在 SQL 查询中查找两天内的新记录

Question

我正在使用查询从两个表中检查更改的叶子：-

with present_day 
as
(
select 1 as emp_id,'Sick Leave' as code , 9 as net_entitlement from dual
union
select 2 as emp_id,'Privilege Leave' as code , 5 as net_entitlement from dual
union
select 3 as emp_id,'Sick Leave' as code , 5 as net_entitlement from dual
union
select 5 as emp_id,'Sick Leave' as code , 10 as net_entitlement from dual
),
yester_day
as
(
select 1 as emp_id,'Sick Leave' as code , 10 as net_entitlement from dual
union
select 2 as emp_id,'Privilege Leave' as code , 5 as net_entitlement from dual
union
SELECT 3 AS EMP_ID,'Sick Leave' AS CODE , 6 AS NET_ENTITLEMENT FROM DUAL
UNION
SELECT 4 AS EMP_ID,'Sick Leave' AS CODE , 6 AS NET_ENTITLEMENT FROM DUAL
)

select a.emp_id,a.code,
a.net_entitlement
from present_day a,
YESTER_DAY B
WHERE A.EMP_ID=B.EMP_ID
AND A.CODE=B.CODE
and a.net_entitlement<>b.net_entitlement

在这个查询中，那些细节即将发生变化，但我需要那些昨天存在而不是现在的行，反之亦然......如何去做？

上面的查询给出了输出：-

emp_id  code          net_entitlement
1     Sick Leave            9
3     Sick Leave            5

需要的输出是：

emp_id  code          net_entitlement
1   Sick Leave  9
3   Sick Leave  5
4   Sick Leave  6
5   Sick Leave  10

我尝试使用 Union，但如果我使用，性能会受到影响：-

UNION 
select a.emp_id,a.code,
a.net_entitlement
from present_day a,
YESTER_DAY B
WHERE A.EMP_ID NOT IN B.EMP_ID
or B.EMP_ID NOT IN A.EMP_ID;

Answer 1

与 Lennart 的想法相同（或至少是他的答案的第一个版本），但您还需要合并选择列表中的值：

...
select coalesce(a.emp_id, b.emp_id) as emp_id,
  coalesce(a.code, b.code) as code,
  coalesce(a.net_entitlement, b.net_entitlement) as net_entitlement
from present_day a
full outer join yester_day b
on a.emp_id = b.emp_id
and a.code = b.code
where a.net_entitlement is null
or b.net_entitlement is null
or a.net_entitlement <> b.net_entitlement;

    EMP_ID CODE            NET_ENTITLEMENT
---------- --------------- ---------------
         1 Sick Leave                    9 
         3 Sick Leave                    5 
         4 Sick Leave                    6 
         5 Sick Leave                   10 

SQL Fiddle.

完全外连接包括在一个或两个表中的记录； where 子句排除了两者中具有相同 net_entitlement 值的那些。 （我更喜欢显式检查 null 而不是依赖幻数，即使 -1 在这里看起来很安全）。

Answer 2

您可以尝试类似（未经测试）：

select a.* from
from present_day a
left join YESTER_DAY B
   on A.EMP_ID=B.EMP_ID
  AND A.CODE=B.CODE
where a.net_entitlement<>coalesce(b.net_entitlement,-1)
union all
select b.*
from present_day a
right join YESTER_DAY B
   on A.EMP_ID=B.EMP_ID
  AND A.CODE=B.CODE
where coalesce(a.net_entitlement,-1)<>b.net_entitlement