Question

我有一个脚本来填充上一个（2016年），当前（2017年）和明年（2018年）。该脚本旨在最初运行以填充表。由于它会填充以前的日期，因此只能运行一次。如何填充将来的日期（2019年）？

insert into my_date
SELECT TO_NUMBER (TO_CHAR (mydate, 'yyyymmdd')) AS my_date_id,
       mydate AS datetime_start,
       mydate + 1 - 1/86400 AS datetime_end,
       TO_CHAR (mydate, 'dd-MON-yyyy') AS date_value,
       TO_NUMBER (TO_CHAR (mydate, 'D')) AS day_of_week,
       TO_CHAR (mydate, 'Day') AS day_of_week_name,
       TO_CHAR (mydate, 'DY') AS day_of_week_name_short,
       TO_NUMBER (TO_CHAR (mydate, 'DD')) AS day_of_month,
       TRUNC (mydate) - TRUNC (mydate, 'Q') + 1 AS day_of_quarter,
       TO_NUMBER (TO_CHAR (mydate, 'DDD')) AS day_of_year, 
       CASE WHEN TO_NUMBER (TO_CHAR (mydate, 'D')) IN (1, 7) THEN 1
            ELSE 0
       END AS weekend_flag,
       TO_NUMBER (TO_CHAR (mydate, 'W')) AS week_in_month,
       TO_NUMBER (TO_CHAR (mydate, 'WW')) AS week_in_year,
       TRUNC(mydate, 'w') AS week_start_date,
       TRUNC(mydate, 'w') + 7 - 1/86400 AS week_end_date,
       TO_CHAR (mydate, 'MM') AS month_value,
       TO_CHAR (mydate, 'Month') AS month_name,
       TO_CHAR (mydate, 'MON') AS month_name_short,
       TRUNC (mydate, 'mm') AS month_start_date,
       LAST_DAY (TRUNC (mydate, 'mm')) + 1 - 1/86400 AS month_end_date,
       TO_NUMBER ( TO_CHAR( LAST_DAY (TRUNC (mydate, 'mm')), 'DD')) AS days_in_month,
       CASE WHEN mydate = LAST_DAY (TRUNC (mydate, 'mm')) THEN 1
            ELSE 0
       END AS last_day_of_month_flag, 
       TO_CHAR (mydate, 'yyyy') AS year_value,
       'YR' || TO_CHAR (mydate, 'yyyy') AS year_name,
       'YR' || TO_CHAR (mydate, 'yy') AS year_name_short,
       TRUNC (mydate, 'Y') AS year_start_date,
       ADD_MONTHS (TRUNC (mydate, 'Y'), 12) - 1/86400 AS year_end_date,
       ADD_MONTHS (TRUNC (mydate, 'Y'), 12) - TRUNC (mydate, 'Y') AS days_in_year
  FROM ( SELECT TRUNC (ADD_MONTHS (SYSDATE, -12), 'yy') - 1 + LEVEL AS mydate
           FROM dual
         CONNECT BY LEVEL <= (SELECT   TRUNC (ADD_MONTHS (SYSDATE, 24), 'yy')
                                   - TRUNC (ADD_MONTHS (SYSDATE, -12), 'yy')
                                FROM DUAL
                             )
       );

在FROM子句中将需要进行更改以避免现有记录。我该如何实现？

Answer 1

管道功能可能会在这里为您提供帮助。

创建一个与您的函数将返回的结果匹配的表类型。
创建一个函数，该函数将精确返回您想要的日期。它可以运行查询以确保日期不在表中，不在所需的日期范围内，等等。
一旦达到标准，就一步一步地返回值。
使用TABLE（）函数从函数中进行选择，以将结果转换为可以查询的表。使用OBJECT_VALUE访问返回的实际值（因为它实际上没有列名）。



create or replace type date_tbl_t as table of date;
/
create or replace function all_dates ( max_year_in in integer ) 
    return date_tbl_t 
    pipelined 
as
    date_l      date;
    offset_l    pls_integer := 0;
    year_l      integer;
begin 
    if date_l is null
    then 
        date_l := sysdate;
    end if;
    year_l := extract ( year from date_l );
    while year_l <= max_year_in 
    loop
        pipe row(date_l);
        date_l := date_l + 1;
        year_l := extract ( year from date_l );
    end loop;
    return;
end all_dates;
/
select 
    to_char(x.object_value, 'yyyymmdd') as my_date_id,
    x.object_value as datetime_start
from table ( all_dates (2019) ) x;
/

create or replace type date_tbl_t as table of date; / create or replace function all_dates ( max_year_in in integer ) return date_tbl_t pipelined as date_l date; offset_l pls_integer := 0; year_l integer; begin if date_l is null then date_l := sysdate; end if; year_l := extract ( year from date_l ); while year_l <= max_year_in loop pipe row(date_l); date_l := date_l + 1; year_l := extract ( year from date_l ); end loop; return; end all_dates; / select to_char(x.object_value, 'yyyymmdd') as my_date_id, x.object_value as datetime_start from table ( all_dates (2019) ) x; /

在Oracle表中填充将来的日期

1 个答案: