Regex -文本匹配

Question

我需要一个正则表达式，它将与字符串S匹配以下条件：

S must be of length: 20
1st character: lowercase letter.
2nd character: word character.
3rd character: whitespace character.
4th character: non word character.
5th character: digit.
6th character: non digit.
7th character: uppercase letter.
8th character: letter (either lowercase or uppercase).
9th character: vowel (a, e, i , o , u, A, E, I, O or U).
10th character: non whitespace character.
11th character: should be same as 1st character.
12th character: should be same as 2nd character.
13th character: should be same as 3rd character.
14th character: should be same as 4th character.
15th character: should be same as 5th character.
16th character: should be same as 6th character.
17th character: should be same as 7th character.
18th character: should be same as 8th character.
19th character: should be same as 9th character.
20th character: should be same as 10th character.

我写了以下文章，但它不起作用。

^([a-z])(\w)(\s)(\W)(\d)([A-Z])([a-zA-Z])([a,e,i,o,u,A,E,I,O,U])(\S)(\D)\1\2\3\4\5\6\7\8\9\10$

样本输入：

ab #1?AZa$ab #1?AZa$

有人能解释一下这是怎么回事吗？

谢谢。

Answer 1

您可以使用这个regex：

^([a-z]\w\s\W\d\D[A-Z][a-zA-Z][aeiouAEIOU]\S)\1$

RegEx演示

1. 如问题所示，第六位是非数字，但您的reges有[A-Z]。
2. 没有必要在字符类中为多个值设置逗号。
3. 没有必要分别捕获10个反向引用的10个值，因为char 1-10与char 11-20相同。我们只需捕获前10个字符在一个组，并使用它作为一个单一的反向引用在最后。

Answer 2

你可能想试试这样的方法：

^([a-z])(\w)(\s)(\W)(\d)(\D)([A-Z])([a-zA-Z])([aeiouAEIOU])(\S)\1\2\3\4\5\6\7\8\9\10$

要捕获单个的类/字符组。但是，如果不需要捕获单个字符，则可以考虑上面发布的解决方案。