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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

regexp to validate a non-zero number

My requirement is to validate a NON-ZERO number. The regular expression that I used is the following.
^[1-9]?\d+(\.\d)?\d*$
VALID VALUES should be
2
22
2222 (up to any number of digits)
2.2222 (up to any number of decimal points)
INVALID VALUES
0
(. without decimal values)
0.1 (any number of decimals, but start digit is 0)
2.4.5 (more than one .)
basically any values starting with 0 or has more than one . or no decimal points provided when . is added, are INCORRECT.

^[1-9]\d*(?:\.\d+)?$
https://regex101.com/r/wHZUoW/1
-Since you don't want the number to start with 0, you shouldn't make the [1-9] at the beginning optional with ?.
-As general good practice, a non-capturing group (?: ... ) is used instead of a capturing group, because the contents do not need to be referenced later.

You can use this regex to validate your numbers:
^(?=[1-9])\d*\.?\d+$
It uses a regex proposed by #WiktorStribiżew as a comment to this question to match decimal numbers (^\d*\.?\d+$), and adds a positive lookahead to ensure that the first character of the number is not 0. Note that if you want to allow numbers such as .3, you should add . to the lookahead character class i.e. ^(?=[1-9.])\d*\.?\d+$
Demo on regex101

[1-9]+[0-9]*(\.[0-9]+)? should work for your case.
[1-9]+ will make sure that the expression starts with a number different than 0.
[0-9]* will make sure to allow the expression to have zeros after the first digit.
(\.[0-9]+)? will allow an extension to the expression, which has to have a . and at least 1 number after it. The ? in the end makes it optional.
By the way, I really like this website to test my regular expressions: https://regexr.com/, you should try it.

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.

UK Currency Regular Expression for javascript

I'm after a regular expression that matches a UK Currency (ie. £13.00, £9,999.99 and £12,333,333.02), but does not allow negative (-£2.17) or zero values (£0.00 or 0).
I've tried to create one myself, but I've got in a right muddle!
Any help greatfully received.
Thanks!

This'll do it (well mostly...)
/^£?[1-9]{1,3}(,\d{3})*(\.\d{2})?$/
Leverages the ^ and $ to make sure no negative or other character is in the string, and assumes that commas will be used. The pound symbol, and pence are optional.
edit: realised you said non-zero so replaced the first \d with [1-9]
Update: it's been pointed out the above won't match £0.01. The below improvement will but now there's a level of complexity where it may quite possibly be better to test /[1-9]/ first and then the above - haven't benchmarked it.
/^£?(([1-9]{1,3}(,\d{3})*(\.\d{2})?)|(0\.[1-9]\d)|(0\.0[1-9]))$/
Brief explanation:
Match beginning of string followed by optional "£"
Then match either:
a >£1 amount with potential for comma separated groupings and optional pence
OR a <£1 >=£0.10 amount
OR a <=£0.09 amount
Then match end of line
The more fractions of pence (zero in the above) you require adding to the regex the less efficient it becomes.

Under Unix/Linux, it's not always possible to type in the '£' sign in a JavaScript file, so I tend to use its hexadecimal representation, thus:
/^\xA3?\d{1,3}?([,]\d{3}|\d)*?([.]\d{1,2})?$/
This seems to take care of all combinations of UK currency amounts representation that I have come across.

/^\xA3?\d{1,}(?:\,?\d+)*(?:.\d{1,2})?$/;
Explanation:
^ Matches the beginning of the string, or the beginning of a line.
xA3 Matches a "£" character (char code 163)
? Quantifier for match between 0 and 1 of the preceding token.
\d Matches any digit character (0-9).
{1,} Match 1 or more of the preceding token.
(?: Groups multiple tokens together without creating a capture group.
\, Matches a "," character (char code 44).
{1,2} Match between 1 and 2 of the preceding token.
$ Matches the end of the string, or the end of a line if the multiline flag (

You could just make two passes:
/^£\d{1,3}(,\d{3})*(\.\d{2})?$/
to validate the format, and
/[1-9]/
to ensure that at least one digit is non-zero.
This is less efficient than doing it in one pass, of course (thanks, annakata, for the benchmark information), but for a first implementation, just "saying what you want" can significantly reduce developing time.