sql查询统计重复项

Question

我得到的结果是:

Code    Declaration 
123     a1 - 2nos
123     a2 - 230nos
123     a3 - 5nos
123     a1 - 100nos
123     a3 - 6nos

是否可以对重复项进行求和，并且我需要像这样显示输出:

Code    Declaration
123     a1 - 102nos
123     a2 - 230nos
123     a3 - 11nos

Answer 1

我的回答基于 previous question你发布的。这是一个 CTE 版本，它将拆分数据并将其汇总到声明中:

;with cte (code, DeclarationItem, Declaration) as
(
  select Code,
    cast(left(Declaration, charindex(',',Declaration+',')-1) as varchar(50)) DeclarationItem,
         stuff(Declaration, 1, charindex(',',Declaration+','), '') Declaration
  from yourtable
  union all
  select code,
    cast(left(Declaration, charindex(',',Declaration+',')-1) as varchar(50)) DeclarationItem,
    stuff(Declaration, 1, charindex(',',Declaration+','), '') Declaration
  from cte
  where Declaration > ''
),
s2 as
(
  select code, 
    ltrim(rtrim(left(declarationitem, charindex('-', declarationitem)-1))) col1
    , reverse(left(reverse(declarationitem), charindex('-', reverse(declarationitem))-1)) col2
  from cte
),
fnl as
(
  select code, col1,
    left(col2, patindex('%[Aa-Zz]%', col2)-1) value,
    substring(col2, patindex('%[Aa-Zz]%', col2), len(col2)) str
  from s2
) 
select code, col1 +' - '+ cast(sum(cast(value as int)) as varchar(50)) + str as declarationitem
from fnl
group by code, col1, str

参见SQL Fiddle with Demo

结果是:

| CODE | DECLARATIONITEM |
--------------------------
|  123 | 123 a1 - 102nos |
|  123 | 123 a2 - 230nos |
|  123 |  123 a3 - 11nos |