SQL一次性查询一个字段不同条件下的统计结果（另一张表的统计数量）

 

做另一张表的统计，比如本部门有多少在职人员。本岗位有多少女生。

有两个表，分别存放了【操作员】和【单据】，要根据单据的不同类型来分类汇总(销售单、销售退货单，笔数和金额)，并且显示在同一张表里，不想用做两次查询再合并的方法，研究了一下，终于搞定：

d_employee表

d_bilndx表

 

代码如下：

select  b.inputid as 开单员编号, 
        e.fullname as 开单员, 

        isnull( ( select count(*)
          from d_bilndx
          where draft=3 and biltype=12 and d_bilndx.inputid=e.id
        ), 0) as '销售开单笔数',

        isnull( ( select sum(d_bilndx.amount)
          from d_bilndx
          where draft=3 and biltype=12 and d_bilndx.inputid=e.id
        ), 0) as '销售开单金额',

        isnull( ( select count(*)
          from d_bilndx
          where draft=3 and biltype=13 and d_bilndx.inputid=e.id
        ), 0) as '销售退单笔数',

        isnull( ( select sum(d_bilndx.amount)
          from d_bilndx
          where draft=3 and biltype=13 and d_bilndx.inputid=e.id
        ), 0) as '销售退单金额',

        count(b.biltype) as 开单总笔数,
        sum(b.Amount) as 开单金额

from d_bilndx as b
left join d_employee as e 
on b.inputid=e.id 
where b.draft=3 and ( b.biltype=12 or b.biltype=13 )
group by b.inputid, e.fullname, e.id

 

得到结果：

 

 

补记：以上代码有一个问题，就是如果没有符合条件的单据，查到的结果为空，而我们可能希望，查不到符合条件的记录，相关字段要显示为0（并且按天来统计），改写代码如下：

 

select  e1.id as ePersonCode, e1.FullName as eFullName, 
        isnull(Bill_Sale_NUm, 0) as Bill_Sale_Num, 
        isnull(Bill_Sale_Amount, 0) as Bill_Sale_Amount, 
        isnull(Bill_SaleReturn_Num, 0) as Bill_SaleReturn_Num, 
        isnull(Bill_SaleReturn_Amount, 0) as Bill_SaleReturn_Amount
from d_employee as e1
left join (

        select  b.inputid as ePersonCode, 
                e.fullname as eFullName, 
        
                isnull( ( select count(*)
                  from d_bilndx
                  where biltype=12 and d_bilndx.inputid=e.id and d_bilndx.date>='2018-06-03' and d_bilndx.date<='2018-06-03'
                ), 0) as Bill_Sale_Num,
        
                isnull( ( select sum(d_bilndx.amount)
                  from d_bilndx
                  where biltype=12 and d_bilndx.inputid=e.id and d_bilndx.date>='2018-06-03' and d_bilndx.date<='2018-06-03'
                ), 0) as Bill_Sale_Amount,
        
                isnull( ( select count(*)
                  from d_bilndx
                  where biltype=13 and d_bilndx.inputid=e.id and d_bilndx.date>='2018-06-03' and d_bilndx.date<='2018-06-03'
                ), 0) as Bill_SaleReturn_Num,
        
                isnull( ( select sum(d_bilndx.amount)
                  from d_bilndx
                  where biltype=13 and d_bilndx.inputid=e.id and d_bilndx.date>='2018-06-03' and d_bilndx.date<='2018-06-03'
                ), 0) as Bill_SaleReturn_Amount,
        
                count(b.biltype) as Bill_Total_Num

        from d_employee as e 
        left join d_bilndx as b
        on b.inputid=e.id 
        where (b.draft=3 or b.draft=2) and ( b.biltype=12 or b.biltype=13 ) and  b.date>='2018-06-03' and b.date<='2018-06-03'
        group by b.inputid, e.fullname, e.id


) as t1
on e1.id = t1.ePersonCode
where e1.id<>'00000'
order by ePersonCode asc

 

得到结果如下：

 

转自：https://www.cnblogs.com/skysowe/p/9117099.html