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Regex to match an aws arn

I am trying to validate aws arn for a connect instance but I am stuck on creating the correct regex.
Below is the string that I want to validate.
arn:aws:connect:us-west-2:123456789011:instance/0533yu22-d4cb-410a-81da-6c9hjhjucec4b9
I want to create a regex which checks below things.
arn:aws:connect:<region_name>:<12 digit account id>:instance/<an alphanumeric instance id>
Can someone please help.
Tried below
^arn:aws:connect:\S+:\d+:instance\/\S+\/queue\/\S+$

There is no /queue/ substring in your example string, and \S+ matches any no whitespace character and will cause backtracking to match the rest of the pattern.
You might update your pattern to ^arn:aws:connect:\S+:\d+:instance\/\S+$ but that will be less precise according to the things you want to check.
A bit more precise pattern could be:
^arn:aws:connect:\w+(?:-\w+)+:\d{12}:instance\/[A-Za-z0-9]+(?:-[A-Za-z0-9]+)+$
^ Start of string
arn:aws:connect: Match literally
\w+(?:-\w+)+: Match 1+ word characters and repeat matching - and 1+ word characters and then match :
\d{12}: Match 12 digits and :
instance\/ Match instance/
[A-Za-z0-9]+(?:-[A-Za-z0-9]+)+ Match 1+ alpha numerics and repeat 1+ times - and 1+ alpha numerics
$ End of string
Regex demo

You need some capture groups to facilitate this. Here I've also used named capture groups for ease of understanding.
const string = "arn:aws:connect:us-west-2:123456789011:instance/0533yu22-d4cb-410a-81da-6c9hjhjucec4b9";
// Regex broken down into parts
const parts = [
'arn:aws:connect:',
'(?<region_name>[^:]+?)', // group 1
':',
'(?<account_id>\\d{12})', // group 2
':instance\\/',
'(?<instance_id>[A-z0-9\\-]+?)', // group 3
'$'
];
// Joined parts into regex expression
const regex = new RegExp(parts.join(''));
// Execute query and assign group values to variables
const { region_name, account_id, instance_id } = regex.exec(string).groups;
console.log("region_name:", region_name);
console.log("account_id:", account_id);
console.log("instance_id:", instance_id);

Regex to Extract Last Part of URL and Before Another Character

In the URLs
https://image/4x/c1/abc/5b026cdb06921e7ca5f7a24aff46512e--wedding-vendors-wedding-receptions.jpg
https://image/4x/c1/abc/5b026cdb06921e7ca5f7a24aff46512e.jpg
I'm trying to capture 5b026cdb06921e7ca5f7a24aff46512e in both of these strings. The string will always happen after the last slash, it will be a random assortment of letters and numbers, it may or may not have --randomtext appended, and it will have .jpg at the end.
I currently have ([^\/]+)$ to extract any string after the last slash, but would like to know how to capture everything before .jpg and --randomtext(if present). I will be using this in javascript.

If what is after the last forward slash is a random assortment of letters and numbers a-z0-9, on option is to use a capturing group.
^.*\/([a-z0-9]+).*\.jpg$
In parts
^ Start of string
.*\/ Match until including the last /
([a-z0-9]+) Capture in group 1 matching 1+ chars a-z or digits 0-9
.* Match any char except a newline 0+ times
\.jpg Match .jpg
$ End of string
Regex demo
const regex = /^.*\/([a-z0-9]+).*\.jpg$/;
["https://image/4x/c1/abc/5b026cdb06921e7ca5f7a24aff46512e--wedding-vendors-wedding-receptions.jpg",
"https://image/4x/c1/abc/5b026cdb06921e7ca5f7a24aff46512e.jpg"
].forEach(s => console.log(s.match(regex)[1]));

You can split by / and take the last part, and then replace anything after -- or .jpg from end with empty string
let arr = ["https://image/4x/c1/abc/5b026cdb06921e7ca5f7a24aff46512e--wedding-vendors-wedding-receptions.jpg","https://image/4x/c1/abc/5b026cdb06921e7ca5f7a24aff46512e.jpg"]
let getText = (url) =>{
return url.split('/').pop().replace(/(--.*|\.jpg)$/g,'')
}
arr.forEach(url=> console.log(getText(url)))
If there are chances to have -- more than one time than instead of replacing you can simply match match(/^[a-z0-9]+/g) and take the first element from matched array

Use:
([^\/]*?)(?:--.*)?\.jpg$
and your desired match will be in $1
https://regex101.com/r/gZ9kSi/1

How to make regex for 3 slashes?

