I need a RegEx for a password that must contain:
at least one uppercase character
at least one lowercase character
at least one digit
at least one symbol character out of: `~!##$%^&*()_-+={}[]|:;"'<>,.?/ and the space character
should have a minimum length of 6 ch.
So far I came up with this:
^\S*(?=\S{6,})(?=.*[A-Z])(?=.*[a-z])(?=.*\d)([`\~\!##\$%\^&\*\(\)_\-\+={}\[\]\|:;"'<>,\.\?\/ ])\S*$
But for some reason it's not working as I hoped. I've always had trouble with these things, so any help would be much appreciated.
Thank you.
That's not very expressible as a single regexp. As you've tagged "javascript", here's how I'd do it in JS.
function validatePassword(password) {
return password.length >= 6 &&
/[A-Z]/.exec(password) &&
/[a-z]/.exec(password) &&
/[0-9]/.exec(password) &&
/[ `~!##$%^&*()_-+={}\[\]|:;"'<>,.?\/]/.exec(password);
}
This should work:
/(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[`\~\!##\$%\^&\*\(\)_\-\+={}\[\]\|:;"'<>,\.\?\/ ]).{6,}/;
The full explanation how i did this is on art of web website
Related
I have tried to create a basic South African contact number regex. Its not working though. The rules are simple.
10 digits or
11 digits if it starts with a +
Examples:
0119879874
0731231234
+27731231234
+27123456789
It must match only digits and length.
My attempt: [+\d]\d{9}\d{0,2}
I tested on the site https://regex101.com/ and it looked like it worked but not when i test it with /[+\d]\d{9}\d{0,2}/.test('12345gjhf6789123456')) then i get a true value.
You should specify ^ - begin of the string and end $
and
/^(\+\d)?\d{10}$/.test('12345gjhf6789123456'))
Rather than use a regex that will not provide any error messaging and is difficult to read (and maintain), I would suggest a simple validation function that lays out the rules explicitly:
function validate(num) {
if (num[0] === '+') {
return num.length === 11;
} else {
return num.length === 10;
}
}
This has a few advantages, including:
faster than a regex
easier to comment and expand later
can be replaced with a full blown predicate library or just an array of rules
I'm using this Wordpress plugin called 'Easy contact form' which offers standard validation methods.
It uses the following regex for phone numbers:
/^(\+{0,1}\d{1,2})*\s*(\(?\d{3}\)?\s*)*\d{3}(-{0,1}|\s{0,1})\d{2}(-{0,1}|\s{0,1})\d{2}$/
Just now it allows the following formats (maybe more):
0612345678
+31612345678
But I want it to allow +316-12345678 and 06-12345678 also ... Is this possible? If so, how?
Thanks in advance!
You can use a less complex regex :
^\+?\d{2}(-?\d){8,9}$
This regex allows a + at the beginning of the phone number, then matches two digits, and after that, digits preceded (or not) by a -, for a total of 10 or 11 digits.
Now you can adapt it if the initial + is only for 11-digits phone numbers :
^\+?\d{3}(-?\d){9}|\d{2}(-?\d){8}$
My regex allow you to use - every digit. If that's an issue, it can be changed :
^\+?\d{3}(-?\d{2}){4}|\d{2}(-?\d{2}){4}$
I think this last regex will answer your needs, and it's quite simple !
When you figure out what patterns you actually want to allow and not allow. Then fill them out in this function and if the statement returns true you have your regex.
var regex = /^\+?\d{3}(-?\d{2}){4}|\d{2}(-?\d{2}){4}$/; //this is your regex, based on #Theox third answer
//allowed patterns
['0612345678', '+31612345678', '+316-12345678', '06-12345678'].every(function(test) {
return regex.exec(test) !== null;
}) &&
//disallowed patterns, none right now
[].every(function(test) {
return regex.exec(test) === null;
});
This question seems to have such an easy answer and an ashaming one for me, that I hope you just comment, then I can delete the thread after solving. ;)
I have a problem with the {n} quantifier in my RegEx. It does not seem to work!
Here my code
document.time.Id.onkeyup = function() {
var that = this.value,
regex = /^[1-9]{1}/
if (that) {
if (!that.match(regex)) {
this.nextSibling.innerHTML="Number must be between '1' and '100'.";
} else {
this.nextSibling.innerHTML="";
}
} else {
this.nextSibling.innerHTML="";
}
}
As you can see, I want to match against 1 till 100 in the end, but I am stuck at the bit, that the quantifier does not work. When I key in 0 there is a match failure, as well with any letter...so it does work "a bit".
Can you please help me?
Your regular expression says to match any string that starts (because it's anchored at the beginning using ^) with any digit between 1 and 9. This is why it doesn't match 0 or letters.
A range validation is something you'd want to check using basic number comparisons:
var numberValue = parseInt(this.value, 10);
if (numberValue >= 1 && numberValue <= 100) {
// valid number
}
For the sake of completeness, you could create a regular expression for that purpose which I don't recommend, though:
^(?:[1-9][0-9]?|100)$
Try using this regex instead:
^[1-9][0-9]?$|^100$
The quantifier you used is actually redundant, since [1-9] and [1-9]{1} mean the same thing.
