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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.

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_\-?=&.]+)$

Negating in /,?(([1-9]-[1-9])|([1-9]))/g

I am trying to match a string containing a mix of digits and hyphenated digits, like a crossword answer specification, for example 1,2-2 or 1-1,3,4,2-2
/,?(([1-9]-[1-9])|([1-9]))/g is what I've come up to match the string
value = value.replace(/,?(([1-9]-[1-9])|([1-9]))/g, '');
replaces ok, and I've checked it out in an online tester.
What I really need is to negate this, so I can use it on a keyup event, examine the contents of a textarea and remove characters that don't fit, so it only allows through characters as in the example.
I've tried ^ where expected, but this it's not doing what I expect, how should I negate the regex so I remove everything that doesn't match?
If there is a better way of doing this I'm open to suggestions too.

var value = 'hello,1,2,3,4-6,1-1,3,test,4,2-2';
var pattern = /,?(([1-9]-[1-9])|([1-9]))/g;
value.replace(pattern, ''); // "hello,test"
You can use String#match. With /g flag, it returns an array of all the matches, then you can use Array#join to join them.
The problem is that String#match returns null when there is no match, so you have to handle that case and use an empty array so that it can join:
(value.match(pattern) || []).join(''); // ",1,2,3,4-6,1-1,3,4,2-2"
Note: It may better to check them on onblur rather than onkeyup. Messing with the text that the user is currently typing will make it annoying. Better to wait for the user to finish typing.

Didn't test it in JS, but this should return the valid string beginning from the left and as long as valid values are encountered (note that I used \d - if you'd like 1-9 only, then use your brackets).
(?:\d(?:-\d)?,)*\d(?:-\d)?
E.g. matching this regular expression with the string "0-1,1,2,3,4-4,2,,1,3--4" will return "0-1,1,2,3,4-4,2" as the first match.

JS/Jquery, Match not finding the PNG = match('/gif|jpg|jpeg|png/')

I have the following code which I use to match fancybox possible elements:
$('a.grouped_elements').each(function(){
var elem = $(this);
// Convert everything to lower case to match smart
if(elem.attr('href').toLowerCase().match('/gif|jpg|jpeg|png/') != null) {
elem.fancybox();
}
});
It works great with JPGs but it isn't matching PNGs for some reason. Anyone see a bug with the code?
Thanks

A couple of things.
Match accepts an object of RegExp, not a string. It may work in some browsers, but is definitely not standard.
"gif".match('/gif|png|jpg/'); // null​​​​​​​​​​​​​​​​​​​​​​​​​​​​
Without the strings
"gif".match(/gif|png|jpg/); // ["gif"]
Also, you would want to check these at the end of a filename, instead of anywhere in the string.
"isthisagif.nope".match(/(gif|png|jpg|jpeg)/); // ["gif", "gif"]
Only searching at the end of string with $ suffix
"isthisagif.nope".match(/(gif|png|jpg|jpeg)$/); // null
No need to make href lowercase, just do a case insensitive search /i.
Look for a dot before the image extension as an additional check.
And some tests. I don't know how you got any results back with using a string argument to .match. What browser are you on?

I guess the fact that it'll match anywhere in the string (it would match "http://www.giftshop.com/" for instance) could be considered a bug. I'd use
/\.(gif|jpe?g|png)$/i

You are passing a string to the match() function rather than a regular expression. In JavaScript, strings are delimited with single quotes, and regular expressions are delimited with forward slashes. If you use both, you have a string, not a regex.

This worked perfectly for me: /.+\.(gif|png|jpe?g)$/i
.+ -> any string
\. -> followed by a point.
(gif|png|jpe?g) -> and then followed by any of these extensions. jpeg may or may not have the letter e.
$ -> now the end of the string it's expected
/i -> case insensitive mode: matches both sflkj.JPG and lkjfsl.jpg

Regex to match part of a string

Regex fun again...
Take for example http://something.com/en/page
I want to test for an exact match on /en/ including the forward slashes, otherwise it could match 'en' from other parts of the string.
I'm sure this is easy, for someone other than me!
EDIT:
I'm using it for a string.match() in javascript

Well it really depends on what programming language will be executing the regex, but the actual regex is simply
/en/
For .Net the following code works properly:
string url = "http://something.com/en/page";
bool MatchFound = Regex.Match(url, "/en/").Success;
Here is the JavaScript version:
var url = 'http://something.com/en/page';
if (url.match(/\/en\//)) {
alert('match found');
}
else {
alert('no match');
}
DUH
Thank you to Welbog and Chris Ballance to making what should have been the most obvious point. This does not require Regular Expressions to solve. It simply is a contains statement. Regex should only be used where it is needed and that should have been my first consideration and not the last.

If you're trying to match /en/ specifically, you don't need a regular expression at all. Just use your language's equivalent of contains to test for that substring.
If you're trying to match any two-letter part of the URL between two slashes, you need an expression like this:
/../
If you want to capture the two-letter code, enclose the periods in parentheses:
/(..)/
Depending on your language, you may need to escape the slashes:
\/..\/
\/(..)\/
And if you want to make sure you match letters instead of any character (including numbers and symbols), you might want to use an expression like this instead:
/[a-z]{2}/
Which will be recognized by most regex variations.
Again, you can escape the slashes and add a capturing group this way:
\/([a-z]{2})\/
And if you don't need to escape them:
/([a-z]{2})/
This expression will match any string in the form /xy/ where x and y are letters. So it will match /en/, /fr/, /de/, etc.
In JavaScript, you'll need the escaped version: \/([a-z]{2})\/.

You may need to escape the forward-slashes...
/\/en\//

Any reason /en/ would not work?

/\/en\// or perhaps /http\w*:\/\/[^\/]*\/en\//

You don't need a regex for this:
location.pathname.substr(0, 4) === "/en/"
Of course, if you insist on using a regex, use this:
/^\/en\//.test(location.pathname)