Question

我正在编写一段代码，使用SQL Developer在Oracle 11gR2中填充时间维度

表格结构

KEY Number(8,0)
HRS Varchar2
MINS Varchar2
SECS Varchar2

我是PL / SQL的新手 - 我通常认为在伪代码中做的方式是，但PL / SQL中的for循环似乎不会这样：

key = 1
for (hrs=0; hr<=23; hr++)
  {
    for (min=0; min<=59; min++)
     {
       for (sec=0; sec<=59; sec++)
         {
           insert (KEY, tochar(hrs:mins:secs), tochar(hr), tochar(min), tochar(sec))
           SK=+1
         }
     }
  }

我从我找到的答案中尝试过这个：

insert into Dim_TIME 
  (Key,
  TIME_IN24_NAME,
    TIME_HOUR24,
    TIME_MINUTE,
    TIME_SECOND)
select
  to_char(t, 'HH24:MI:SS'),
  to_number(to_char(t, 'HH24'), '00'),
  to_number(to_char(t, 'MI'), '00'),
  to_number(to_char(t, 'SS'), '00')
from (
  select trunc(sysdate) + (level - 1)/86400 as t
  from dual
  connect by level <= 86400
);

但不确定我如何递增键（列字段不会自动递增）。

我是如何实现这一目标的？

谢谢

Answer 1

您可以创建SEQUENCE来增加字段KEY。

CREATE SEQUENCE dim_time_seq;

并在你的代码中使用它，

INSERT INTO dim_time 
  (KEY,
  time_in24_name,
  time_hour24,
  time_minute,
  time_second)
SELECT
  dim_time_seq.nextval,   
  to_char(t, 'HH24:MI:SS'),
  to_number(to_char(t, 'HH24'), '00'),
  to_number(to_char(t, 'MI'), '00'),
  to_number(to_char(t, 'SS'), '00')
FROM (
  SELECT trunc(SYSDATE) + (LEVEL - 1)/86400 AS t
  FROM dual
  CONNECT BY LEVEL <= 86400
);

Oracle SQL填充时间维度

1 个答案: