解析日期后从 Oracle Select 语句返回编号

2021-12-13 00:00:00 parsing sql date oracle case

我想编写一个 Oracle SQL select 语句，通过返回代码来判断它是否可以解析给定格式的日期字符串 - 失败时为零(异常)，成功时返回正数:

I want to write a Oracle SQL select statement that tells if it could parse a date string in the given format by returning a code - zero on failure (exception) and a positive number in case of success :

    SELECT
     CASE
      WHEN PARSING SUCCESSFUL (ie. to_date('1-Jan-2001','dd-mon-yy') succeeds) THEN 1
      ELSE 0
     END  
    FROM DUAL;

我该怎么写?如果解析失败，ELSE 条件会返回一个值吗?我需要在 SELECT 语句本身中进行所有这些检查.请帮忙.

How do I write this ? If the parsing fails, will the ELSE condition return a value ? I need to do all these checks in the SELECT statement itself. Please help.

谢谢忧郁

推荐答案

如果您可以创建一个函数，那么您可以执行以下操作:

If you can create a function then you can do something like:

SQL 小提琴

Oracle 11g R2 架构设置:

CREATE OR REPLACE FUNCTION is_Valid_Date (
  p_date   IN VARCHAR2,
  p_format IN VARCHAR2 DEFAULT 'DD-MON-YY'
) RETURN NUMBER
IS
  d DATE;
BEGIN
  d := TO_DATE( p_date, p_format );
  RETURN 1;
EXCEPTION
  WHEN OTHERS THEN
    RETURN 0;
END;
/

查询 1:

SELECT is_Valid_Date( '12-Feb-13' ),
       is_Valid_Date( 'XX-Feb-13' )
FROM DUAL

结果:

| IS_VALID_DATE('12-FEB-13') | IS_VALID_DATE('XX-FEB-13') |
|----------------------------|----------------------------|
|                          1 |                          0 |

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