Question

我有以下数据

empid date        amount
1     12-FEB-2017  10
1     12-FEB-2017  10
1     13-FEB-2017  10
1     14-FEB-2017  10

我需要一个查询来返回给定id和日期的总金额，即结果集

empid date         amount
1     12-FEB-2017  20
1     13-FEB-2017  10
1     14-FEB-2017  10

但是想想，从UI我将获得日期作为输入..如果他们通过日期返回该日期的结果..如果他们没有通过日期返回最近日期的结果。

下面是我写的查询..但它部分工作..

SELECT sum(amount),empid,date
FROM employee emp,
  where
  ((date= :ddd) OR aum_valutn_dt = (select max(date) from emp))
  AND emp.id = '1'
group by (empid,date)

请帮助..

Answer 1

我认为你可以做这样的事情

但是你应该尝试以其他方式做到这一点非常糟糕为了获得最新的日期，它正在做额外的工作

select amt, empid, date 
from 
(
   select amt, empid, date, rank() over (order by date desc) date_rank
   from
   (SELECT sum(amount) amt,empid,date
    FROM employee emp
    where emp.id = '1'
    and (date = :ddd or :ddd is null)
    group by empid, date)
)
where date = :ddd or (:ddd is null and date_rank=1)

Answer 2

这是另一种选择;扫描TEST表两次以便......记住性能。

SQL> with test (empid, datum, amount) as
  2    (select 1, date '2017-02-12', 10 from dual union all
  3     select 1, date '2017-02-12', 10 from dual union all
  4     select 1, date '2017-02-13', 10 from dual union all
  5     select 1, date '2017-02-14', 10 from dual
  6    )
  7  select t.empid, t.datum, sum(t.amount) sum_amount
  8  from test t
  9  where t.datum = (select max(t1.datum)
 10                   from test t1
 11                   where t1.empid = t.empid
 12                     and (t1.datum = to_date('&&par_datum', 'dd.mm.yyyy')
 13                          or '&&par_datum' is null)
 14                  )
 15  group by t.empid, t.datum;
Enter value for par_datum: 13.02.2017

     EMPID DATUM      SUM_AMOUNT
---------- ---------- ----------
         1 13.02.2017         10

SQL> undefine par_datum
SQL> /
Enter value for par_datum:

     EMPID DATUM      SUM_AMOUNT
---------- ---------- ----------
         1 14.02.2017         10

SQL>

Answer 3

SELECT sum（amount），empid，date 来自员工emp，

其中date = nvl（（：ddd，（从emp选择max（date）））

AND emp.id ='1' group by（empid，date）

Answer 4

我的解决方案如下：

with t (empid, datum, amount) as
      (select 1, date '2017-02-12', 10 from dual union all
       select 1, date '2017-02-12', 10 from dual union all
       select 1, date '2017-02-13', 10 from dual union all
       select 1, date '2017-02-14', 10 from dual
      )
select empid, datum, s
  from (select empid, datum, sum(amount) s, max(datum) over (partition by empid) md
          from t
         group by empid, datum)
 where datum = nvl(to_date(:p, 'yyyy-mm-dd'), md);

计算子查询中的最大日期，然后在外部子查询中，将日期与nvl(to_date(:p, 'yyyy-mm-dd'), md)进行比较。如果参数为null，则将日期字段与最大日期进行比较。

执行oracle pl / sql查询并根据运行时日期值返回结果集

4 个答案: