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Javascript: Regex to exclude whitespace and special characters

I need a regex to validate,
Should be of length 18
First 5 characters should be either (xyz34|xyz12)
Remaining 13 characters should be alphanumeric only letters and numbers, no whitespace or special characters is allowed.
I have a pattern like here, '/^(xyz34|xyz12)((?=.*[a-zA-Z])(?=.*[0-9])){13}/g'
But this is allowing whitespace and special characters like ($,% and etc) which is violating the rule #3.
Any suggestion to exclude this whitespace and special characters and to strictly check that it must be letters and numbers?

You should not quantify lookarounds. They are non-consuming patterns, i.e. the consecutive positive lookaheads check the presence of their patterns but do not advance the regex index, they check the text at the same position. It makes no sense repeating them 13 times. ^(xyz34|xyz12)((?=.*[a-zA-Z])(?=.*[0-9])){13} is equal to ^(xyz34|xyz12)(?=.*[a-zA-Z])(?=.*[0-9]), and means the string can start with xyz34 or xyz12 and then should have at least 1 letter and at least 1 digits.
You may consider fixing the issue by using a consuming pattern like this:
If you do not care if the last 13 chars contain only digits or only letters, use the patterns suggested by other users, like /^(?:xyz34|xyz12)[a-zA-Z\d]{13}$/ or /^xyz(?:34|12)[a-zA-Z0-9]{13}$/
If there must be at least 1 digit and at least 1 letter among those 13 alphanumeric chars, use /^xyz(?:34|12)(?=[a-zA-Z]*\d)(?=\d*[a-zA-Z])[a-zA-Z\d]{13}$/.
See the regex demo #1 and the regex demo #2.
NOTE: these are regex literals, do not use them inside single- or double quotes!
Details
^ - start of string
xyz - a common prefix
(?:34|12) - a non-capturing group matching 34 or 12
(?=[a-zA-Z]*\d) - there must be at least 1 digit after any 0+ letters to the right of the current location
(?=\d*[a-zA-Z]) - there must be at least 1 letter after any 0+ digtis to the right of the current location
[a-zA-Z\d]{13} - 13 letters or digits
$ - end of string.
JS demo:
var strs = ['xyz34abcdefghijkl1','xyz341bcdefghijklm','xyz34abcdefghijklm','xyz341234567890123','xyz14a234567890123'];
var rx = /^xyz(?:34|12)(?=[a-zA-Z]*\d)(?=\d*[a-zA-Z])[a-zA-Z\d]{13}$/;
for (var s of strs) {
console.log(s, "=>", rx.test(s));
}

.* will match any string, for your requirment you can use this:
/^xyz(34|12)[a-zA-Z0-9]{13}$/g
regex fiddle

/^(xyz34|xyz12)[a-zA-Z0-9]{13}$/
This should work,
^ asserts position at the start of a line
1st Capturing Group (xyz34|xyz12)
1st Alternative xyz34 matches the characters xyz34 literally (case sensitive)
2nd Alternative xyz12 matches the characters xyz12 literally (case sensitive)
Match a single character present in the list below [a-zA-Z0-9]{13}
{13} Quantifier — Matches exactly 13 times

Regex-How can I filter out any grouping of 3 numbers? [duplicate]

I'm attempting to string match 5-digit coupon codes spread throughout a HTML web page. For example, 53232, 21032, 40021 etc... I can handle the simpler case of any string of 5 digits with [0-9]{5}, though this also matches 6, 7, 8... n digit numbers. Can someone please suggest how I would modify this regular expression to match only 5 digit numbers?

>>> import re
>>> s="four digits 1234 five digits 56789 six digits 012345"
>>> re.findall(r"\D(\d{5})\D", s)
['56789']
if they can occur at the very beginning or the very end, it's easier to pad the string than mess with special cases
>>> re.findall(r"\D(\d{5})\D", " "+s+" ")

Without padding the string for special case start and end of string, as in John La Rooy answer one can use the negatives lookahead and lookbehind to handle both cases with a single regular expression
>>> import re
>>> s = "88888 999999 3333 aaa 12345 hfsjkq 98765"
>>> re.findall(r"(?<!\d)\d{5}(?!\d)", s)
['88888', '12345', '98765']

full string: ^[0-9]{5}$
within a string: [^0-9][0-9]{5}[^0-9]

Note: There is problem in using \D since \D matches any character that is not a digit , instead use \b.
\b is important here because it matches the word boundary but only at end or beginning of a word .
import re
input = "four digits 1234 five digits 56789 six digits 01234,56789,01234"
re.findall(r"\b\d{5}\b", input)
result : ['56789', '01234', '56789', '01234']
but if one uses
re.findall(r"\D(\d{5})\D", s)
output : ['56789', '01234']
\D is unable to handle comma or any continuously entered numerals.
\b is important part here it matches the empty string but only at end or beginning of a word .
More documentation: https://docs.python.org/2/library/re.html
More Clarification on usage of \D vs \b:
This example uses \D but it doesn't capture all the five digits number.
This example uses \b while capturing all five digits number.
Cheers

A very simple way would be to match all groups of digits, like with r'\d+', and then skip every match that isn't five characters long when you process the results.

You probably want to match a non-digit before and after your string of 5 digits, like [^0-9]([0-9]{5})[^0-9]. Then you can capture the inner group (the actual string you want).

You could try
\D\d{5}\D
or maybe
\b\d{5}\b
I'm not sure how python treats line-endings and whitespace there though.
I believe ^\d{5}$ would not work for you, as you likely want to get numbers that are somewhere within other text.

I use Regex with easier expression :
re.findall(r"\d{5}", mystring)
It will research 5 numerical digits. But you have to be sure not to have another 5 numerical digits in the string

How does the following code mean two consecutive numbers?

This is from an exercise on FCC beta and i can not understand how the following code means two consecutive numbers seeing how \D* means NOT 0 or more numbers and \d means number, so how does this accumulate to two numbers in a regexp?
let checkPass = /(?=\w{5,})(?=\D*\d)/;

This does not match two numbers. It doesn't really match anything except an empty string, as there is nothing preceding the lookup.
If you want to match two digits, you can do something like this:
(\d)(\d)
Or if you really want to do a positive lookup with the (?=\D*\d) section, you will have to do something like this:
\d(?=\D*\d)
This will match against the last digit which is followed by a bunch of non-digits and a single digit. A few examples (matched numbers highlighted):
2 hhebuehi3
^
245673
^^^^^
2v jugn45
^ ^
To also capture the second digit, you will have to put brackets around both numbers. Ie:
(\d)(?=\D*(\d))
Here it is in action.
In order to do what your original example wants, ie:
number
5+ \w characters
a non-number character
a number
... you will need to precede your original example with a \d character. This means that your lookups will actually match something which isn't just an empty string:
\d(?=\w{5,})(?=\D*\d)
IMPORTANT EDIT
After playing around a bit more with a JavaScript online console, I have worked out the problem with your original Regex.
This matches a string with 5 or more characters, including at least 1 number. This can match two numbers, but it can also match 1 number, 3 numbers, 12 numbers, etc. In order to match exactly two numbers in a string of 5-or-more characters, you should specify the number of digits you want in the second half of your lookup:
let regex = /(?=\w{5,})(?=\D*\d{2})/;
let string1 = "abcd2";
let regex1 = /(?=\w{5,})(?=\D*\d)/;
console.log("string 1 & regex 1: " + regex1.test(string1));
let regex2 = /(?=\w{5,})(?=\D*\d{2})/;
console.log("string 1 & regex 2: " + regex2.test(string1));
let string2 = "abcd23";
console.log("string 2 & regex 2: " + regex2.test(string2));
My original answer was about Regex in a vacuum and I glossed over the fact that you were using Regex in conjunction with JavaScript, which works a little differently when comparing Regex to a string. I still don't know why your original answer was supposed to match two numbers, but I hope this is a bit more helpful.

