Javascript 正则表达式匹配符号：!$%^&*()_+|~-=`{}[]:";'<>?,./
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Regex to Match Symbols: !$%^&*()_+|~-=`{}[]:";'<>?,./
提问by pixelbobby
I'm trying to create a Regex test in JavaScript that will test a string to contain any of these characters:
我正在尝试在 JavaScript 中创建一个 Regex 测试，该测试将测试包含以下任何字符的字符串：
!$%^&*()_+|~-=`{}[]:";'<>?,./
More Info If You're Interested :)
更多信息，如果您有兴趣:)
It's for a pretty cool password change application I'm working on. In case you're interested here's the rest of the code.
这是我正在开发的一个非常酷的密码更改应用程序。如果您有兴趣，这里是其余的代码。
I have a table that lists password requirements and as end-users types the new password, it will test an array of Regexes and place a checkmark in the corresponding table row if it... checks out :) I just need to add this one in place of the 4th item in the validationarray.
我有一个列出密码要求的表格，当最终用户输入新密码时，它将测试一组正则表达式并在相应的表格行中放置一个复选标记，如果它...签出:) 我只需要添加这个代替validation数组中的第 4 项。
var validate = function(password){
valid = true;
var validation = [
RegExp(/[a-z]/).test(password), RegExp(/[A-Z]/).test(password), RegExp(/\d/).test(password),
RegExp(/\W|_/).test(password), !RegExp(/\s/).test(password), !RegExp("12345678").test(password),
!RegExp($('#txtUsername').val()).test(password), !RegExp("cisco").test(password),
!RegExp(/([a-z]|[0-9])/).test(password), (password.length > 7)
]
$.each(validation, function(i){
if(this)
$('.form table tr').eq(i+1).attr('class', 'check');
else{
$('.form table tr').eq(i+1).attr('class', '');
valid = false
}
});
return(valid);
}
Yes, there's also corresponding server-side validation!
是的，还有相应的服务器端验证！
回答by Jeff Hillman
The regular expression for this is really simple. Just use a character class. The hyphen is a special character in character classes, so it needs to be first:
这个的正则表达式非常简单。只需使用字符类。连字符是字符类中的一个特殊字符，因此它需要排在第一位：
/[-!$%^&*()_+|~=`{}\[\]:";'<>?,.\/]/
You also need to escape the other regular expression metacharacters.
您还需要转义其他正则表达式元字符。
Edit:The hyphen is special because it can be used to represent a range of characters. This same character class can be simplified with ranges to this:
编辑：连字符很特殊，因为它可用于表示一系列字符。可以使用以下范围来简化相同的字符类：
/[$-/:-?{-~!"^_`\[\]]/
There are three ranges. '$' to '/', ':' to '?', and '{' to '~'. the last string of characters can't be represented more simply with a range: !"^_`[].
有三个范围。'$' 到 '/'，':' 到 '?'，'{' 到 '~'。最后一串字符不能用范围更简单地表示：!"^_`[]。
Use an ACSII tableto find ranges for character classes.
使用ACSII 表查找字符类的范围。
回答by AmirZpr
The most simple and shortest way is to use this:
最简单和最短的方法是使用这个：
/[\W\S]/
It means: All characters that are not a digit or an English letter (\W) or a white-space character (\S).
它的意思是：所有不是数字或英文字母 ( \W) 或空白字符 ( \S) 的字符。
It maybe is not as perfect as Jeff's solution, but it's much simpler and I don't think it differs in practicality.
它可能不像杰夫的解决方案那么完美，但它更简单，我认为它在实用性上没有什么不同。
回答by MikeSchem
Answer
回答
/[^\w\s]/
Explanation
解释
This creates a negated character class removing the word characters and space characters. All that is left is the special characters.
这将创建一个否定字符类，删除单词字符和空格字符。剩下的就是特殊字符了。
\wwill select all "word" characters equivalent to [^a-zA-Z0-9_]\swill select all "whitespace" characters equivalent to [ \t\n\r\f\v]
\w将选择等效于的所有“单词”字符[^a-zA-Z0-9_]\s将选择等效于的所有“空白”字符[ \t\n\r\f\v]
adding the ^at the beginning of the class says to not select any of the following.
在^课程开头添加表示不要选择以下任何一项。
回答by Arni Gudjonsson
// The string must contain at least one special character, escaping reserved RegEx characters to avoid conflict
const hasSpecial = password => {
const specialReg = new RegExp(
'^(?=.*[!@#$%^&*"\[\]\{\}<>/\(\)=\\\-_′+`~\:;,\.\|])',
);
return specialReg.test(password);
};
回答by Harfel Jaquez
A simple way to achieve this is the negative set [^\w\s]. This essentially catches:
实现此目的的一个简单方法是负集 [^\w\s]。这基本上抓住了：
- Anything that is not an alphanumeric character (letters and numbers)
- Anything that is not a space, tab, or line break (collectively referred to as whitespace)
- 任何非字母数字字符（字母和数字）
- 任何非空格、制表符或换行符（统称为空格）
For some reason [\W\S] does not work the same way, it doesn't do any filtering. A comment by Zael on one of the answers provides something of an explanation.
出于某种原因， [\W\S] 的工作方式不同，它不做任何过滤。Zael 对其中一个答案的评论提供了某种解释。
回答by Yaron Landau
Replace all latters from any language in 'A', and if you wish for example all digits to 0:
替换“A”中任何语言的所有后者，例如，如果您希望所有数字都为 0：
return str.replace(/[^\s!-@[-`{-~]/g, "A").replace(/\d/g, "0");
