JavaScript在字符串中查找复杂正则表达式的第一个匹配项

Question

I'm trying to write a script that deals with passwords. I want to find the first match on a regular expression in a long string. For example, if I have the string testingtestingPassWord12#$testingtesting and I want to find the first instance that has 2 uppercase, 2 lowercase, 2 numbers, 2 special characters and a minimum length of 12, it should return PassWord12#$ . If I want the same criteria but with a length of 16 it should return tingPassWord12#$ .

This is what I have for a regular expression: (?=.*[AZ].*[AZ])(?=.*[!@#$&*])(?=.*[0-9].*[0-9])(?=.*[az].*[az].*[az]).{12}

That regex is based on this SO: Regex to validate password strength

I tried the following but it just returns the first 12 characters in the string instead of the matched string:

var str = 'testingtestingPassWord12#$testingtesting',
    re = /(?=.*[A-Z].*[A-Z])(?=.*[!@#$&*])(?=.*[0-9].*[0-9])(?=.*[a-z].*[a-z].*[a-z]).{12}/;

console.log(str.match(re));


// Output: ["testingtestingPa"]

What am I missing?

Answer 1

Solution using a simple for loop.

Basically it iterates through each substring of the desired length and checks if it matches the password conditions:

 var str = 'testingtestingPassWord12#$testingtesting', re = /(?=(?:.*[AZ]){2})(?=(?:.*[!@#$&*]){2})(?=(?:.*[0-9]){2})(?=(?:.*[az]){2}).{12,}/; var substring, res = ''; var passwordLength = 12; for (var i = 0; i < str.length - passwordLength; i++) { substring = str.substr(i, passwordLength); if(substring.match(re)) { res = substring.match(re); break; } } document.body.textContent = res;