I was just looking for a regex expression to check and see if both numbers and letters exist.
Just to clarify the query, the regex is going to be written in javascript and used to validate an address.
I would use a regular expression which matches any letter followed by any digit (with any possible characters in between) or digit then letter (with anything in between):
var hasNumbersAndLetters = function(str) {
var regex = /(?:[A-Za-z].*?\d|\d.*?[A-Za-z])/;
return !!str.match(regex);
};
Much easier to run two checks.
/\pL/ && /\pN/
To do both checks in one pattern, you need something like
/\pL.*\pN|\pN.*\pL/s
Languages supporting zero-width lookaheads can eliminate the redundancy:
/^(?=.*\pL/)(?=.*\pN/)/s ( or /^(?=.*\pL/).*\pN/s )
But it's harder to read.
Pardon me for not using JS's match function, but the question is really about regular expressions, and I'm not familiar with JS's match function.
if it is a single word you are matching without spaces, with both numbers and letters, it can be assumed they touch somewhere - so if it matches letter then number or number then letter we have a match - so:
([a-zA-Z][0-9]|[0-9][a-zA-Z])
Edit: where there may be spaces then you can use lookahead assertions like this
([a-zA-Z](?=.*[0-9])|[0-9](?=.*[a-zA-Z]))
Saw you wanted to validate an address
Removed by regex answer as javascript has no Unicode support, except for matching single characters http://www.regular-expressions.info/unicode.html
This should do it:
^[\w]*[^\W_][\w]*$
Related
I want to match all words which are starting with dollar sign but not slash and dollar sign.
I already try few regex.
(?:(?!\\)\$\w+)
\\(\\?\$\w+)\b
String
$10<i class="">$i01d</i>\$id
Expected result
*$10*
*$i01d*
but not this
*$id*
After find all expected matching word i want to replace this my object.
One option is to eliminate escape sequences first, and then match the cleaned-up string:
s = String.raw`$10<i class="">$i01d</i>\$id`
found = s.replace(/\\./g, '').match(/\$\w+/g)
console.log(found)
The big problem here is that you need a negative lookbehind, however, JavaScript does not support it. It's possible to emulate it crudely, but I will offer an alternative which, while not great, will work:
var input = '$10<i class="">$i01d</i>\\$id';
var regex = /\b\w+\b\$(?!\\)/g;
//sample implementation of a string reversal function. There are better implementations out there
function reverseString(string) {
return string.split("").reverse().join("");
}
var reverseInput = reverseString(input);
var matches = reverseInput
.match(regex)
.map(reverseString);
console.log(matches);
It is not elegant but it will do the job. Here is how it works:
JavaScript does support a lookahead expression ((?>)) and a negative lookahead ((?!)). Since this is the reverse of of a negative lookbehind, you can reverse the string and reverse the regex, which will match exactly what you want. Since all the matches are going to be in reverse, you need to also reverse them back to the original.
It is not elegant, as I said, since it does a lot of string manipulations but it does produce exactly what you want.
See this in action on Regex101
Regex explanation Normally, the "match x long as it's not preceded by y" will be expressed as (?<!y)x, so in your case, the regex will be
/(?<!\\)\$\b\w+\b/g
demonstration (not JavaScript)
where
(?<!\\) //do not match a preceding "\"
\$ //match literal "$"
\b //word boundary
\w+ //one or more word characters
\b //second word boundary, hence making the match a word
When the input is reversed, so do all the tokens in order to match. Furthermore, the negative lookbehind gets inverted into a negative lookahead of the form x(?!y) so the new regular expression is
/\b\w+\b\$(?!\\)/g;
This is more difficult than it appears at first blush. How like Regular Expressions!
If you have look-behind available, you can try:
/(?<!\\)\$\w+/g
This is NOT available in JS. Alternatively, you could specify a boundary that you know exists and use a capture group like:
/\s(\$\w+)/g
Unfortunately, you cannot rely on word boundaries via /b because there's no such boundary before '\'.