If you input 1000 with your current code and regex, the number will pass because a match counts as long as the regex matches any part of the string. Using $ (end of line anchor) forces the regex to check the whole string.
But you should probably be using a simple if check for that.
if (that > 0 && that <= 100 && that % 1 == 0) {
...
}
Looking for some help for validating password with the following rules:
8+ characters
contains at least 1 upper case letter
contains at least 1 lower case letter
contains at least 1 number
Cannot start with a number
contains no special characters
I had gotten as far as:
(?=.*\d.*)(?=.*[a-z].*)(?=.*[A-Z].*)(?=.*[!#\$%&\?].*).{8,}
but can't seem to figure out how to get the first digit to not match a digit, and set the special character class to not match as well. Any help would be greatly appreciated.
I find that breaking this down into individual tests is:
easier to code
easier to read
easier to maintain
and more flexible when requirements change
Try something like this:
var testPassword = function (password) {
var minLengthMet = password.length >= 8,
hasUpper = (/[A-Z]+/).test(password),
hasLower = (/[a-z]+/).test(password),
hasNumber = (/[0-9]+/).test(password),
letterBegin = (/^[A-Za-z]/).test(password),
noSpecials = !(/[^A-Za-z0-9]+/).test(password);
return minLengthMet && hasUpper && hasLower && hasNumber && letterBegin && noSpecials;
};
See it in action here: http://jsfiddle.net/H9twa/
Here is what I would go with:
(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?!.*[!#\$%&\?])^\D.{7}
Note that the .* after each look-ahead term was superfluous.
(?!...) is a negative look-ahead, to make sure there are no special characters.
^\D requires that the first character be a non-digit. Then I simply require 7 characters after that, because the end is not enforced.
But why exclude special characters from passwords? Usually just the opposite is encouraged.
How about:
pwd.length >= 8 &&
pwd.match(/[A-Z]/) &&
pwd.match(/[a-z]/) &&
pwd.match(/\d/) &&
!pwd.match(/^\d/) &&
!pwd.match(/[!#\$%&\?]/);
Just in case you need to maintain this code ever?
I want to test a passwprd which must have at least 6 characters and 1 number in it. What regex string I can use with JS to get this done?
UPDATED
I forgot to write it must have at least 6 alpha characters and 1 numeric character but it should also allow special characters or any other character. Can you please modify your answers? I greatly appreciated your responses
This does smell a little like a homework question, but oh well. You can actually accomplish this concisely using a single regular expression and the "look ahead" feature.
/(?=.{6}).*\d.*/.test("helelo1")
The first bit in the brackets says "peek ahead to see if there's 6 characters". Following this we check for any number of characters, followed by a number, followed by any number of characters.
It is even possible to accomplish your goal in a single regex without having the faculty of look ahead... It's just a little hard to look at the solution and not wince:
new RegExp("[0-9].....|" +
".[0-9]....|" +
"..[0-9]...|" +
"...[0-9]..|" +
"....[0-9].|" +
".....[0-9]").test("password1")
Try this:
password.match(/(?=.*\d).{6}/);
More info here.
As far as I know this is best done with a combination of string functions and regex:
if( myPass.match(/[a-zA-Z]/).length >= 6 && myPass.match(/\d/g).length ) {
// Good passwords are good!
}
EDIT: Updated to include the new stipulations. Special characters are allowed, but not required.
if (/.{6,}/.test(password) && /\d/.test(password)) {
// success
} else {
// fail
}
/^(?=[\w\d]{6,}$)(?=.*\d)/.test(password)
requires 6 or more characters (letters, numbers or _)
requires at least one digit
won't allow any special characters
This is a js to check password,
it checks min 7 chars, contains 1 Upper case and 1 digit and 1 special character and must not contain a space, hope it will help you.
pwLength = this.value.length;
if (pwLength > 7 && pwLength < 21) {
charLengthIcon.removeClass("fail").addClass("pass");
}
else charLengthIcon.removeClass("pass").addClass("fail");
if (this.value.match(/[A-Z]/g)) {
capLetterIcon.removeClass("fail").addClass("pass");
}
else capLetterIcon.removeClass("pass").addClass("fail");
if (this.value.match(/[0-9]/g)) {
numberIcon.removeClass("fail").addClass("pass");
}
else numberIcon.removeClass("pass").addClass("fail");
if (this.value.match(/[##$%!$&~*^(){}?><.,;:"'-+=|]/g)) {
splcharIcon.removeClass("fail").addClass("pass");
}
else splcharIcon.removeClass("pass").addClass("fail");
if (this.value.match(/[\s/]/g)) {
whiteSpce.removeClass("pass").addClass("fail");
}
else whiteSpce.removeClass("fail").addClass("pass");
confirmPW();
});