sql - Oracle REGEXP_INSTR() 和 "a-z"字符范围与预期不匹配

Question

我想在 oracle 数据库中使用 REGEXP_INSTR() 来检查小写/大写字符。我知道 [:upper:] 和 [:lower:] POSIX 字符类，但我使用了 a-z 这给了我真正奇怪的结果我不明白。谁能解释一下？

SELECT REGEXP_INSTR('abc','[A-Z]',1,1,0,'c') FROM DUAL
-- Got 2, expected 0

SELECT REGEXP_INSTR('zyx','[A-Z]',1,1,0,'c') FROM DUAL
-- Got 1, expected 0

SELECT REGEXP_INSTR('ABC','[a-z]',1,1,0,'c') FROM DUAL
-- Got 1, expected 0

SELECT REGEXP_INSTR('ZYX','[a-z]',1,1,0,'c') FROM DUAL
-- Got 2, expected 0

SELECT REGEXP_INSTR('a3','[A-F0-9]',1,1,0,'c') FROM DUAL
-- Got 2, expected 2

SELECT REGEXP_INSTR('b3','[A-F0-9]',1,1,0,'c') FROM DUAL
-- Got 1, expected 2

SELECT REGEXP_INSTR('b3','[A-F0-9]') FROM DUAL
-- Got 1, expected 1 or 2

SELECT REGEXP_INSTR('a3','[A-F0-9]') FROM DUAL
-- Got 2, expected same as above

Answer 1

行为的原因是归类规则。查看NLS_SORT documentation :

• If the value is BINARY, then the collating sequence for ORDER BY queries is based on the numeric value of characters (a binary sort that requires less system overhead).
  
• If the value is a named linguistic sort, sorting is based on the order of the defined linguistic sort. Most (but not all) languages supported by the NLS_LANGUAGE parameter also support a linguistic sort with the same name.

将 NLS_SORT 设置为 BINARY 以便 [A-Z] 可以按照与 ASCII 表中相同的顺序进行解析，

alter session set nls_sort = 'BINARY'

然后，您将获得一致的结果。

参见 online demo .