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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.

regular expression javascript or jquery

I have the following types of query string. It could be anyone of the below.
index.html?cr=33&m=prod&p=ded
index.html?cr=33&m=prod&p=ded&c=ddl&h=33&mj=ori
From the above query string, i wanted to extract only m & p "m=prod&p=ded".
Otherway to do is split and get, that i have already done it.
But I wanted to achieve it using regular expression.
Any help is highly appreciated.

You can use something like this :
(?<=&)m=[^& ]*\&p=[^& ]*
In javascript , It doesn't support ?<= positive Lookbehind.
So you can use the following code ( in javascript ):
var s="index.html?cr=33&m=prod&p=ded"
s.match(/m=[^& ]*\&p=[^& ]*/g)
Refer Regexp101

If both query parameters come one after another then you you can use:
url.match(/[?&]m=([^&]*)&p=([^&]*)/i);

Regular expression for "-" in javascript

I want to know how can I get a regular expression for matching number, ., and - only.
I am using this:
/^[0-9\.'S]+$/
by this it working fine but not working for symbol "-".

You simply haven't used the literal dash - (or minus) in the regex. Try:
/^[0-9\.-]+$/
But if you want a proper number, you might want to use a more proper regex:
/^-?[0-9]+(?:\.[0-9]+)?$/
The first regex can accept things such as 3987----.... while the second will not accept it, but will accept things like -87.983274.

That's because - is not part of your character class. You are only using - in the class range (which only includes digits). Also, I don't know what the S and the ' are doing there:
/^[0-9.-]+$/
Also, I can promise you that after taking the time to read through this tutorial regular expressions will seem a lot less confusing to you.

Try the below regex.
/^-?[0-9\.]+$/

What's wrong with this regular expression to find URLs?

I'm working on a JavaScript to extract a URL from a Google search URL, like so:
http://www.google.com/search?client=safari&rls=en&q=thisisthepartiwanttofind.org&ie=UTF-8&oe=UTF-8
Right now, my code looks like this:
var checkForURL = /[\w\d](.org)/i;
var findTheURL = checkForURL.exec(theURL);
I've ran this through a couple regex testers and it seems to work, but in practice the string I get returned looks like this:
thisisthepartiwanttofind.org,.org
So where's that trailing ,.org coming from?
I know my pattern isn't super robust but please don't suggest better patterns to use. I'd really just like advice on what in particular I did wrong with this one. Thanks!

Remove the parentheses in the regex if you do not process the .org (unlikely since it is a literal). As per #Mark comment, add a + to match one or more characters of the class [\w\d]. Also, I would escape the dot:
var checkForURL = /[\w\d]+\.org/i;

What you're actually getting is an array of 2 results, the first being the whole match, the second - the group you defined by using parens (.org).
Compare with:
/([\w\d]+)\.org/.exec('thisistheurl.org')
→ ["thisistheurl.org", "thisistheurl"]
/[\w\d]+\.org/.exec('thisistheurl.org')
→ ["thisistheurl.org"]
/([\w\d]+)(\.org)/.exec('thisistheurl.org')
→ ["thisistheurl.org", "thisistheurl", ".org"]
The result of an .exec of a JS regex is an Array of strings, the first being the whole match and the subsequent representing groups that you defined by using parens. If there are no parens in the regex, there will only be one element in this array - the whole match.

You should escape .(DOT) in (.org) regex group or it matches any character. So your regex would become:
/[\w\d]+(\.org)/
To match the url in your example you can use something like this:
https?://([0-9a-zA-Z_.?=&\-]+/?)+
or something more accurate like this (you should choose the right regex according to your needs):
^https?://([0-9a-zA-Z_\-]+\.)+(com|org|net|WhatEverYouWant)(/[0-9a-zA-Z_\-?=&.]+)$

Javascript Regular Expressions Lookbehind Failing

I am hoping that this will have a pretty quick and simple answer. I am using regular-expressions.info to help me get the right regular expression to turn URL-encoded, ISO-8859-1 pound sign ("%A3"), into a URL-encoded UTF-8 pound sign ("%C2%A3").
In other words I just want to swap %A3 with %C2%A3, when the %A3 is not already prefixed with %C2.
So I would have thought the following would work:
Regular Expression: (?!(\%C2))\%A3
Replace With: %C2%A3
But it doesn't and I can't figure out why!
I assume my syntax is just slightly wrong, but I can't figure it out! Any ideas?
FYI - I know that the following will work (and have used this as a workaround in the meantime), but really want to understand why the former doesn't work.
Regular Expression: ([^\%C2])\%A3
Replace With: $1%C2%A3
TIA!

Why not just replace ((%C2)?%A3) with %C2%A3, making the prefix an optional part of the match? It means that you're "replacing" text with itself even when it's already right, but I don't foresee a performance issue.

Unfortunately, the (?!) syntax is negative lookahead. To the best of my knowledge, JavaScript does not support negative lookbehind.
What you could do is go forward with the replacement anyway, and end up with %C2%C2%A3 strings, but these could easily be converted in a second pass to the desired %C2%A3.

You could replace
(^.?.?|(?!%C2)...)%A3
with
$1%C2%A3

I would suggest you use the functional form of Javascript String.replace (see the section "Specifying a function as a parameter"). This lets you put arbitrary logic, including state if necessary, into a regexp-matching session. For your case, I'd use a simpler regexp that matches a superset of what you want, then in the function call you can test whether it meets your exact criteria, and if it doesn't then just return the matched string as is.
The only problem with this approach is that if you have overlapping potential matches, you have the possibility of missing the second match, since there's no way to return a value to tell the replace() method that it isn't really a match after all.