This question already has answers here:
Regular expression to limit number of characters to 10
(6 answers)
Closed 6 years ago.
I need regex for validating alphanumeric String with length of 6 chars. I tried with following regex And done it for allowing alphanumeric chars but don't know how to stop exceeding more than six chars.
var regex = new RegExp("^[a-zA-Z0-9\b]+$");
Fiddle here.
you just have to add {6} in the end
var regex = new RegExp('[a-zA-Z0-9\b]{6}$'); You can test it out here https://regex101.com/
Related
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 2 years ago.
Hi I'm trying to remove the letters and special characters from javascript.
Example:
var id = "Parameter[0].Category"
I only need the "0" from this. Thank you
Try this
var numbers = id.match(/\d+/)[0];
console.log(numbers);
In general this is called filtering using regular expressions.
If you want to filter out letters and special characters you can use regular expression in JS, like this:
id = id.replace(/\D/g, "")
You replace every (g option) character in your string that is not a digit \D with blank ""
This question already has answers here:
Regex to match only letters
(20 answers)
Closed 3 years ago.
I need to find regex pattern for url and use regex.test() so only string like this:
http://*.margonem.pl/
so it's exactly like above string and where * must appear and can be string which only contains a-z letters without any signs.
That would be http:\/\/[a-z]+\.margonem\.pl\/:
Matches
http://a.margonem.pl/
http://foo.margonem.pl/
Does not match
http://hello-world.margonem.pl/
http://abcq443435u4531.margonem.pl/
This question already has answers here:
Negating a backreference in Regular Expressions
(6 answers)
Closed 4 years ago.
Lets take:
stringi = 'xnxx xnnx xnnxn'
My regex is: (n)[^n]
I want to make my regex a little more dynamic like that:
(n)[^\1] -\1 beeing the capt. grp 1
My desired result would be that:
(n)[^\1] would be equal (n)[^n]
(x)[^\1] would be equal (x)[^x]
How can I not match a NOT-\1 character?
using a negative lookahead, the . is to match any character as n length is one
(n)(?!\1).
This question already has an answer here:
Learning Regular Expressions [closed]
(1 answer)
Closed 6 years ago.
How can I create a regex for that returns true if it has only numbers and '+' basically 0-9 & +. Using javascript or jQuery.
Regex for plus anywhere: /^[0-9+]+$/
Regex for plus only infront: /^\+?[0-9]+$/
What it does:
^ Matches the beginning of the string
[0-9+] Matches 0123456789+
+ Matches one or more
$ Matches the end of the string
Other version:
\+? Matches zero or one plus signs in the front
Maybe try regexr for future regex development.
How to test in code:
function isOnlyNumber(str) {
return /^[0-9+]+$/.test(str);
}
This question already has answers here:
How to validate phone numbers using regex
(43 answers)
Closed 8 years ago.
I have this regex expression var re = /(?:\d{3}|\(\d{3}\))([\w-\/\.]?)\d{3}\1\d{4}/;, however, the \w whitespace doesn't work on this test console.log(re.test('123 456 7890'));
Here is my jsfiddle: http://jsfiddle.net/Bqb22/
You shouldn't use \w for whitespace. Use \s instead. (\w is word character, the same as [0-9A-Za-z_], and should not be used to indicate whitespace).