Question

我有一张表格列出了所有供应商以及我付款的时间。我想估算每个供应商支付的总金额。目前，特定供应商可能有25行，在不同日期支付不同金额。我只想为每个vendor_code，供应商名称以及支付给他们的总金额显示一行。

到目前为止，我有这个：

function addDefinition(object, payload) {
    var newObject = Object.assign({}, object);
    if(!newObject[payload.guid]){
    newObject[payload.guid] = [];
    }
    newObject[payload.guid].push(payload);
    return newObject;
}

目前，我收到此错误：select s.vendor_code, a.address_name, a.addr2, a.addr3, a.city, a.state, a.postal_code, a.vend_class_code, date_applied, SUM(amt_net) as 'Total' from appyhdr_all s, apmaster a where s.vendor_code = a.vendor_code and s.pay_to_code = a.pay_to_code and void_flag=0 and s.payment_type <> 4

Answer 1

这应该有效：

select s.vendor_code 
       , a.address_name 
       , a.addr2
       , a.addr3 
       , a.city
       , a.state 
       , a.postal_code
       , a.vend_class_code 
       , SUM(amt_net) as 'Total'
from    appyhdr_all s
join    apmaster a
        on s.vendor_code = a.vendor_code
where   1 = 1
    and s.pay_to_code = a.pay_to_code
    and void_flag=0
    and s.payment_type <> 4
group 
by      s.vendor_code 
       , a.address_name 
       , a.addr2
       , a.addr3 
       , a.city
       , a.state 
       , a.postal_code
       , a.vend_class_code

Answer 2

是的，您应该包含group by子句并将所有选定的非聚合属性放在那里：

select s.vendor_code, 
       a.address_name, 
       a.addr2, a.addr3, 
       a.city, a.state, 
       a.postal_code, a.vend_class_code, 
       date_applied, SUM(amt_net) as 'Total'
from appyhdr_all s, apmaster a
where s.vendor_code = a.vendor_code
  and s.pay_to_code = a.pay_to_code
  and void_flag=0
  and s.payment_type <> 4
group by s.vendor_code,       --<--here it is 
       a.address_name, 
       a.addr2, a.addr3, 
       a.city, a.state, 
       a.postal_code, a.vend_class_code, 
       date_applied

顺便说一下，您可能想要移动最新的连接语法：

...
from appyhdr_all s
inner join apmaster a
    on s.vendor_code = a.vendor_code
   and s.pay_to_code = a.pay_to_code
where void_flag=0
...

已编辑到期OP评论：

“当我实现它时，我仍然会获得相同供应商代码的多行。”

这意味着每个供应商的apmaster上有几行。您必须为每个供应商决定一行apmaster行。这是每个供应商的总数：

with total_by_vendor as (
 select s.vendor_code, SUM(amt_net) as 'Total'
 from appyhdr_all s inner join apmaster a
    on s.vendor_code = a.vendor_code
   and s.pay_to_code = a.pay_to_code
 where void_flag=0
   and s.payment_type <> 4
 group by s.vendor_code
)
select *
from total_by_vendor;

只需加入apmaster的最新数据：

with total_by_vendor as (
 select s.vendor_code, SUM(amt_net) as 'Total'
 from appyhdr_all s inner join apmaster a
    on s.vendor_code = a.vendor_code
   and s.pay_to_code = a.pay_to_code
 where void_flag=0
   and s.payment_type <> 4
 group by s.vendor_code
),
recent_apmaster as (
 select vendor_code, a.address_name, 
       a.addr2, a.addr3, 
       a.city, a.state, 
       a.postal_code, a.vend_class_code,
       row_number() OVER ( PARTITION BY vendor_code 
                           ORDER BY a.address_name desc ) --put here criteria
       as n
 from apmaster a
)
select *
from total_by_vendor T inner join recent_apmaster a
  on T.vendor_code = a.vendor_code
where n = 1;

Answer 3

当您使用聚合函数时：

SUM（amt_net）为'总计'

确保列出现在GROUP BY子句中

Answer 4

先尝试一些简单的事情。如果appyhdr_all s是供应商，这应该有用。

select s.vendor_code, 
       SUM(amt_net) as 'Total'
from appyhdr_all s
inner join apmaster a
    on s.vendor_code = a.vendor_code
   and s.pay_to_code = a.pay_to_code
where void_flag=0
and s.payment_type <> 4
group by s.vendor_code

SQL特定代码花费的总和金额

4 个答案: