如何加入来自多个 SQL SELECTS 的聚合结果？答案

【问题标题】：How do I JOIN aggregations results from several SQL SELECTS?如何加入来自多个 SQL SELECTS 的聚合结果？
【发布时间】：2010-11-17 22:43:33
【问题描述】：

我有一个MEMBERS 表，其中包含以下相关列：

 Name  
 JoinDate  
 Level   --1=Gold,2=Silver,3=Bronze**

我想创建一个查询来返回一个会员摘要，其中列出了按年份和会员级别加入的总数。基本上，我的结果集中的列是这样的：

| YEAR | GOLD | SILVER | BRONZE | TOTAL |

我可以使用以下 3 个查询分别获得 Gold、Silver 和 Bronze 会员每年的不同计数：

SELECT YEAR(JoinDate) AS YEAR, COUNT(*) AS GOLD FROM Members  
WHERE Level=1 GROUP BY YEAR(JoinDate) ORDER BY YEAR(JoinDate)  

SELECT YEAR(JoinDate) AS YEAR, COUNT(*) AS SILVER FROM Members  
WHERE Level=2 GROUP BY YEAR(JoinDate) ORDER BY YEAR(JoinDate)  

SELECT YEAR(JoinDate) AS YEAR, COUNT(*) AS BRONZE FROM Members  
WHERE Level=3 GROUP BY YEAR(JoinDate) ORDER BY YEAR(JoinDate)

我也可以使用类似的查询获得总数：

SELECT YEAR(JoinDate) AS YEAR, COUNT(*) AS TOTAL FROM Members  
GROUP BY YEAR(JoinDate) ORDER BY YEAR(JoinDate)

我的问题是我还没有找到将所有这些简化为单个查询的方法。这是怎么做到的？

【问题讨论】：

标签： sql sql-server join group-by

【解决方案1】：

要将总数添加到朱丽叶的答案中，只需添加 COunt(*)

SELECT YEAR(JoinDate) AS YEAR,    
     SUM(case when Level = 1 then 1 else 0 end) AS GoldCount,    
     SUM(case when Level = 2 then 1 else 0 end) AS SilverCount,    
     SUM(case when Level = 3 then 1 else 0 end) AS BronzeCount,
     Count(*) TotalCount
FROM Members  
GROUP BY YEAR(JoinDate) 
ORDER BY YEAR(JoinDate)

【讨论】：

【解决方案2】：

您正在寻找所谓的交叉表查询或数据透视表。

这应该会为你做的..

SELECT      YEAR(JoinDate) YEAR,  
            SUM(CASE [Level] WHEN 1 THEN 
                    1 ELSE 0 END) Gold, 
            SUM(CASE [Level] WHEN 2 THEN 
                    1 ELSE 0 END) Silver, 
            SUM(CASE [Level] WHEN 3 THEN 
                    1 ELSE 0 END) Bronze,
        COUNT([Level]) Total
FROM        members
GROUP BY    YEAR(JoinDate) 
ORDER BY    YEAR(JoinDate)

更多关于交叉表查询here。

【讨论】：

【解决方案3】：

最简单的方法是：

SELECT YEAR(JoinDate) AS YEAR,
    SUM(case when Level = 1 then 1 else 0 end) AS GoldCount,
    SUM(case when Level = 2 then 1 else 0 end) AS SilverCount,
    SUM(case when Level = 3 then 1 else 0 end) AS BronzeCount
FROM Members  
GROUP BY YEAR(JoinDate) ORDER BY YEAR(JoinDate)

【讨论】：