SQL GROUP BY 与 FILTER IN SELECT答案

【问题标题】：SQL GROUP BY with FILTER IN SELECTSQL GROUP BY 与 FILTER IN SELECT
【发布时间】：2014-11-26 18:04:18
【问题描述】：

如果我有以下示例数据：

╔══════════════╦══════════════════╦════════════╦═══════╗
║    Client    ║      con_id      ║ mat1_07_03 ║ Ccode ║
╠══════════════╬══════════════════╬════════════╬═══════╣
║ Clients Name ║ C13109BBFD511534 ║ $1,062.00  ║ NOFL  ║
║ Clients Name ║ C11AC9BBF74D6882 ║ $879.73    ║ NOFL  ║
║ Clients Name ║ C12A69BBF1ACB578 ║ $2,790.29  ║ NOFA  ║
║ Clients Name ║ C12A69BBF1ACB578 ║ $912.00    ║ NOFL  ║
║ Clients Name ║ C6B0CA1A767C9744 ║ $2,180.11  ║ NOFL  ║
║ Clients Name ║ C11AC9BBF74D6882 ║ $878.67    ║ NOFA  ║
║ Clients Name ║ C13B79BBF4F1F450 ║ $300.00    ║ NOFL  ║
║ Clients Name ║ C12A69BBF1ACB578 ║ $1,790.67  ║ NOFL  ║
║ Clients Name ║ CA6869E2FE38A449 ║ $240.00    ║ NOFA  ║
║ Clients Name ║ C46439FB0D847140 ║ $3,392.66  ║ NOFL  ║
║ Clients Name ║ C12A69BBF1ACB578 ║ $1,791.73  ║ NOFA  ║
║ Clients Name ║ C13B49BBF12ED236 ║ $0.00      ║ NOFL  ║
║ Clients Name ║ C12A69BBF1ACB578 ║ $879.73    ║ NOFL  ║
╚══════════════╩══════════════════╩════════════╩═══════╝

并应用以下查询：

SELECT 
     [Client]=MAX(m.Client)
    ,[CaseCount]=COUNT(m.con_id)
    ,[AmtInDispute]=CONVERT(char, SUM(Convert(money, m.mat1_07_03)), 101)
FROM lntmu11.matter m
GROUP BY m.con_id
ORDER BY COUNT(m.Client) DESC

如何进一步对 Ccode 列进行分组以获得 COUNT 的行数 NOFA 或 NOFL

我想要的输出将显示为：

╔══════════╦═══════════╦═══════════════╦═══════════════╦══════════════╗
║  Client  ║ CaseCount ║ NOFACaseTotal ║ NOFLCaseTotal ║ AmtInDispute ║
╠══════════╬═══════════╬═══════════════╬═══════════════╬══════════════╣
║ Client A ║      3548 ║          2000 ║          1548 ║ 5,658,307.60 ║
║ Client B ║      3366 ║           100 ║          3266 ║ 2,885,649.48 ║
║ Client C ║      3014 ║           800 ║          2214 ║ 2,851,507.13 ║
║ Client D ║      2340 ║           340 ║          2000 ║ 3,467,207.12 ║
╚══════════╩═══════════╩═══════════════╩═══════════════╩══════════════╝

【问题讨论】：

+1 感谢您花时间将您的问题排版得这么好。

标签： sql sql-server tsql pivot

【解决方案1】：

只使用条件聚合：

SELECT m.Client, CaseCount = COUNT(m.con_id),
       CONVERT(varchar(255), SUM(Convert(money, m.mat1_07_03)), 101) as AmtInDispute
       sum(case when cCode = 'NOFA' then 1 else 0 end) as NOFACaseTotal,
       sum(case when cCode = 'NOFL' then 1 else 0 end) as NOFLCaseTotal,
FROM lntmu11.matter m
GROUP BY m.client
ORDER BY COUNT(m.Client) DESC;

注意：在进行字符转换时，始终包括长度（在 T-SQL 中）。

【讨论】：

【解决方案2】：

您可以使用条件聚合来获取其他 2 列。使用带有 CASE 表达式的聚合函数来计算每列所需的 Ccode 值。

SELECT 
     [Client]=MAX(m.Client)
    ,[CaseCount]=COUNT(m.con_id)
    ,NOFACaseTotal = sum(case when m.Ccode = 'NOFA' then 1 else 0 end)
    ,NOFLCaseTotal = sum(case when m.Ccode = 'NOFL' then 1 else 0 end)
    ,[AmtInDispute]=CONVERT(char(50), SUM(Convert(money, m.mat1_07_03)), 101)
FROM lntmu11.matter m
GROUP BY m.con_id
ORDER BY COUNT(m.Client) DESC;

见SQL Fiddle with Demo。

【讨论】：

【解决方案3】：

就像 Gordon 和 Bluefeet 所说的条件聚合 FTW！

SELECT 
     [Client]=MAX(m.Client)
    ,[CaseCount]=COUNT(m.con_id)
    ,[AmtInDispute]=CONVERT(char, SUM(Convert(money, m.mat1_07_03)), 101)
    ,[NOFACaseTotal]=sum(case when Ccode = 'NOFA' then 1 else 0 end)
    ,[NOFLCaseTotal]=sum(case when Ccode = 'NOFL' then 1 else 0 end)
FROM lntmu11.matter m
GROUP BY m.con_id
ORDER BY COUNT(m.Client) DESC

【讨论】：