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What Regex would capture both the beginning and end from of a string?

I am trying to edit a DateTime string in typescript file.
The string in question is 02T13:18:43.000Z.
I want to trim the first three characters including the letter T from the beginning of a string AND also all 5 characters from the end of the string, that is Z000., including the dot character. Essentialy I want the result to look like this: 13:18:43.
From what I found the following pattern (^(.*?)T) can accomplish only the first part of the trim I require, that leaves the initial result like this: 13:18:43.000Z.
What kind of Regex pattern must I use to include the second part of the trim I have mentioned? I have tried to include the following block in the same pattern (Z000.)$ but of course it failed.
Thanks.
Any help would be appreciated.

There is no need to use regular expression in order to achieve that. You can simply use:
let value = '02T13:18:43.000Z';
let newValue = value.slice(3, -5);
console.log(newValue);
it will return 13:18:43, assumming that your string will always have the same pattern. According to the documentation slice method will substring from beginIndex to endIndex. endIndex is optional.

as I see you only need regex solution so does this pattern work?
(\d{2}:)+\d{2} or simply \d{2}:\d{2}:\d{2}
it searches much times for digit-digit-doubleDot combos and digit-digit-doubleDot at the end
the only disadvange is that it doesn't check whether say there are no minutes>59 and etc.
The main reason why I didn't include checking just because I kept in mind that you get your dates from sources where data that are stored are already valid, ex. database.

Solution
This should suffice to remove both the prefix from beginning to T and postfix from . to end:
/^.*T|\..*$/g
console.log(new Date().toISOString().replace(/^.*T|\..*$/g, ''))
See the visualization on debuggex
Explanation
The section ^.*T removes all characters up to and including the last encountered T in the string.
The section \..*$ removes all characters from the first encountered . to the end of the string.
The | in between coupled with the global g flag allows the regular expression to match both sections in the string, allowing .replace(..., '') to trim both simultaneously.

Get all the WORDS except one specific word

I want to get all the words, except one, from a string using JS regex match function. For example, for a string testhello123worldtestWTF, excluding the word test, the result would be helloworldWTF.
I realize that I have to do it using look-ahead functions, but I can't figiure out how exactly. I came up with the following regex (?!test)[a-zA-Z]+(?=.*test), however, it work only partially.
http://refiddle.com/refiddles/59511c2075622d324c090000

IMHO, I would try to replace the incriminated word with an empty string, no?

Lookarounds seem to be an overkill for it, you can just replace the test with nothing:
var str = 'testhello123worldtestWTF';
var res = str.replace(/test/g, '');

Plugging this into your refiddle produces the results you're looking for:
/(test)/g
It matches all occurrences of the word "test" without picking up unwanted words/letters. You can set this to whatever variable you need to hold these.

WORDS OF CAUTION
Seeing that you have no set delimiters in your inputted string, I must say that you cannot reliably exclude a specific word - to a certain extent.
For example, if you want to exclude test, this might create a problem if the input was protester or rotatestreet. You don't have clear demarcations of what a word is, thus leading you to exclude test when you might not have meant to.
On the other hand, if you just want to ignore the string test regardless, just replace test with an empty string and you are good to go.

Dot at start and end using regex

Hello I am trying to validate an input using regex in Javascript, what my requirement is that I can have at most one dot ('.') in the string and that can't be at the start and end.
I got a solution in
/^[^\.].*[^\.]$/;
But the issue is input "x" is considered as invalid
valid inputs are like
"x", "x.x", "xx.x" , "x.xx" like so
invalid like ".x" and "x."

How about
/^(?!\.)[^\.]*\.?[^\.]*(?!\.).$/

The correct regex for
my requirement is that I can have at most one dot ('.') in the string
and that can't be at the start and end
is
/^([^\.]|([^\.]*.?[^\.]))$/
/^([^\.]|([^\.].*[^\.]))$/ or /^[^\.].*[^\.]$/ accepts String
containing more than 1 dot . Hence it will also accept X..X too.
Please check working snippet also
validateString("XX.X");
validateString("X.X");
validateString("X...X");
validateString("X");
validateString("X.X.X");
validateString(".XX");
validateString("XX.");
function validateString(str){
console.log(/^([^\.]|([^\.]*.?[^\.]))$/.test(str));
}

With your current regex, you are targeting a string that should be at least 2 characters long, as both [^\.] parts are a mandatory character.
Your regex should include an extra check in case there is just one character, which you can do like this:
^([^\.]|([^\.]+\.?[^\.]+))$

How to get the valid part of a regex match

I want to test if a user string is "ok so far", in that it might not be valid as a whole but it is a subset of a valid one.
I have a regex say ^[0-9]{4}-[0-9]{4}-[0-9]{4}-[0-9]$
such that "1234-1234-5678-5678" is valid
"1234-12" or even "1" does not match pattern but its a valid subset of a valid format, in other words the input is ok so far.
is there a neat way of doing this without making many many regexes, its friday.

Not sure if I understood well your problem, but I think you want to have something like this:
^([0-9]{4}-){1,3}[0-9]{1,4}$
Working demo
This will match set of 4 digits and can have the last set from 1 to 4 digits
You can also shorten your regex with:
^(\d{4}-){1,3}\d{1,4}$

You could possibly use one final regex for validation of the form you currently have, and a on the fly regex for the user input being valid for each subset.
My idea would be to have ([0-9]{1,4}-)+

For your case this will check as one types:
/^(\d(\d(\d(\d(-(\d(\d(\d(\d(-(\d(\d(\d(\d(-(\d)?)?)?)?)?)?)?)?)?)?)?)?)?)?)?)?$/

This regex will match key for key as you type, although it is a little cumbersome.
^([0-9]{1,4}|[0-9]{4}-[0-9]{0,4}|[0-9]{4}-[0-9]{4}-[0-9]{0,4}|[0-9]{4}-[0-9]{4}-[0-9]{4}-[0-9]{0,4})$
Here is a live example

JavaScript: How to get specific text in a string using RegEx

I tummbled into this RegEx and I googled it. A lot. But unfortunately didn't quite understand how RegEx works...
So to make this quick since only a tiny winny part of my work requires it so I will be needing you guys. again :))
So here it goes...
All I want is to retrieve a specific string with a format of 0000x0000. For example:
Input:NameName975x945NameName
Output:
975x945
Must also consider string like this:
NameNameName9751x9451NameNameName
(the integer and string are longer...)

Use regex in String.prototype.match() to get specific part of string.
str.match(/\d+x\d+/)[0]
var str = "NameName975x945NameName";
var match = str.match(/\d+x\d+/)[0];
console.log(match)

We need a bit more detail, but I'll go in order:
Assuming there can be any number of digits before and after the x, and these can be of different lengths:
[\d]+x[\d]+
Assuming the number of digits before the x needs to be equal to the number of digits after the x (as in your example) and this number is finite (and small enough so that your regex isn't obscenely long):
[\d]{1}x[\d]{1}|[\d]{2}x[\d]{2}|[\d]{3}x[\d]{3} (and so on)
Check out this related answer for more details on handling this as the length of the number gets longer.
Then you can use String.prototype.match() with your regex to grab the matches within your string.