如何根据另一列的值在 SQL select 查询中 create/add 作为 return 输出的列？

Question

我使用 Case 语句将一列 (industry_type) 动态添加到 SQL 视图。现在我想使用同一列 name/values 添加另一个具有 industry_group 输出的动态列。我尝试使用相同的方法创建 industry_group 列。但我无法这样做，因为在视图中创建的列在“main table”中不可用。

我是否需要在源中创建一个带有 industry_group 的新列？或者有没有办法在同一视图和 return industry_group 值中动态创建它？

如果你能帮助我解决这个问题，那将是一个很大的帮助。

示例代码：

select 
    column 1, column 2, date,
    case
        when (column 1 like '%12321%' or column 2 like'%poultry%' 
           then 'Poultry Facilities'
        when (column 1  like '%43434%') or column 2 like'%Hospi%' 
           then 'Hospitals'
        else null
    end as [industry_type],
from table

输出：

Date        Industry Type
------------------------------
7/02/2021   Poultry facilities
7/02/2021   Poultry facilities
7/02/2021   Poultry facilities
10/02/2021  Poultry facilities
11/01/2021  Hospitals
10/01/2021  Hospitals
5/02/2021   Hospitals

试过这个脚本1：

select 
    column 1, column 2, date,
    case
       when (column 1 like '%12321%' or column 2 like'%poultry%' 
          then 'Poultry Facilities'
       when (column 1  like '%43434%') or column 2 like'%Hospi%' 
          then 'Hospitals'
       else null
    end as [industry_type],
    'PF' as industry_group,
from table

输出：

Date        Industry Type     Industry Group
----------------------------------------------
7/02/2021   Poultry facilities  PF
7/02/2021   Poultry facilities  PF
7/02/2021   Poultry facilities  PF
10/02/2021  Poultry facilities  PF
11/01/2021  Hospitals           PF
10/01/2021  Hospitals           PF
5/02/2021   Hospitals           PF

试过脚本 2

select 
    column 1, column 2, date,
    case
       when (column 1 like '%12321%' or column 2 like'%poultry%' 
          then 'Poultry Facilities'
    end as [industry_type],
    'PF' as industry_group,
    case
       when (column 1  like '%43434%') or column 2 like'%Hospi%' 
          then 'Hospitals'
          else null
    end as [industry_type],
    'HOS' as industry_group,
from table

很遗憾，“脚本 2”无法运行。

期望的输出：

Date        Industry Type   Industry Group
---------------------------------------
7/02/2021   Poultry facilities  PF
7/02/2021   Poultry facilities  PF
7/02/2021   Poultry facilities  PF
10/02/2021  Poultry facilities  PF
11/01/2021  Hospitals          HOS
10/01/2021  Hospitals          HOS
5/02/2021   Hospitals          HOS

我需要根据 10 多个搜索条件return 正确的行业组

请注意：industry_type 在“master table”中不可用

谢谢大家！

Answer 1

您的输出似乎与查询不匹配：查询未显示任何列日期，我希望输出包含 3 列：

Column1, column2, industry_type

相反它只有 2

如果您粘贴实际代码而不是来自 excel

的图片，将会有所帮助

无论如何要在 select 中添加一列，只需将其添加到字段中：

SELECT column1,
column2,
case
...
END AS industry_type,
"PF" as industry_group
FROM table;

问题增强后更新

每一列本身就是一个规则，如果你想使它依赖于以前的你应该重写规则两次，或者进行嵌套查询

第一个例子

select 
    column 1, column 2, date,
    case
       when (column 1 like '%12321%' or column 2 like'%poultry%' 
          then 'Poultry Facilities'
       when (column 1  like '%43434%') or column 2 like'%Hospi%' 
          then 'Hospitals'
       else null
    end as [industry_type],
    case
       when (column 1 like '%12321%' or column 2 like'%poultry%' 
          then 'PF'
       when (column 1  like '%43434%') or column 2 like'%Hospi%' 
          then 'HOS'
       else null
    end as [industry_group],
from table

第二个例子

Select
Date, industry_type,
Case
When industry_type = "Hospitals" then "HOS"
When industry_type = "Poultry facilities" then "POL"
...
End
As industry_group
FROM (
    select 
    date,
    case
       when (column 1 like '%12321%' or column 2 like'%poultry%' 
          then 'Poultry Facilities'
    end as [industry_type],
    case
       when (column 1  like '%43434%') or column 2 like'%Hospi%' 
          then 'Hospitals'
          else null
    end as [industry_type],
    from table)

如何根据另一列的值在 SQL select 查询中 create/add 作为 return 输出的列？

How to create/add a column as a return output in an SQL select query based on another column's values?

mysql

sql

select

case

calculated-columns