Javascript正则表达式:^ ^ / \ s /]+[a - z]{ 1,}[0 - 9]*[\ - \ _]*[/ ^ \]^ / \ s /美元

I want to create a regular expression which will:

我想要创建一个正则表达式

not contain any space and special characters except "-" and "_"
除“-”和“_”外，不包含任何空格和特殊字符
it should contain at least one alphabet character
它应该包含至少一个字母字符。

The regular expression I created is:

我创建的正则表达式是:

^[^/\s/]+[a-z]{1,}[0-9]*[\-\_]*[^\/][^/\s/]$

It only only matches if my string contains at least 4 characters, including 1 alphabet. I tried it on https://regex101.com/#javascript Can someone help me what I am doing wrong here.

它只匹配如果我的字符串包含至少4个字符，包括1个字母。我在https://regex101.com/#javascript上尝试过，可以有人帮助我在这里做错了什么。

2 个解决方案

#1

You need to learn about lookaround. One solution to your problem is:

你需要了解周围的情况。你的问题的一个解决办法是:

/(?=^[\w-]{4,}$)(.*[a-z].*)/gmi

(?=^[\w-]{4,}$) will assert that you input will contains only chars in the range a-z, digit, _ and , - with a length of at least 4.
{4}$)将断言您的输入将只包含a-z、数字、_和-的范围内的字符，长度至少为4。
(.*[a-z].*) ensure that there will be at least one char in the range a-z.
(.*)确保在a-z范围内至少有一个字符。

See Demo

看到演示

#2

Using fundamental regex primitives, you can use this:

使用基本的regex原语，您可以使用以下方法:

/^[0-9_-]*[a-z]+[0-9a-z_-]*$/i

It works correctly with these sample input strings:

它可以正确地使用这些示例输入字符串:

999c123-
999 c123 -
a123-88asd
a123 - 88 asd
9923--_b
9923年,_b
B
B
99-luftballoons
99 - luftballoons
Z8f
Z8f

And does not match these strings:

与这些字符串不匹配:

999
999年
-51-
-51 -
---_-
——_

It's fast and will work in pretty much every regex engine, even non-standard (non-extended) grep.

它的速度很快，而且将在几乎所有的regex引擎，甚至非标准(非扩展)grep中工作。

#1