问题描述
您好我正在尝试在JS中使用regEx来识别3个相同的连续字符（可能是字母，数字以及所有非字母数字字符）
Hi I am trying to use regEx in JS for identifying 3 identical consecutive characters (could be alphabets,numbers and also all non alpha numeric characters)
这标识3相同的连续字母和数字：'（（[0-9a-zA-Z]）\\\）'
这标识了3个相同的连续非字母数字：'（（[^ 0-9a -zA-Z]）\1\1）'
This identifies 3 identical consecutive alphabets and numbers : '(([0-9a-zA-Z])\1\1)'
This identifies 3 identical consecutive non alphanumerics : '(([^0-9a-zA-Z])\1\1)'
我试图将两者结合起来，如下：'（（[0-9a-zA-Z ]）\ 1 \ 1）|（（[^ 0-9a-zA-Z]）\\\）'
I am trying to combine both, like this : '(([0-9a-zA-Z])\1\1)|(([^0-9a-zA-Z])\1\1)'
但我在做错误的东西，它不工作..（'88aa3BBdd99 @@'返回true）
But I am doing something wrong and its not working..(returns true for '88aa3BBdd99@@')
编辑：找到3个相同的字符，这似乎是错的/（ ^（[0-9a-zA-Z] | [^ 0-9a-zA-Z]）\1\ 1）/ - >
Edit : And to find NO 3 identical characters, this seems to be wrong /(^([0-9a-zA-Z]|[^0-9a-zA-Z])\1\1)/ --> RegEx in JS to find No 3 Identical consecutive characters
谢谢
Nohsib
thanksNohsib
推荐答案
问题是在整个正则表达式中，反向引用从左到右计算。因此，如果你将它们组合起来，你的数字就会改变：
The problem is that backreferences are counted from left to right throughout the whole regex. So if you combine them your numbers change:
(([0-9a-zA-Z])\2\2)|(([^0-9a-zA-Z])\4\4)
你也可以删除外部的parens：
You could also remove the outer parens:
([0-9a-zA-Z])\1\1|([^0-9a-zA-Z])\2\2
或你可以在一组parens中一起捕获替代品，并在结尾附加一个反向引用：
Or you could just capture the alternatives in one set of parens together and append one back-reference to the end:
([0-9a-zA-Z]|[^0-9a-zA-Z])\1\1
但是，既然你的角色类匹配所有字符，你也可以这样：
But since your character classes match all characters anyway you can have that like this as well:
([\s\S])\1\1
如果你激活DOTALL或SINGLELINE选项，您可以使用。代替：
And if you activate the DOTALL or SINGLELINE option, you can use a . instead:
(.)\1\1
这篇关于如何结合这些正则表达式的JavaScript的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！