Question

使用下面的代码我可以在本月和过去6个月返回。我的代码：

SELECT TO_CHAR(add_months(TRUNC(to_date( sysdate),'Month'), -rownum+1), 'Month') mon,
      rownum month_order
FROM dual
CONNECT BY rownum <=
  (SELECT COUNT(mon)
   FROM
       (SELECT TO_CHAR( add_months( start_date, level-1 ), 'fmMonth' ) AS mon
        FROM
          (SELECT to_date( add_months(TRUNC(sysdate),-6)) start_date,
            to_date( sysdate) end_date
          FROM dual)
          CONNECT BY level <= months_between( TRUNC(end_date,'MM'), TRUNC(start_date,'MM') ) + 1) dual);

同样地，我必须返回此日期和过去9个日期。请帮助我使用connect by返回10个日期。提前谢谢。

Answer 1

试试这个 -

SELECT ( TRUNC( SYSDATE ) + 1 ) - ROWNUM
FROM   DUAL
CONNECT BY ROWNUM <= 10

Answer 2

这是过去十天的查询。

 select sysdate - (level-1)
        , level as day_order
 from dual
 connect by level <= 10;

您的月份查询似乎极度过度设计。这会做同样的事情：

select to_char(add_months(trunc(sysdate, 'MM'), 1 - level), 'Month')
       , level as month_order
from dual
connect by level <= 7;

“假设数据在今天的日期可用，剩余的9个日期没有任何数据，但必须将计数显示为零”

在外部联接中使用生成的结果集：

with q as (
 select sysdate - (level-1) as dt
        , level as day_order
 from dual
 connect by level <= 10 
)
select q.dt as txn_date
      , sum(t42.col1)
from q
     left outer join t42
      on t42.transaction_date = q.dt
group by q.dt;

Answer 3

以同样的方式，我必须返回此日期和过去的9个日期。

这将是今天和过去9天：

SELECT TRUNC( SYSDATE ) - LEVEL + 1 AS day
FROM   DUAL
CONNECT BY LEVEL <= 10

本月和过去6个月

SELECT TO_CHAR( ADD_MONTHS( TRUNC( SYSDATE, 'MM' ), 1 - LEVEL ), 'Month' ) AS month
FROM   DUAL
CONNECT BY LEVEL <= 7

需要返回sysdate和过去9个日期

3 个答案: