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Match all numbers in regex except for phrase foo2020

In a jumbled up string such as:
dfnjvqifoo2020o43e25w54p32n5qvto4325432543nvgn4325gn2020repw
I want to match all numbers from the string except the phrase foo2020
I have tried /(?<!foo)\d+/g
The problem is that the 020 in foo2020 gets matched when it's not supposed to, I know it's because \d means matching a single digit but I don't know how to get around this

Your regex, /(?<!foo)\d+/g, matches any one or more digits that are not immediately preceded with foo. That means, if a leftmost digit is not immediately preceded with foo, it will be matched with this regex.
All you need to do is add another restriction: do not start matching digits if there is a digit immediately on the left, i.e. add (?<!\d), a left-hand digit boundary construct.
/(?<!foo)(?<!\d)\d+/g
// ^^^^^^^
See the regex demo.
Note the two consecutive lookbehinds, (?<!foo)(?<!\d) are executed at the same position one after another, which means there will be no match if there is foo or a digit immediately to the left of the current location.
If you need to make sure there is no digit immediately on the right, append (?!\d), a right-hand digit boundary construct, after \d+ pattern. It is not required here, but if you need to match 50 in abc50def and not in abc500def, you would need this negative lookahead.

If you want the match only, you could add \d* to the lookbehind. Note that this is not widely supported by all browsers. I will work in for example chrome or nodejs.
(?<!foo\d*)\d+
Explanation
(?<!foo\d*) Negative lookbehind, assert what is on the left is not foo followed by 0+ digits
\d+ Match 1+ digits
Regex demo
Another option is to use a capturing group, match what you don't want and capture what you do want.
foo\d+|(\d+)
Regex demo

Match the alphanumeric words(NOT NUMERIC-ONLY words) which have unique digits

Using regular expression, I want to select only the words which:
are alphanumeric
do not contain only numbers
do not contain only alphabets
have unique numbers(1 or more)
I am not really good with the regex but so far, I have tried [^\d\s]*(\d+)(?!.*\1) which takes me nowhere close to the desired output :(
Here are the input strings:
I would like abc123 to match but not 123.
ab12s should also match
Only number-words like 1234 should not match
Words containing same numbers like ab22s should not match
234 should not match
hel1lo2haha3hoho4
hel1lo2haha3hoho3
Expected Matches:
abc123
ab12s
hel1lo2haha3hoho4

You can use
\b(?=\d*[a-z])(?=[a-z]*\d)(?:[a-z]|(\d)(?!\w*\1))+\b
https://regex101.com/r/TimjdW/3
Anchor the start and end of the pattern at word boundaries with \b, then:
(?=\d*[a-z]) - Lookahead for an alphabetical character somewhere in the word
(?=[a-z]*\d) - Lookahead for a digit somewhere in the word
(?:[a-z]|(\d)(?!\w*\1))+ Repeatedly match either:
[a-z] - Any alphabetical character, or
(\d)(?!\w*\1) - A digit which does not occur again in the same word

Here is a bit shorter & faster regex to make it happen since it doesn't assert negative lookahead for each character:
/\b(?=[a-z]*\d)(?=\d*[a-z])(?!\w*(\d)\w*\1)[a-z\d]+\b/ig
RegEx Demo
RegEx Details:
\b: Word boundary
(?=[a-z]*\d): Make sure we have at least a digit
(?=\d*[a-z]): Make sure we have at least a letter
(?!\w*(\d)\w*\1): Make sure digits are not repeated anywhere in the word
[a-z\d]+: Match 1+ alphanumericals
\b: Word boundary

You could assert all the conditions using one negative lookahead:
\b(?![a-z]+\b|\d+\b|\w*(\d)\w*\1)[a-z\d]+\b
See live demo here
The important parts are starting match from \b and immediately looking for the conditions:
[a-z]+\b Only alphabetic
\d+\b Only numeric
\w*(\d)\w*\1 Has a repeating digit

You can use this
\b(?!\w*(\d)\w*\1)(?=(?:[a-z]+\d+)|(?:\d+[a-z]+))[a-z0-9]+\b
\b - Word boundary.
(?!\w*(\d)\w*\1) - Condition to check unique digits.
(?=(?:[a-z]+\d+)|(?:\d+[a-z]+)) - Condition to check alphanumeric words.
[a-z0-9]+ - Matches a to z and 0 to 9
Demo

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

/(\S)\1(\1)+/g matching all occurrences of three equal non-whitespace characters following each other

Its given: /(\S)\1(\1)+/g matches all occurrences of three equal non-whitespace characters following each other.
I don't understand why there is () around (\S) and 2nd (\1), but not around 1st (\1). Can anyone help in explaining how above regex works?
src: http://www.javascriptkit.com/javatutors/redev2.shtml
Thnx in advance.

The \S needs parentheses to capture its value, so you can refer back to the captured value with \1. \1 means "match the same text which capturing group #1 matched".
I believe there is a problem with this regex. You said you want to match "three equal non-whitespace characters". But the + will make this match 3 or more equal, consecutive non-whitespace characters.
The g on the end means "apply this regex over the entire input string, or globally".

The second set of parentheses is not necessary. It needlessly captures the repeated character a second time, while matching the same strings as this regex:
/(\S)\1\1+/g
Also, as #AlexD pointed out, the description should say that it matches at least three characters. If you replaced that regex with BONK in the string fooxxxxxxbar:
'fooxxxxxxbar'.replace(/(\S)\1\1+/g, 'BONK')
..you might expect the result to be fooBONKBONKbar from their description, because there are two sets of three 'x's. But in fact the result would be fooBONKbar; the first \1 matches the second 'x', and the \1+ matches the third 'x' and any 'x's that follow it. If they wanted to match just three characters, they should have left the + off.
I noticed several other sloppy descriptions like that, plus at least one outright error: \B is equivalent to (?!\b) (a position that's not a word boundary), not [^\b] (a character that's not a backspace). For that matter, their description of word boundaries--"the position between a word and a space"--is wrong, too. A word boundary isn't defined by any particular character, like a space--in fact, it can just as well be the absence of any character that creates one. The string:
Word
...starts with a word boundary because 'W' is a word character and, being first, it's not preceded by another word character. Similarly, the 'd' is not followed by another word character, so the end of the string is also a word boundary.
Also, a regex doesn't know from words, only word characters. The definition of a word character can vary depending on the regex flavor and Unicode or locale settings, but it always includes [A-Za-z0-9_] (ASCII letters and digits plus the underscore). A word boundary is simply a position that's between one of those characters and any other character (or no other character, as I explained earlier).
If you want to learn about regexes, I suggest you forget that site and start here instead: regular-expressions.info.

regex for single lower case word

I am looking for a regex pattern that ensures the user puts in a single lower case word with only letters of the alphabet. Basically they are picking a subdomain. Thanks in advance

The character class [a-z] describes one single character of the alphabet of lowercase letters a–z. If you want if an input does only contain characters of that class, use this:
^[a-z]+$
^ and $ mark the start and end of the string respectively. And the quantifier + allows one or more repetitions of the preceding expression.

^[a-z]+$ Will find one and only one lower-case word, with no spaces before or after the word.

/^[a-z]+$/
make sure you aren't using 'i' after the last slash
/[a-z]+/
if you are searching for any words within the context

If you want to find all occurrences of lowercase-only ASCII-char words, you can use
text.match(/\b[a-z]+\b/g)
See the regex demo.
Details:
\b - a word boundary
[a-z]+ - one or more (+) lowercase ASCII letters
\b - a word boundary
The g flag makes it extract all occurrences.
See the JavaScript demo:
const text = "123456789 Ticket number (CO2) text";
console.log(text.match(/\b[a-z]+\b/g));