Question

我有两个表，每个表都有2列（名称[字符串]，值[整数]）我试图将两者结合起来以显示每个名称的sum（value）之间的差异。例如

表1（t1）：

表（t2）：

N1  1
N2  1
N2  1

结果应如下所示：

N1 4  (5-1)
N2 2  (4-2)
N3 5  (5-0)

如果其中一个表中缺少N，我们想假设为0

如果两个表中都存在名称，我只知道如何进行联接，但是如果其中一个表中都缺少名称，我不知道如何处理将空结果替换为0的情况。

我当前的查询：

select t1.name, t2.name, t1.Sum-t2.Sum as "Diff"
from (select t1.name, sum(t1.value) as Sum from t1 group by t1.name) t1 
inner join
(select t2.name, sum(t2.value) as Sum from t2 group by t2.name) t2 
on t1.name = t2.name

结果忽略N3，因为t1中缺少N3。

谢谢

Answer 1

您可以使用CTE（公用表表达式）聚合两个表，然后使用FULL OUTER JOIN将它们联接起来，如下所示：

with
l as (
  select name, sum(value) as total
  from table1
  group by name
),
r as (
  select name, sum(value) as total
  from table2
  group by name
)
select
  coalesce(l.name, r.name) as name,
  coalesce(l.total, 0) - coalesce(r.total, 0) as difference
from l
full outer join r on l.name = r.name
order by coalesce(l.name, r.name)

Answer 2

使用FULL OUTER JOIN和ISNULL

SELECT ISNULL(x.name, y.name) as Name,
       ISNULL(x.sumValues, 0) - ISNULL(y.sumValues, 0) as ValueDiff
FROM
(SELECT name, SUM(value) AS sumValues FROM t1 GROUP BY name) x
    FULL OUTER JOIN
    (SELECT name, SUM(value) AS sumValues FROM t2 GROUP BY name) y
        ON x.name = y.name;

Answer 3

一个非常简单的方法是union all和聚合：

select name, sum(value)
from ((select name, value
       from t1
      ) union all
      (select name, -value
       from t2
      )
     ) t
group by name;

join有点混乱，因为除非非常小心，否则它可能导致值重复和丢失。

Answer 4

这将起作用：

输入：

create table t1 (name varchar(20),value number);
create table t2 (name varchar(20),value number);
//do inserts

select name1,nvl(sum1-sum2,sum1) from(select sum(t1.value) as sum1,t1.name name1 from t1 group by t1.name)
,(select sum(t2.value) as sum2,t2.name name2 from t2 group by t2.name) where name1=name2(+);

输出：

N1  4
N2  2
N3  5

SQL：这不是一个简单的左右联接问题

4 个答案: