Sql server SQL Server错误的结果

我在两个不同的表上使用了select，返回值必须是相同的，但不是，我无法计算出来，所以我们在这里

update oper
set OdenecekTutar = 
      (select 
          case 
             when round(sum(rezervasyon.odeme_satis_kalan), 2) is not null
                then round(sum(rezervasyon.odeme_satis_kalan), 2) 
                else 0 
          end
      from rezervasyon 
      where rezervasyon.oper = oper.operID)

    select round(sum(oper.OdenecekTutar),2)  as borc_from_oper from oper
//borc_from_oper => 1372283,38

    select round(sum(odeme_satis_kalan),2) as borc_from_oper2 from rezervasyon
//borc_from_oper2 => 1372283,35

表2中的关系

oper.operID one rezervasyon.oper有很多

oper.total/\*total

price of all rezervasyons price this case this column

'OdenecekTutar'\*/

sum（rezervasyon.price/\*本例中此列为'odeme\u satis\u kalan'\*/）

---

那么，我的代码有什么问题吗？

如果不使用round函数，结果是否相同？如果是这样，就不要在update语句中取整

您还可以将您的案例陈述更改为：

round(sum(ISNULL(rezervasyon.odeme_satis_kalan, 0)),2)

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差异仅仅是因为[oper]中的更新值是四舍五入的

但是[rezervasyon]中的值不是四舍五入的。
而最后一个查询对所有记录进行求和。
不超过分组的四舍五入总数

下面是一个示例片段，以演示该效果：

就个人而言，我不会在更新中对其进行舍入，并使用具有足够精度的[OdenecekTutar]类型。
但我猜这是必需的，因为[OdenecekTutar]的类型是十进制（N，2）

---

此外，如果[rezervasyon]中有operId与[oper]不匹配的记录，结果仍然会不同

一,。你想通过案例陈述2实现什么。JOIN 3在哪里。样本输入和结果4。错误代码我有两个不同的表和两个不同的案例1案例是保存我的总价另一个是全部售出的服务详细信息，但我不能用相同的数据得到相同的结果请阅读一些改进问题的提示。我的代码只是您案例陈述的一个紧凑形式。。。但是你应该尽量不要在更新中舍入

declare @oper table (operID int primary key, OdenecekTutar decimal(10,2));
insert into @oper (operID) values (1),(2),(3);

declare @rezervasyon table (rezervasyonID int identity(1,1) primary key, oper int, odeme_satis_kalan decimal(10,4));
insert into @rezervasyon (oper, odeme_satis_kalan) values (1,10.018),(1,20.028),(2,30.018),(2,40.028);

select * from @rezervasyon;

update o
set OdenecekTutar = coalesce(round(r.sum_odeme_satis_kalan, 2) ,0)
from @oper as o
left join (select oper, sum(odeme_satis_kalan) as sum_odeme_satis_kalan from @rezervasyon group by oper) as r 
  on (r.oper = o.operID);

select * from @oper;

select round(sum(OdenecekTutar),2) as round_sum_OdenecekTutar from @oper;
-- Result: round_sum_OdenecekTutar = 100.10

select round(sum(odeme_satis_kalan),2) as round_sum_odeme_satis_kalan, sum(odeme_satis_kalan) as sum_odeme_satis_kalan from @rezervasyon;
-- Result: round_sum_odeme_satis_kalan = 100.0900

--
-- Group first, then sum
--
select sum(round_sum_odeme_satis_kalan) as total_round_sum_odeme_satis_kalan from (select oper, round(sum(odeme_satis_kalan),2) as round_sum_odeme_satis_kalan from @rezervasyon group by oper) q;
--Result: total_round_sum_odeme_satis_kalan = 100.1000