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how to do regex date format change?

Currently below regex is working fine with dates but I want it to accept those date and month also which has single digit. How can I do that?
Regex should accept below formats also:
'11/4/2021' or '1/4/2021' or '1/04/2021'
dateString = '11/04/2021'
let dateformat = /^(((0[1-9]|[12]\d|3[01])\/(0[13578]|1[02])\/((19|[2-9]\d)\d{2}))|((0[1-9]|[12]\d|30)\/(0[13456789]|1[012])\/((19|[2-9]\d)\d{2}))|((0[1-9]|1\d|2[0-8])\/02\/((19|[2-9]\d)\d{2}))|(29\/02\/((1[6-9]|[2-9]\d)(0[48]|[2468][048]|[13579][26])|(([1][26]|[2468][048]|[3579][26])00))))$/g;
if(dateString.match(dateformat)){
let operator = dateString.split('/');
console.log(operator)
}

Don't. Use Date.parse() instead.

For month: ^0{1}[1-9]$|^[1-9]{1}$|^1[012]{1}$
The first part is with 0, the second part is without 0, and the last one is for 10, 11 and 12.
For days: ^0{1}[1-9]{1}$|^[1-9]{1}$|^[12]{1}[0-9]{1}$|^3[01]{1}$
The first one is for days with from 1-9 starting with 0 and the second one is for the same but without the 0.
About the if the max day is 31, 30 or 28 I would use javascript for that.

You are able to achieve the requested by adding to the 8th group an OR statement with the 1 to 9 characters. That's for the days. The same goes for the months.
Let me give you an example.
Your matching group for the days right now is looking like this:
(0[1-9]|[12]\d|30)
Which means that you accept all numbers which start with 0 and a digit from 1 to 9 afterwards, a number starting with either 1 or 2 and any digit afterwards, or 30.
In order to accept the digits from 1 to 9 you have to add another condition to your matching group and that is the 1 to 9 digits. So your group will look something like the following:
(0[1-9]|[12]\d|30|[1-9])
This is the most basic thing you can do. There are plenty of ways to optimize this regex and do it in a better way. You can think about the 31st day of the month, since right now it is not capturing it.
The same way I shown in the example for the days' matching group, you can do it for the months' matching group.

Pad day and time in MomentJS without extra logic

I have a date formatted in Moment.JS but when I call
currentDay.month()
//Or
currentDay.date()
I get a 1 digit number (for example Jan 1st would be 0 for month and 1 for day). I need to get this with a leading 0 for date and month if less than 10. I can do this with something like...
currentDay.date() >= 10 ? currentDay.date() : '0'+currentDay.date()
But I was looking for something a bit nicer. Can I do this easily with MomentJS? I don't see anything in docs.

You can do this, eg for the month, with
moment().format('MM')

Regex for timestamp string

I'm looking to target the following string using jQuery. Specifically, I need to wrap it in a strong tag for styling purposes. I can't change the source data. My regex-fu is pathetic. Any suggestions?
Nov 18, 2013, 4pm CST:
Thanks guys - these are excellent answers. I should have been slightly more specific - I need to match all occurrences of this format within a collection, e.g.:
$('.admin-comments').match(/[A-Z]{1}[a-z]{2}\s[0-9]{1,2},\s[0-9]{4},\s[0-9]{1,2}[a|p]m\s[A-Z]{3}/)
(I have a log of comments and I'm trying to wrap the timestamp in a strong element.)
Edit: Final Working Solution
var adminComment = $('.admin-comments');
if (adminComment.length) {
var adminCommentTxt = adminComment.text();
var formatCommentTimestamp = adminCommentTxt.replace(/([A-Z]{1}[a-z]{2}\s[0-9]{1,2},\s[0-9]{4},\s[0-9\s]{1,2}[ap]m\s[A-Z]{3}\:)/g, "<strong>$1</strong>");
adminComment.html(formatCommentTimestamp);
}

