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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.

regex replace all backward slashes before '\?'

I have a string such as
'frontend\less\defaults\layout.css?file=\foo'
I want a regex that replaces it with
'frontend/less/defaults/layout.css?file=\foo'
I tried /\\/g, but it keeps matching stuff after a \?, which I want to avoid somehow

Following will work, use a lookahead in your regexp:
var myString="path\\to\\file.php?query=\\something"
var r=(/\?/g.test(myString))?/(\\)(?=.+[\?])/ig:/\\/ig;
.replace(r,"/")

You can do this with String.replace, with a replacement function:
str.replace(/^([^?]*)/, function (_, $1) {
return $1.replace(/\\/g, '/');
});
This will work regardless of whether the query string exists or not.
Explanation
/^([^?]*)/
([^?]*) will match and capture everything before ? (if any).
I assume the URL is valid, so there is no validation done here.
(Thanks to #Pumbaa80 for the suggestion. There is no need to match the query string part if it is going to stay the same after the replacement)

Unless you know the number of \'s in advance, I doubt you can do this with a comprehensible regex. I would:
split the string in two parts: the part before the ?, and after it
use your regex on the first part
put the two strings back together.

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.

Regular expression to remove a file's extension

I am in need of a regular expression that can remove the extension of a filename, returning only the name of the file.
Here are some examples of inputs and outputs:
myfile.png -> myfile
myfile.png.jpg -> myfile.png
I can obviously do this manually (ie removing everything from the last dot) but I'm sure that there is a regular expression that can do this by itself.
Just for the record, I am doing this in JavaScript

Just for completeness: How could this be achieved without Regular Expressions?
var input = 'myfile.png';
var output = input.substr(0, input.lastIndexOf('.')) || input;
The || input takes care of the case, where lastIndexOf() provides a -1. You see, it's still a one-liner.

/(.*)\.[^.]+$/
Result will be in that first capture group. However, it's probably more efficient to just find the position of the rightmost period and then take everything before it, without using regex.

The regular expression to match the pattern is:
/\.[^.]*$/
It finds a period character (\.), followed by 0 or more characters that are not periods ([^.]*), followed by the end of the string ($).
console.log(
"aaa.bbb.ccc".replace(/\.[^.]*$/,'')
)

/^(.+)(\.[^ .]+)?$/
Test cases where this works and others fail:
".htaccess" (leading period)
"file" (no file extension)
"send to mrs." (no extension, but ends in abbr.)
"version 1.2 of project" (no extension, yet still contains a period)
The common thread above is, of course, "malformed" file extensions. But you always have to think about those corner cases. :P
Test cases where this fails:
"version 1.2" (no file extension, but "appears" to have one)
"name.tar.gz" (if you view this as a "compound extension" and wanted it split into "name" and ".tar.gz")
How to handle these is problematic and best decided on a project-specific basis.

/^(.+)(\.[^ .]+)?$/
Above pattern is wrong - it will always include the extension too. It's because of how the javascript regex engine works. The (\.[^ .]+) token is optional so the engine will successfully match the entire string with (.+)
http://cl.ly/image/3G1I3h3M2Q0M
Here's my tested regexp solution.
The pattern will match filenameNoExt with/without extension in the path, respecting both slash and backslash separators
var path = "c:\some.path/subfolder/file.ext"
var m = path.match(/([^:\\/]*?)(?:\.([^ :\\/.]*))?$/)
var fileName = (m === null)? "" : m[0]
var fileExt = (m === null)? "" : m[1]
dissection of the above pattern:
([^:\\/]*?) // match any character, except slashes and colon, 0-or-more times,
// make the token non-greedy so that the regex engine
// will try to match the next token (the file extension)
// capture the file name token to subpattern \1
(?:\. // match the '.' but don't capture it
([^ :\\/.]*) // match file extension
// ensure that the last element of the path is matched by prohibiting slashes
// capture the file extension token to subpattern \2
)?$ // the whole file extension is optional
http://cl.ly/image/3t3N413g3K09
http://www.gethifi.com/tools/regex
This will cover all cases that was mentioned by #RogerPate but including full paths too

another no-regex way of doing it (the "oposite" of #Rahul's version, not using pop() to remove)
It doesn't require to refer to the variable twice, so it's easier to inline
filename.split('.').slice(0,-1).join()

This will do it as well :)
'myfile.png.jpg'.split('.').reverse().slice(1).reverse().join('.');
I'd stick to the regexp though... =P

return filename.split('.').pop();
it will make your wish come true. But not regular expression way.

In javascript you can call the Replace() method that will replace based on a regular expression.
This regular expression will match everything from the begining of the line to the end and remove anything after the last period including the period.
/^(.*)\..*$/
The how of implementing the replace can be found in this Stackoverflow question.
Javascript regex question

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)