Question 209428 in Javascript Question posted in
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javascript regex for string

I need to check that a string input contains exactly 2 special characters and 1 number.

is this the correct regex for this job?

(?=(.*[`!@#$%^&*-_=+'/.,]){2})

this is only checking special characters and not numbers.

I tried the regex above and it is not checking numbers

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2

Answers


  1. 0

    Please find below your regex:

    const regex = /^(?=.*[!@#$%^&*()-=_+{}[]|\;:'",.<>/?])(?=.*d).*[!@#$%^&*()-=_+{}[]|\;:'",.<>/?].*[!@#$%^&*()-=_+{}[]|\;:'",.<>/?].*d.*$/;
const input = "Your str**ing input here1";
const isValid = regex.test(input);
console.log(isValid)

    (?=.*[!@#$%^&*()-=_+{}[]|\;:'",.<>/?]) asserts that there is at least one special character present, and (?=.*d) asserts that there is at least one digit present. The rest of the regular expression ensures that there are exactly 2 special characters and 1 number present in the string.

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  2. 0

    You can use this instead:

    /^(?=.*[!@#$%^&*()-_=+'/.\,].*[!@#$%^&*()-_=+'/.\,])(?=.*d).*$/

    and here is the test case to test the regex as per your requirement.

    function verifyString(inputString) {
  const regexPattern = /^(?=.*[!@#$%^&*()-_=+'/.\,].*[!@#$%^&*()-_=+'/.\,])(?=.*d).*$/;
  return regexPattern.test(inputString);
}

// Example usage:
const testString1 = "ab@cde$1";
const testString2 = "abcde123";

console.log(verifyString(testString1)); // true
console.log(verifyString(testString2)); // false

    I hope it works for you!

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