Question

我有字符串'KK12340000'，我只想在KK之后的前4位数按升序排列时验证它，所以我使用类似的东西：

select case when REGEXP_LIKE('KK12340000', '^KK(?=\d{4})(?:(.)\\1*|0?1?2?3?4?5?6?7?8?9?)\d{4}$') then 1 else 0 end as valid from dual;

但它返回0.所以在oracle中甚至可以按升序验证数字吗？

Answer 1

试试这个：

  SELECT CASE
          WHEN REGEXP_LIKE (
                  'KK12230000',
                  '^KK(?:(.)\\1*|0?1?2?3?4?5?6?7?8?9?)\d{4}$')
          THEN
             1
          ELSE
             0
       END
          AS valid
  FROM DUAL;

Answer 2

您也可以尝试这样的事情

SELECT 'KK12340000' LIKE 'KK%' AND 
SUBSTRING('KK12340000', 3, 1) - SUBSTRING('KK12340000', 4, 1) < 0 AND
SUBSTRING('KK12340000', 4, 1) - SUBSTRING('KK12340000', 5, 1) < 0 AND
SUBSTRING('KK12340000', 5, 1) - SUBSTRING('KK12340000', 6, 1) < 0

REGEXP_LIKE ASC号码

2 个答案: