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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.

javascript regex replace multiline strings [duplicate]

var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre.*?<\/pre>/gm );
alert(arr); // null
I'd want the PRE block be picked up, even though it spans over newline characters. I thought the 'm' flag does it. Does not.
Found the answer here before posting. SInce I thought I knew JavaScript (read three books, worked hours) and there wasn't an existing solution at SO, I'll dare to post anyways. throw stones here
So the solution is:
var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre[\s\S]*?<\/pre>/gm );
alert(arr); // <pre>...</pre> :)
Does anyone have a less cryptic way?
Edit: this is a duplicate but since it's harder to find than mine, I don't remove.
It proposes [^] as a "multiline dot". What I still don't understand is why [.\n] does not work. Guess this is one of the sad parts of JavaScript..

DON'T use (.|[\r\n]) instead of . for multiline matching.
DO use [\s\S] instead of . for multiline matching
Also, avoid greediness where not needed by using *? or +? quantifier instead of * or +. This can have a huge performance impact.
See the benchmark I have made: https://jsben.ch/R4Hxu
Using [^]: fastest
Using [\s\S]: 0.83% slower
Using (.|\r|\n): 96% slower
Using (.|[\r\n]): 96% slower
NB: You can also use [^] but it is deprecated in the below comment.

[.\n] does not work because . has no special meaning inside of [], it just means a literal .. (.|\n) would be a way to specify "any character, including a newline". If you want to match all newlines, you would need to add \r as well to include Windows and classic Mac OS style line endings: (.|[\r\n]).
That turns out to be somewhat cumbersome, as well as slow, (see KrisWebDev's answer for details), so a better approach would be to match all whitespace characters and all non-whitespace characters, with [\s\S], which will match everything, and is faster and simpler.
In general, you shouldn't try to use a regexp to match the actual HTML tags. See, for instance, these questions for more information on why.
Instead, try actually searching the DOM for the tag you need (using jQuery makes this easier, but you can always do document.getElementsByTagName("pre") with the standard DOM), and then search the text content of those results with a regexp if you need to match against the contents.

You do not specify your environment and version of JavaScript (ECMAScript), and I realise this post was from 2009, but just for completeness:
With the release of ECMA2018 we can now use the s flag to cause . to match \n (see https://stackoverflow.com/a/36006948/141801).
Thus:
let s = 'I am a string\nover several\nlines.';
console.log('String: "' + s + '".');
let r = /string.*several.*lines/s; // Note 's' modifier
console.log('Match? ' + r.test(s)); // 'test' returns true
This is a recent addition and will not work in many current environments, for example Node v8.7.0 does not seem to recognise it, but it works in Chromium, and I'm using it in a Typescript test I'm writing and presumably it will become more mainstream as time goes by.

Now there's the s (single line) modifier, that lets the dot matches new lines as well :)
\s will also match new lines :D
Just add the s behind the slash
/<pre>.*?<\/pre>/gms

[.\n] doesn't work, because dot in [] (by regex definition; not javascript only) means the dot-character. You can use (.|\n) (or (.|[\n\r])) instead.

I have tested it (Chrome) and it's working for me (both [^] and [^\0]), by changing the dot (.) with either [^\0] or [^] , because dot doesn't match line break (See here: http://www.regular-expressions.info/dot.html).
var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre[^\0]*?<\/pre>/gm );
alert(arr); //Working

In addition to above-said examples, it is an alternate.
^[\\w\\s]*$
Where \w is for words and \s is for white spaces

[\\w\\s]*
This one was beyond helpful for me, especially for matching multiple things that include new lines, every single other answer ended up just grouping all of the matches together.

Greedy Regex for varying number of new line [duplicate]

var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre.*?<\/pre>/gm );
alert(arr); // null
I'd want the PRE block be picked up, even though it spans over newline characters. I thought the 'm' flag does it. Does not.
Found the answer here before posting. SInce I thought I knew JavaScript (read three books, worked hours) and there wasn't an existing solution at SO, I'll dare to post anyways. throw stones here
So the solution is:
var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre[\s\S]*?<\/pre>/gm );
alert(arr); // <pre>...</pre> :)
Does anyone have a less cryptic way?
Edit: this is a duplicate but since it's harder to find than mine, I don't remove.
It proposes [^] as a "multiline dot". What I still don't understand is why [.\n] does not work. Guess this is one of the sad parts of JavaScript..

DON'T use (.|[\r\n]) instead of . for multiline matching.
DO use [\s\S] instead of . for multiline matching
Also, avoid greediness where not needed by using *? or +? quantifier instead of * or +. This can have a huge performance impact.
See the benchmark I have made: https://jsben.ch/R4Hxu
Using [^]: fastest
Using [\s\S]: 0.83% slower
Using (.|\r|\n): 96% slower
Using (.|[\r\n]): 96% slower
NB: You can also use [^] but it is deprecated in the below comment.

[.\n] does not work because . has no special meaning inside of [], it just means a literal .. (.|\n) would be a way to specify "any character, including a newline". If you want to match all newlines, you would need to add \r as well to include Windows and classic Mac OS style line endings: (.|[\r\n]).
That turns out to be somewhat cumbersome, as well as slow, (see KrisWebDev's answer for details), so a better approach would be to match all whitespace characters and all non-whitespace characters, with [\s\S], which will match everything, and is faster and simpler.
In general, you shouldn't try to use a regexp to match the actual HTML tags. See, for instance, these questions for more information on why.
Instead, try actually searching the DOM for the tag you need (using jQuery makes this easier, but you can always do document.getElementsByTagName("pre") with the standard DOM), and then search the text content of those results with a regexp if you need to match against the contents.

You do not specify your environment and version of JavaScript (ECMAScript), and I realise this post was from 2009, but just for completeness:
With the release of ECMA2018 we can now use the s flag to cause . to match \n (see https://stackoverflow.com/a/36006948/141801).
Thus:
let s = 'I am a string\nover several\nlines.';
console.log('String: "' + s + '".');
let r = /string.*several.*lines/s; // Note 's' modifier
console.log('Match? ' + r.test(s)); // 'test' returns true
This is a recent addition and will not work in many current environments, for example Node v8.7.0 does not seem to recognise it, but it works in Chromium, and I'm using it in a Typescript test I'm writing and presumably it will become more mainstream as time goes by.

Now there's the s (single line) modifier, that lets the dot matches new lines as well :)
\s will also match new lines :D
Just add the s behind the slash
/<pre>.*?<\/pre>/gms

[.\n] doesn't work, because dot in [] (by regex definition; not javascript only) means the dot-character. You can use (.|\n) (or (.|[\n\r])) instead.

I have tested it (Chrome) and it's working for me (both [^] and [^\0]), by changing the dot (.) with either [^\0] or [^] , because dot doesn't match line break (See here: http://www.regular-expressions.info/dot.html).
var ss= "<pre>aaaa\nbbb\nccc</pre>ddd";
var arr= ss.match( /<pre[^\0]*?<\/pre>/gm );
alert(arr); //Working

In addition to above-said examples, it is an alternate.
^[\\w\\s]*$
Where \w is for words and \s is for white spaces

[\\w\\s]*
This one was beyond helpful for me, especially for matching multiple things that include new lines, every single other answer ended up just grouping all of the matches together.

javascript regexp to match path depth

Been struggling for the last hour to try and get this regexp to work but cannot seem to crack it.
It must be a regexp and I cannot use split etc as it is part of a bigger regexp that searches for numerous other strings using .test().
(public\/css.*[!\/]?)
public/css/somefile.css
public/css/somepath/somefile.css
public/css/somepath/anotherpath/somefile.css
Here I am trying to look for path starting with public/css followed by any character except for another forward slash.
so "public/css/somefile.css" should match but the other 2 should not.
A better solution may be to somehow specify the number of levels to match after the prefix using something like
(public\/css\/{1,2}.*)
but I can't seem to figure that out either, some help with this would be appreciated.
edit
No idea why this question has been marked down twice, I have clearly stated the requirement with sample code and test cases and also attempted to solve the issue, why is it being marked down ?

You can use this regex:
/^(public\/css\/[^\/]*?)$/gm
^ : Starts with
[^/] : Not /
*?: Any Characters
$: Ends with
g: Global Flag
m: Multi-line Flag

Something like this?
/public\/css\/[^\/]+$/
This will match
public/css/[Any characters except for /]$
$ is matching the end of the string in regex.

Transform mixed-case word into (multiple) Capitalized words?

How do I "normalise" caps & spaces?
Example: coreControllerC4a automatically turns into Core Controller C4a when function is called?

Here is a starting example
"coreControllerC4a".split(/(?=[A-Z])/)
which results in
["core", "Controller", "C4a"]
Of course, there are many counter-examples, and this doesn't cover Unicode, but it does show using a regular expression to split a string. (Note the use of a look-ahead to avoid consuming any data at the split.) Now it's just a matter of transforming ("map"ing) each element to the correct case.
Happy coding.