Here you go: /^[A-Z]{1}[a-z]{2}\s[0-9]{1,2},\s[0-9]{4},\s[0-9]{1,2}[a|p]m\s[A-Z]{3}\:$/
'Nov 18, 2013, 4pm CST'.match(/^[A-Z]{1}[a-z]{2}\s[0-9]{1,2},\s[0-9]{4},\s[0-9]{1,2}[a|p]m\s[A-Z]{3}\:$/)
["Nov 18, 2013, 4pm CST"]
Keep in mind that this regex is expecting the line to start and end with this date, if the date is contained within other text, remove the ^ from the start and the $ from the end.
Hope this helps.
To further explain the regex and hopefully ++ your "regex-fu"
[A-Z]{1} - match one upper case letter
[a-z]{2} - match two lower case letters
So far we are at Nov, Oct, Jan, etc.
\s - space
[0-9]{1,2} - a 1 (min) or 2 (max) digit number
, - literal comma
\s - space
[0-9]{4} - a 4 digit number (the year)
So now we have matched: Nov 18, 2013
, - literal comma
\s - space
[0-9]{1,2} - just like before, a one or two digit number
[a|p]m - 'a' or 'p' followed by an 'm'
Now we've matched: Nov 18, 2013, 4pm
[A-Z]{3} An upper case three character string
\: literal colon
That is the entire string.
Putting ^ at the beginning of the regex means the text we are matching against MUST begin with the pattern; similarly, the $ states that the text we are matching MUST end with the pattern.
Good luck!

Rob M's answer is perfectly valid, and a more generic version than mine and may be exactly what you're looking for. However, if you want to be more specific with your months and time zones this may be useful for you:
(Oct)?(Nov)?\s\d{1,2},\s\d{4},\s\d{1,2}(pm)?(am)?\s(CST)?(PST)?:
This will match all of:
Nov 18, 2013, 4pm CST:
Oct 8, 2011, 11am PST:
Nov 02, 1981, 2am PST:
Oct 31, 1843, 12pm CST:
If you needed more months, you simply add each one like so:
(Mmm)? where Mmm corresponds to the month you want to match.
Similarly if you need more time zones, you'd add them like so:
(ZZZ)? where ZZZ corresponds to the timezone you want to match.
Similar to Rob's answer, if your date string is the only thing on the line, you could add the /^ and &/ prefix & suffix.

how to adjust date constructor in substr javascript

if you look at my date validation when i come to test if it is in the past it works although the date constructor expects a zero date month so how do i subtract one from the substring value representing the month (one from the result, not the position)
//start of datefield
var dateformat=/^(?:(?:31\/(?:0[13578]|1[02])|(?:29|30)\/(?:0[13-9]|1[012])|(?:0[1-9]|1\d|2[0-8])\/(?:0[1-9]|1[0-2]))\/[2-9]\d{3}|29\/02\/(?:[2-9]\d(?:0[48]|[2468][048]|[13579][26])|(?:[2468][048]|[3579][26])00))$/;
if (!date.match(dateformat))
{
errors.push("format incorrect use dd/mm/yyyy make sure you are entering correct days to the month remember 30 days have september, april, june & november, only 28 days in february unless leap year next is 2016");
}
var today = new Date();
var courseYear =date.substr(6,4) // use substr or substring to capture the last four digits
var courseMonth =date.substr(3,2) // use substr or substring to capture the four and fifth digits
var courseDay = date.substr(0,2)//e the first and second digits
var dateToCompare = new Date(courseYear, courseMonth, courseDay);
if (dateToCompare < today) {
errors.push("this date is in the past");
}

so how do i subtract one
With the subtraction operator "-" and the number literal "1". It also has the benefit of converting the string to a number before. For year and day, you might use the unary plus to do that conversion explicitly (though the Date constructor does it implicitly):
new Date(+courseYear, courseMonth-1, +courseDay);

regex javascript to add to expression

I have to write a few regexes. I did all except the following. I don't know how to add something via regex only.
Change this  9/28/2005  to  09.28.2005  (Am able to replace slashes with period but how to add leading zero to 9)
Change September 21, 2006 to 21. September 2006 (am able to replace , with . but how to change the order?)
Change 5:00 PM to i7:00 Uhr   (12 to 24 hour)?

1st Case (padding with zero):
result = subject.replace(/\b(\d)(?=[\/.])/ig, "0$1");
2nd Case (change order):
result = subject.replace(/\b(january|february|march|april|may|june|july|august|september|october|november|december) +(\d{2}), +(\d{2,4})\b/ig, "$2. $1 $3");
3rd case (12-24 hr)
As JavaScript RegEx engine does not support If-Then-Else Conditionals. So, it is hard and time-taking job to create a pure RegEx pattern for that purpose.
Hope this helps.

Just do the math manually... e.g. make a function GetTimeZoneOffset and go from there...
Return the timezone difference between UTC and Local Time:
var d = new Date()
var n = d.getTimezoneOffset();
The resulting value of n will be: 240
Add the difference and viola you have a real date which you can format what way you want..