Also, this is a cool site for testing your regex expressions. And this explains the word boundary anchor.
If you're using a language that supports negative lookback assertions you can use something like this.
(?<!\\)\$\w+
I think this is the cleanest approach, but unfortunately it's not supported by all languages.
This is a hackier implementation that may work as well.
(?:(^\$\w+)|[^\\](\$\w+))
This matches either
A literal $ at the beginning of a line followed by multiple word characters. Or...
A literal $ this is preceded by any character except a backslash.
Here is a working example.
I have constructed the following Regex, which allows strings that only satisfy all three conditions:
Allows alphanumeric characters.
Allows special characters defined in the Regex.
String length must be min 8 and max 20 characters.
The Regex is:
"^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[$#$!%*?&])[A-Za-z\d$#$!%*?&]$"
I use the following Javascript code to verify input:
var regPassword = new RegExp("^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[$#$!%*?&])[A-Za-z\d$#$!%*?&]$");
regPassword.test(form.passwordField.value);
The test() method returns false for such inputs as abc123!ZXCBN. I have tried to locate the problem in the Regex without any success. What causes the Regex validation to fail?
I see two major problems. One is that inside a string "...", backslashes \ have a special meaning, independent of their special meaning inside a regex. In particular, \d ends up just becoming d — not what you want. The best fix for that is to use the /.../ notation instead of new RegExp("..."):
var regPassword = /^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[$#$!%*?&])[A-Za-z\d$#$!%*?&]$/;
The other problem is that your regex doesn't match your requirements.
Actually, the requirements that you've stated don't really make sense, but I'm guessing you want something like this:
Must contain at least one lowercase letter, at least one uppercase letter, at least one digit, and at least one of the special characters $#$!%*?&.
Can only contain lowercase letters, uppercase letters, digits, and the special characters $#$!%*?&.
Total length must be between 8 and 20 characters, inclusive.
If so, then you've managed #1 and #2, but forgot about #3. Right now your regex demands that the length be exactly 1. To fix this, you need to add {8,20} after the [A-Za-z\d$#$!%*?&] part:
var regPassword = /^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[$#$!%*?&])[A-Za-z\d$#$!%*?&]{8,20}$/;
Is there a way to negate any regular expression? I'm using regular expressions to validate input on a form. I'm now trying to create a button that sanitizes my input. Is there a way so I can use the regular expression used for the validating also for stripping the invalid characters?
I'm using this regex for validation of illegal characters
<input data-val-regex-pattern="[^|<>:\?'\*\[\]\=%\$\+,;~&\{\}]*" type="text" />
When clicking on a button next to it, I'm calling this function:
$('#button').click(function () {
var inputElement = $(this).prev();
var regex = new RegExp(inputElement.attr('data-val-regex-pattern'), 'g');
var value = inputElement.val();
inputElement.val(value.replace(regex, ''));
});
At the moment the javascript is doing the exact opposite of what I'm trying to accomplish. I need to find a way to 'reverse' the regex.
Edit: I'm trying to reverse the regex in the javascript function. The regex in the data-val-regex-pattern-attribute is doing his job for validation.
To find the invalid characters, just take the ^ off from your regex. The carret is the negative of everything that is inside the brackets.
data-val-regex-pattern="[|<>:\?'\*\[\]\=%\$\+,;~&\{\}]*"
This will return the undesired characters so you can replace them.
Also, as you want to take off a lot of non-word characters, you could try a simpler regex. If you want only word characters and spaces, you could use something like this:
data-val-regex-pattern="[\W\S]*"
Your reges is as so:
[^|<>:\?'\*\[\]\=%\$\+,;~&\{\}]*
That means, it matches any non-invalid character multiple times.
Then you replace this for empty, so you leave only the bad characters.
Try this instead, without the negation (hat moved somewhere else):
[|^<>:\?'\*\[\]\=%\$\+,;~&\{\}]*
The following answer is to the general question of negating a regular expression. In your specific case you just need to negate a character group, or more precisely remove the negation of a character group - which is detailed in other answers.
Regular languages – those consisting of all strings entirely by matched some RE – are in fact closed under negation: there is another RE which matches exactly those strings the original RE does not. It is however not trivial to construct, which perhaps explains why RE implementations often do not offer a negation operator.
However the Javascript regexp language has extensions that make it more expressive than regular languages; in particular there is the construct of negative lookahead.
If R1 is a regexp then
^(?!.*(R1))
matches precisely the strings that does not contain a match for R1.
And
^(?!R1$)
matches precisely the strings where the whole string is not a match for R1.
Ie. negation.
For rewriting any substring not matching a given regexp, the above is insufficient. One would have to do something like
((?!R1).)*
Which would catch any substring not containing a subsubstring that matches R1. - But consideration of the edge cases show that this does not quite do what we are after. For example ((?!ab).)* matches "b" in "ab", because "ab" is not a substring of "b".
One can cheat, and make your regexp like;
(.*)(R1|$)
And rewrite to T1$2
Where T1 is the target string you want to rewrite to.
This should rewrite any portion of the string not matching R1 to T1. However I would be very careful about any edge cases for this. So much so that it might be better to write the regexp from scratch rather than trying a general approach.
Im working on a password validation that should only allow a-z 0-9 and these characters "!"#$%&'()*+,-./:;<=>?#[\]^_{|}~`
I tried using a regex but I'm not too good with them and I wasnt sure if this is even possible or if Im not escaping the correct characters.
var allowedCharacters = /^[A-Za-Z0-9!"#$%&'()*+,-.\/:;<=>?#[\\]^_`{|}~]+$/;
if (!s.value.match(allowedCharacters)){
displayIllegalTextError();
return false;
}
You need to place the dash at the start or end of the regex, or it will try to create a character range (,-.). Then, a-Z isn't a valid range, you probably meant a-z. Also, you need to escape the closing brackets:
/^[A-Za-z0-9!"#$%&'()*+,.\/:;<=>?#[\\\]^_`{|}~-]+$/
Looking over the ascii chart here I see your regex could be reduced to this character range:
/^[\x21-\x7e]+$/
If you just want to learn special behavior of character classes, you should read up
on it via regex basic tutorials.
Note that class behavior differs amongst the different flavors.
Simpler and more to the point using unicode: ^[\u0021-\u007E]+$.
/^[\u0021-\u007E]+$/.test('MyPassword!') // returns true
/^[\u0021-\u007E]+$/.test('MyPassword™') // returns false
Now if you would like to go a few steps further and actually create a more complex validation such as: minimum length 8 characters and at least one lowercase, one uppercase, one digit and one special character:
^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])(?=.*[^a-zA-Z0-9])[\u0021-\u007E]{8,}$
I need help with a RegEx for a password. The password must contain at least one special char (like "§$&/!) AND a number.
E.g. a password like "EdfA433&" must be valid whereas "aASEas§ö" not as it contains not a number.
I have the following RegEx so far:
^(?=.*[0-9])(?=.*[a-zA-Z]).{3,}$
But this one is obviously checking only for a number. Can anyone help?
You're better off just using multiple more simple regular expressions: any code checking anything like this won't be performance sensitive, and the additional complexity of maintenance given a more complex regexp probably isn't justifiable.
So, what I'd go for:
var valid = foo.match(/[0-9]/) && foo.match(/["§$&/!]/);
I wonder if you really want to define special characters like that: Does é count as a special character? Does ~ count as a special character?
^(?=.*\d)(?=.*\W).{3,}$
checks for at least one digit (\d) and one non-alphanumeric character (\W). \W is the inverse of \w which matches digits, letters and the underscore.
If you want to include the underscore in the list of "special characters", use
^(?=.*\d)(?=.*[\W_]).{3,}$
I would divide function that checks if password is "hard" into some parts and in each part I would check one condition. You can see some complicated regex on Daily WTF with password reset: http://thedailywtf.com/Articles/The-Password-Reset-Facade.aspx