?= Positive lookahead
w{5,} matches any word character (equal to [a-zA-Z0-9_])
{5,}. matches between 5 and unlimited
\D* matches any character that\'s not a digit (equal to [^0-9])
* matches between zero and unlimited
\d matches a digit (equal to [0-9])
This expression is global - so tries to match all
You can always check your expression using regex101

Javascript regex for extracting certain part of string

I need to extract certain part of Javascript string. I was thinking to do it with regex, but couldn't come up with one which does it correctly.
String can have variable length & can contain all possible characters in all possible combinations.
What I need to extract from it, is 10 adjacent characters, that match one of next two possible combinations:
9 numbers & 1 letter "X" (capital letter "X", not X as variable letter!)
10 numbers
So, if input string is this: "[1X,!?X22;87654321X9]ddee", it should return only "87654321X9".
I hope I've explained it good enough. Thanks in advance!

This Regex will work:
\d{9}X|\d{8}X\d|\d{7}X\d{2}|\d{6}X\d{3}|\d{5}X\d{4}|\d{4}X\d{5}|\d{3}X\d{6}|\d{2}X\d{7}|\d{1}X\d{8}|\d{10}|X\d{9}
As described, It need to match 9 digits and any letter, and the letter can be at any position of the sequence.
\d{9}X # will match 9 digits and a letter in the end
\d{8}X\d # will match 8 digits a lettter then a digit again
...
\d{1}X\d{8} # will match 1 digits a lettter then 8 digits
\{10} # will match 10 digits
Edited to match only X

You can use this much simpler regex:
/(?!\d*X\d*X)[\dX]{10}/
RegEx Breakup:
(?!\d*X\d*X) # negative lookahead to fail the match if there are 2 X ahead
[\dX]{10} # match a digit or X 10 times
Since more than one X is not allowed due to use of negative lookahead, this regex will only allow either 10 digits or ekse 9 digits and a single X.
RegEx Demo
This regex has few advantages over the other answer:
Much simpler regex that is easier to read and maintain
Takes less than half steps to complete which can be substantial difference on larger text.

Is there a difference between \d and \d+? [duplicate]

This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 6 years ago.
https://www.freecodecamp.com/challenges/find-numbers-with-regular-expressions
I was doing a lesson in FCC, and they mentioned that the digit selector \d finds one digit and adding a + (\d+) in front of the selector allows it to search for more than one digit.
I experimented with it a bit, and noticed that its the g right after the expression that searches for every number, not the +. I tried using \d+ without the g after the expression, and it only matched the first number in the string.
Basically, whether I use \d or \d+, as long as I have the g after the expression, It will find all of the numbers. So my question is, what is the difference between the two?
// Setup
var testString = "There are 3 cats but 4 dogs.";
var expression = /\d+/g;
var digitCount = testString.match(expression).length;

The g at the end means global, ie. that you want to search for all occurrences. Without it, you'll just get the first match.
\d, as you know, means a single digit. You can add quantifiers to specify whether you want to match all the following, or a certain amount of digits afterwards.
\d means a single digit
\d+ means all sequential digits
So let's say we have a string like this:
123 456
7890123
/\d/g will match [1,2,3,4,5,6,7,8,9,0,1,2,3]
/\d/ will match 1
/\d+/ will match 123
/\d+/g will match [123,456,7890123]
You could also use /\d{1,3}/g to say you want to match all occurrences where there are from 1 to 3 digits in a sequence.
Another common quantifier is the star symbol, which means 0 or more. For example /1\d*/g would match all sequences of digits that start with 1, and have 0 or more digits after it.

Counting the occurrences of \d will find the number of digits in the string.
Counting the occurrences of \d+ will find the number of integers in the string.
I.E.
123 456 789
Has 9 digits, but 3 integers.

\d means any digit from 0 to 9, the + says "one or more times".
As long as your numbers are single digit there is no difference, but in the string "I have 23 cows" and \d would match 2 alone whereas \d+ would match 23.