使用硬编码的列名进行 SQL 更新答案

【问题标题】：SQL update with hard-coded column names使用硬编码的列名进行 SQL 更新
【发布时间】：2015-12-20 15:30:39
【问题描述】：

我需要帮助为以下示例创建 SQL 更新（带有示例数据）：

表 1：详细信息结构：VENDOR CHAR(10)、EMPLOYEE CHAR(10)、WEEK INT

VENDOR   EMPLOYEE      WEEK
VEN01       EMP01         1
VEN01       EMP01         1
VEN02       EMP03         1
VEN03       EMP02         1
VEN01       EMP01         2
VEN01       EMP01         2
VEN01       EMP03         2
VEN03       EMP02         2
VEN02       EMP01         3
VEN02       EMP01         3
VEN02       EMP03         3
VEN03       EMP03         3

表 2：SUMMARTRAN（需要更新）

结构：WEEK01 INT、WEEK02 INT、WEEK03 INT、VENDOR CHAR(10)、EMPLOYEE CHAR(10) SQL 更新此表 (SUMMARTRAN) 的结果应如下所示：

WEEK01     WEEK02    WEEK03    VENDOR   EMPLOYEE
2          3         0         VEN01
1          0         3         VEN02
1          1         1         VEN03
2          2         2                  EMP01
1          1         0                  EMP02
1          1         2                  EMP03

【问题讨论】：

您要添加新记录还是更新现有记录？
@DanBracuk 将新记录添加到空表中。也就是说，我认为我会先截断 SUMMARTRAN，然后从 DETAILTRAN 中添加记录

标签： sql sql-server

【解决方案1】：

您可以使用grouping sets 和条件聚合来解决此问题：

select vendor, employee,
       sum(case when week = 1 then 1 else 0 end) as week01,
       sum(case when week = 2 then 1 else 0 end) as week02,
       sum(case when week = 3 then 1 else 0 end) as week03
from DETAILTRAN
group by grouping sets ((vendor), (employee));

您可以将其合并到insert，summartran 的声明中：

insert into summertran(vendor, employee, week01, week02, week03)
    select vendor, employee,
           sum(case when week = 1 then 1 else 0 end) as week01,
           sum(case when week = 2 then 1 else 0 end) as week02,
           sum(case when week = 3 then 1 else 0 end) as week03
    from DETAILTRAN
    group by grouping sets ((vendor), (employee));

【讨论】：

从未听说过Grouping set。感谢分享：）这里是Demo
戈登，谢谢。我也从未听说过分组集（但我是 SQL 新手）。 VR46，谢谢你的演示。

【解决方案2】：

试试这个

with VENDOR as
(
select isnull(count(case when WEEK = 1 then WEEK end),0) WEEK01,
       isnull(count(case when WEEK = 2 then WEEK end),0) WEEK02 ,  
       isnull(count(case when WEEK = 3 then WEEK end),0) WEEK03 ,
VENDOR
from DETAILTRAN 
group by VENDOR
),EMPLOYEE as
(
select isnull(count(case when WEEK = 1 then WEEK end),0) WEEK01,
       isnull(count(case when WEEK = 2 then WEEK end),0) WEEK02 ,  
       isnull(count(case when WEEK = 3 then WEEK end),0) WEEK03 ,
EMPLOYEE
from DETAILTRAN 
group by EMPLOYEE
)
select WEEK01, WEEK02, WEEK03, VENDOR, EMPLOYEE = '' from VENDOR 
union all
select WEEK01, WEEK02, WEEK03, VENDOR='', EMPLOYEE from EMPLOYEE

SQL FIDDLE DEMO

要将结果更新为SUMMARTRAN 表，请使用此

with VENDOR as
(
select isnull(count(case when WEEK = 1 then WEEK end),0) WEEK01,
       isnull(count(case when WEEK = 2 then WEEK end),0) WEEK02 ,  
       isnull(count(case when WEEK = 3 then WEEK end),0) WEEK03 ,
VENDOR
from DETAILTRAN 
group by VENDOR
),EMPLOYEE as
(
select isnull(count(case when WEEK = 1 then WEEK end),0) WEEK01,
       isnull(count(case when WEEK = 2 then WEEK end),0) WEEK02 ,  
       isnull(count(case when WEEK = 3 then WEEK end),0) WEEK03 ,
EMPLOYEE
from DETAILTRAN 
group by EMPLOYEE
)
insert into  SUMMARTRAN (WEEK01, WEEK02, WEEK03, VENDOR, EMPLOYEE)
select WEEK01, WEEK02, WEEK03, VENDOR, EMPLOYEE = '' from VENDOR 
union all
select WEEK01, WEEK02, WEEK03, VENDOR='', EMPLOYEE from EMPLOYEE

【讨论】：

VR46，首先，谢谢。我不明白的是您的代码中第二个表（SUMMARTRAN）的更新在哪里？
Insert 就是我所说的更新。我在第二个查询中添加了insert into..
VR46，非常感谢。现在我只需要弄清楚上述哪种方法更快。
@Hidalgo - 毫无疑问，Gordan 的方法会更快
VR46，谢谢。但是你的例子对我来说是一个很好的学习。我对您如何快速创建 SQL Fiddle 演示印象深刻（也需要学习这个工具）。