sql - 为什么这个 regexp_like 没有得到我想要的结果

Question

我正在尝试使用以下 regexp_like 进行更大的查询，但它不起作用，我做错了什么？

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:]](double [...]) 使用以下解决方案：

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？

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

Answer 2

@Sebastion Brosch 的替代品

您可以将字符类 ( [:digit:]) 替换为明确的数字范围，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

为了完整起见，您也可以使用\d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}$');