I want to write regex for following
students/ad34567-06c1-498c-9b15-cdbac695c1f2/data/sessions
Where students, data and sessions should be exact match.
i have tried this
[students]\[a-z]\[a-z]\[a-z]

You can try this regex, although your question is not clear to me.
^students\/([\w\-\d]+)\/data\/sessions$
Check here https://regex101.com/r/xnxwCX/1
you can grab the data in between students/, /data/session.

In your regex [students]\\[a-z]\\[a-z]\\[a-z] you are trying to match with word students in a character class [students] which will match one of the specified characters instead of matching the whole word.
To match a forward slash you have to use \/ instead of //. [a-z] is specified without a quantifier and will match 1 character from a-z.
To match your example string you might use
^students\/[a-z0-9]+(?:-[a-z0-9]+)+\/data\/sessions$
Regex demo
This part [a-z0-9]+(?:-[a-z0-9]+)+ matches one or more times a lowercase character or a digit [a-z0-9]+
Following a non capturing group repeated one or more times that will match a hyphen followed by matching one or more times a lowercase character or a digit (?:-[a-z0-9]+)+
You might also use [a-f0-9] if your characters are a -f

Grab full regex word if pattern inside it matches

How do I retrieve an entire word that has a specific portion of it that matches a regex?
For example, I have the below text.
Using ^.[\.\?\!:;,]{2,} , I match the first 3, but not the last. The last should be matched as well, but $ doesn't seem to produce anything.
a!!!!!!
n.......
c..,;,;,,
huhuhu..
I want to get all strings that have an occurrence of certain characters equal to or more than twice. I produced the aforementioned regex, but on Rubular it only matches the characters themselves, not the entire string. Using ^ and $
I've read a few stackoverflow posts similar, but not quite what I'm looking for.

Change your regex to:
/^.*[.?!:;,]{2,}/gm
i.e. match 0 more character before 2 of those special characters.
RegEx Demo

If I understand well you are trying to match an entire string that contains at least the same punctuation character two times:
^.*?([.?!:;,])\1.*
Note: if your string has newline characters, change .* to [\s\S]*
The trick is here:
([.?!:;,]) # captures the punct character in group 1
\1 # refers to the character captured in group 1

Regex to match '-' delimited alphanumeric words

I would like to test if user type only alphanumeric value or one "-".
hello-world -> Match
hello-first-world -> match
this-is-my-super-world -> match
hello--world -> NO MATCH
hello-world-------this-is -> NO MATCH
-hello-world -> NO MATCH (leading dash)
hello-world- -> NO MATCH (trailing dash)
Here is what I have so far, but I dont know how to implement the "-" sign to test it if it is only once without repeating.
var regExp = /^[A-Za-z0-9-]+$/;

Try this:
/^[A-Za-z0-9]+(?:-[A-Za-z0-9]+)*$/
This will only match sequences of one or more sequences of alphanumeric characters separated by a single -. If you do not want to allow single words (e.g. just hello), replace the * multiplier with + to allow only one or more repetitions of the last group.

Here you go (this works).
var regExp = /^[A-Za-z0-9]+([-]{1}[A-Za-z0-9]+)+$/;
letters and numbers greedy, single dash, repeat this combination, end with letters and numbers.

(^-)|-{2,}|[^a-zA-Z-]|(-$) looks for invalid characters, so zero matches to that pattern would satisfy your requirement.

I'm not entirely sure if this works because I haven't done regex in awhile, but it sounds like you need the following:
/^[A-Za-z0-9]+(-[A-Za-z0-9]+)+$/
You're requirement is split up in the following:
One or more alphanumeric characters to start (that way you ALWAYS have an alphanumeric starting.
The second half entails a "-" followed by one or more alphanumeric characters (but this is optional, so the entire thing is required 0 or more times). That way you'll have 0 or more instances of the dash followed by 1+ alphanumeric.
I'm just not sure if I did the regex properly to follow that format.

The expression can be simplified to: /^[^\W_]+(?:-[^\W_]+)+$/
Explanation:
^ match the start of string
[^\W_]+ match one or more word(a-zA-Z0-9) chars
(?:-[^\W_]+)+ match one or more group of '-' follwed by word chars
$ match the end of string
Test: https://regex101.com/r/MODQxw/1