Question

我正在尝试使用以下regexp_like进行较大的查询，但它不起作用，我在做什么错？

with xx as
  (select '333-22-234223' as a
   from dual)
select xx.a
from xx
where
  regexp_like(xx.a,'^[:digit:]{3}-[:digit:]{2}-[:digit:]{6}$');

Answer 1

您可以使用[[:digit:]]（双[...]）以下解决方案：

WITH xx AS (
    SELECT '333-22-234223' AS a FROM dual
)
SELECT xx.a
FROM xx
WHERE REGEXP_LIKE(xx.a, '^[[:digit:]]{3}-[[:digit:]]{2}\-[[:digit:]]{6}$');

...或使用[0-9]代替[[:digit:]]：

WITH xx AS (
    SELECT '333-22-234223' AS a FROM dual
)
SELECT xx.a
FROM xx
WHERE REGEXP_LIKE(xx.a, '^[0-9]{3}-[0-9]{2}\-[0-9]{6}$');

演示： http://sqlfiddle.com/#!4/3149e4/120/1

为什么需要双括号？

这些字符类仅在带括号的表达式内有效。
来源： https://docs.oracle.com/cd/B12037_01/server.101/b10759/ap_posix001.htm

Answer 2

@Sebastion Brosch的替代方法

您可以使用明确的数字范围替换字符类（[:digit:]），如下所示：

with xx as
(select '333-22-234223' as a
from dual)
select xx.a
from xx
where
regexp_like(xx.a,'^[0-9]{3}-[0-9]{2}-[0-9]{6}$');

Answer 3

出于完整性考虑，您也可以将\d用作数字：

with xx as
  (select '333-22-234223' as a
   from dual)
select xx.a
from xx
where regexp_like(xx.a,'^\d{3}-\d{2}-\d{6}$');

为什么这个regexp_like无法得到我想要的结果

3 个答案: