This question already has answers here:
Regex to find not start and end with dot and allow some special character only not all
(3 answers)
Closed 2 years ago.
I am using following
ng-pattern="/^[a-zA-Z][a-zA-Z0-9._](.*[a-zA-Z0-9])?$/"
The matching String should
not start with a special character,
not end with special character, and
not include consecutive symbols except . (dot) and _ (underscore).
But it is not working.
Please, any suggestion.
Try using the word character class as a start ([\w] = [a-zA-Z0-9_]):
I'm not sure what you mean by consecutive symbols. But this might help:
/^[a-zA-Z]([\w.]*[a-zA-Z0-9])?$/
Maybe, have a look at the JavaScript RegExp Reference
Related
This question already has answers here:
Why do regex constructors need to be double escaped?
(5 answers)
Difference between regex [A-z] and [a-zA-Z]
(6 answers)
Regular expression works on regex101.com, but not on prod
(1 answer)
Closed 2 years ago.
I know this question has been asked a few times and the answers are unique to the specific regexes in question, but aside from that, I've tried to make sure that I'm escaping characters that have special meaning. Whilst this regex plays ball on https://regex101.com (in JavaScript mode), in my app it's having other ideas!
^([A-z0-9][A-z0-9 '\-,\.:\&]{0,245}[A-z0-9]|[A-z0-9][A-z0-9 '\-,\.:\&]{1,244}[A-z0-9])$
This is what I tell the user: 2-247 characters, start and end with A-z 0-9, permitted special characters: ' - , . : &
...but as you see, I'm actually also ensuring that the string starts with:
a) Two non-special characters, or
b) One non-special character followed by a special character, as long as that special character is followed by one or more non-special characters.
This is how I'm implementing the regex:
var nameRegex = new RegExp("^([A-z0-9][A-z0-9 '\-,\.:&]{0,245}[A-z0-9]|[A-z0-9][A-z0-9 '\-,\.:&]{1,244}[A-z0-9])$");
if (!nameRegex.test(formElements[i].value)) {
// validation stuff here
}
Everything the regex intends on doing, it does. I've tested every condition that I'm checking for. But it does more. Regex 101 disallows a string like d*d, but my app? Perfectly fine.
I'll try .match instead of .test, maybe .test isn't the tool I think I need for this job?
This question already has answers here:
Regex lazy vs greedy confusion
(2 answers)
Why does a simple .*? non-greedy regex greedily include additional characters before a match?
(3 answers)
Closed 3 years ago.
I'm trying this in javascript
/\/.*?$/.exec('foo/bar/tar')[0]
I was expecting to get /tar as result but getting /bar/tar. As far as I understand non-greed regex would take the smallest match.
I'm circumventing this with myvar.split('/').reverse()[0] but I couldn't understand what is going wrong with the regex.
There is nothing wrong with the regex but the pattern \/.*?$ matches from the first forward slash until the end of the string non greedy.
The dot matches any character except a newline and does not take a forward slash into account, so that will result in /bar/tar.
If you want to match /tar, you could match a forward slash, followed by not matching anymore forward slashes using a negated character class and then assert the end of the string.
\/[^\/]+$
Pattern demo
console.log(/\/[^\/]+$/.exec('foo/bar/tar')[0]);
This question already has answers here:
How do I use a Regex to replace non-alphanumeric characters with white space?
(1 answer)
Check for special characters in string
(12 answers)
Closed 3 years ago.
I have to use regex for my password validation that include special characters at least one.
https://en.wikipedia.org/wiki/ASCII
export const passwordValidation = password => {
const regPassword = /^(?=.*?[#?!#$%^&*-]).{8,}$/
return regPassword.test(password)
}
I tried this way but I think this isn't good way.
Is there other way to check all special characters by ascii code except alphanumeric ?
First, you need to define what a "special" character is. Do you mean anything not in the range A-Z (English alphabet)? A-Z and 0-9? Something else? Then you either use a character class listing the ones you want, which is what you've done, or a negated class saying you want something other than what's in the class:
return /^(?=.*?[^a-z0-9]).{8,}$/i.test(password);
// ^---- negated
This question already has answers here:
Using $ anchor inside a character class does not work
(2 answers)
Closed 3 years ago.
In js regex, I have
[\.\?!][\s$]
what I want to do is match
literal dot, or literal question mark or explanation mark
then
either 1 whitespace character or, be at the end of the string.
However the regex above, tries to match the literal $.
Does anyone know how to fix this?
Thanks
Try this Regex:
[.?!](?:\s|$)
Click for Demo
Explanation:
[.?!] - matches either a . or a ? or a ! literally
(?:\s|$) - matches either a white-space or the End-of-line
This question already has answers here:
How can I match a pipe character followed by whitespace and another pipe?
(5 answers)
What special characters must be escaped in regular expressions?
(13 answers)
Closed 5 years ago.
I'm using this JS regex text.match(/|(https:.+?path.+?)|/)[1] to get a regex of a URL that is in between pipe | characters but it's not working.
The text is ||https://url.com/path/value|| but I can't seem to extract the URL from it. I need to have path in the middle to identify this particular URL since there are other URLs in the file.
It doesn't have to be a URL that I'm extracting. I mainly would like to know how to extract something from between a pair of characters (| in this case).
You need to escape the pipe ("|") characters:
text.match(/\|(https:.+?path.+?)\|/)[1]
Pipe is a special character that basically means "or". https://www.regular-expressions.info/alternation.html
To grab everything between the two sets of || then you could use this regex:
text.match(/\|\|(.*)\|\|/)
The first part \|\| matches the characters || literally.
The next part (.*)matches any character zero or more and groups the result.
The last part \|\| matches the closing characters || literally.