUM(Quoted)
from        #Olegggg
group by    EffectiveDate

union all 
select      EffectiveDate,
            'Rated',
            COUNT(QuotedPremium)
from        #Olegggg
group by    EffectiveDate

---

JamieD您的解决方案简单而正确，除了

1.ActualStatus不会在组中用于Rated和QuotedReal。它只能按生效日期分组和聚合，以便每个日期有一行

---

不需要有所有的联合，因为我们没有任何重复的数据拉

select      EffectiveDate,
            ActualStatus,
            COUNT(*) as CountActual
 from        #Olegggg
where       ActualStatus IN ('Bound','Declined',...'Indicated')
group by    EffectiveDate,
        ActualStatus
union 
select      EffectiveDate,
       'Quoted Real',
        SUM(Quoted)
from        #Olegggg
group by    EffectiveDate,
   //removed ActualStatus
union 
  select      EffectiveDate,
            'Rated',
             COUNT(QuotedPremium)
 from        #Olegggg
  group by    EffectiveDate,
    //removed ActualStatus

---

您可以将表格和样本数据的ddl作为插入来发布，而不是图片吗？这可能与使用DISTINCT运算符有关，选择DISTINCT EffectiveDate，“Quoted Real”，SUMGQUOTED from OLEGGGYES，这将提供与group BY相同的输出，无需使用union all，因为我们没有任何重复的数据可提取。所有工会都比工会成本低得多。如果您知道您的集合之间不可能有重复项，则UNION ALL是合适的运算符。@rmehra76 SELECT DISTINCT决不能生成一个结果集，该结果集在DISTINCT应用到的列中包含一个以上的值实例。@russ，我不明白您的意思。我在这篇文章中在哪里使用了DISTINCT？