We Keep Coding

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 for is empty space

I'm suresomeone of you know how to do this easily, I am trying to validate my inputs that only allows numbers, characters and/or an empty space which mean nothing in a string "", so I have /^[0-9a-zA-Z]+$/ only missing that empty space in there. Seems I can't find the express for it, so I tried to create a variabel var emptyStr = "" and put it in there but don't really know how to write the correct syntax.

Use * rather than +. * is 0 or more times. + is 1 or more times.
/^[0-9a-zA-Z]*$/

/^[A-z0-9\s]*$/
The \s will also capture spaces

Fastest way to remove hyphens from a string

I have IDs that look like: 185-51-671 but they can also have letters at the end, 175-1-7b
All I want to do is remove the hyphens, as a pre-processing step. Show me some cool ways to do this in javascript? I figure there are probably quite a few questions like this one, but I'm interested to see what optimizations people will come up with for "just hyphens"
Thanks!
edit: I am using jQuery, so I guess .replace(a,b) does the trick (replacing a with b)
numberNoHyphens = number.replace("-","");
any other alternatives?
edit #2:
So, just in case anyone is wondering, the correct answer was
numberNoHyphens = number.replace(/-/g,"");
and you need the "g" which is the pattern switch or "global flag" because
numberNoHyphens = number.replace(/-/,"");
will only match and replace the first hyphen

You need to include the global flag:
var str="185-51-671";
var newStr = str.replace(/-/g, "");

This is not faster, but
str.split('-').join('');
should also work.
I set up a jsperf test if anyone wants to add and compare their methods, but it's unlikely anything will be faster than the replace method.
http://jsperf.com/remove-hyphens-from-string

var str='185-51-671';
str=str.replace(/-/g,'');

Gets much easier in String.prototype.replaceAll(). Check out the browser support for the built-in method.
const str = '185-51-671';
console.log(str.replaceAll('-', ''));

Som of these answers, prior to edits, did not remove all of the hyphens. You would need to use .replaceAll("-","")

In tidyverse, there are multiple functions that could suit your needs. Specifically, I would use str_remove, which will replace in a string, the giver character by an empty string (""), effectively removing it (check here the documentation). Example of its usage:
str_remove(x, '-')

Can someone tell me the purpose of the second capture group in the jQuery rts regular expression?

In Jeff Roberson's jQuery Regular Expressions Review he proposes changing the rts regular expression in jQuery's ajax.js from /(\?|&)_=.*?(&|$)/ to /([?&])_=[^&\r\n]*(&?)/. In both versions, what is the purpose of the second capture group? The code does a replacement of the current random timestamp with a new random timestamp:
var ts = jQuery.now();
// try replacing _= if it is there
var ret = s.url.replace(rts, "$1_=" + ts + "$2");
Doesn't it only replace what it matches? I am thinking this does the same:
var ret = s.url.replace(/([?&])_=[^&\r\n]*/, "$1_=" + ts);
Can someone explain the purpose of the second capture group?

It's to pick up the next delimiter in the query string on the URL, so that it still works properly as a query string. Thus if the url is
http://foo.bar/what/ever?blah=blah&_=12345&zebra=banana
then the second group picks up the "&" before "zebra".
That's an awesome blog post by the way and everybody should read it.
edit — now that I think about it, I'm not sure why it's necessary to bother with replacing that second delimiter. In the "fixed" expression, that greedy * will pick up the whole parameter value and stop at the delimiter (or the end of the string) anyway.

I think you're right. It was needed in the original because matching the ampersand or end-of-string was how the .*? knew when to stop. In Jeff's version that's no longer necessary.

As the author of the article I can't tell you the reason for the second capture group. My intent with the article was to take existing regexes and simply make them more efficient - i.e. they should all match the same text - just do it faster. Unfortunately I did not have time to delve deeply into the code to see exactly how each and every one of them was being used. I assumed that the capture group for this one was there for a reason so I did not mess with it.

Processing Javascript RegEx submatches

I am trying to write some JavaScript RegEx to replace user inputed tags with real html tags, so [b] will become <b> and so forth. the RegEx I am using looks like so
var exptags = /\[(b|u|i|s|center|code){1}]((.){1,}?)\[\/(\1){1}]/ig;
with the following JavaScript
s.replace(exptags,"<$1>$2</$1>");
this works fine for single nested tags, for example:
[b]hello[/b] [u]world[/u]
but if the tags are nested inside each other it will only match the outer tags, for example
[b]foo [u]to the[/u] bar[/b]
this will only match the b tags. how can I fix this? should i just loop until the starting string is the same as the outcome? I have a feeling that the ((.){1,}?) patten is wrong also?
Thanks

The easiest solution would be to to replace all the tags, whether they are closed or not and let .innerHTML work out if they are matched or not it will much more resilient that way..
var tagreg = /\[(\/?)(b|u|i|s|center|code)]/ig
div.innerHTML="[b][i]helloworld[/b]".replace(tagreg, "<$1$2>") //no closing i
//div.inerHTML=="<b><i>helloworld</i></b>"

AFAIK you can't express recursion with regular expressions.
You can however do that with .NET's System.Text.RegularExpressions using balanced matching. See more here: http://blogs.msdn.com/bclteam/archive/2005/03/15/396452.aspx
If you're using .NET you can probably implement what you need with a callback.
If not, you may have to roll your own little javascript parser.
Then again, if you can afford to hit the server you can use the full parser. :)
What do you need this for, anyway? If it is for anything other than a preview I highly recommend doing the processing server-side.

You could just repeatedly apply the regexp until it no longer matches. That would do odd things like "[b][b]foo[/b][/b]" => "<b>[b]foo</b>[/b]" => "<b><b>foo</b></b>", but as far as I can see the end result will still be a sensible string with matching (though not necessarily properly nested) tags.
Or if you want to do it 'right', just write a simple recursive descent parser. Though people might expect "[b]foo[u]bar[/b]baz[/u]" to work, which is tricky to recognise with a parser.

The reason the nested block doesn't get replaced is because the match, for [b], places the position after [/b]. Thus, everything that ((.){1,}?) matches is then ignored.
It is possible to write a recursive parser in server-side -- Perl uses qr// and Ruby probably has something similar.
Though, you don't necessarily need true recursive. You can use a relatively simple loop to handle the string equivalently:
var s = '[b]hello[/b] [u]world[/u] [b]foo [u]to the[/u] bar[/b]';
var exptags = /\[(b|u|i|s|center|code){1}]((.){1,}?)\[\/(\1){1}]/ig;
while (s.match(exptags)) {
s = s.replace(exptags, "<$1>$2</$1>");
}
document.writeln('<div>' + s + '</div>'); // after
In this case, it'll make 2 passes:
0: [b]hello[/b] [u]world[/u] [b]foo [u]to the[/u] bar[/b]
1: <b>hello</b> <u>world</u> <b>foo [u]to the[/u] bar</b>
2: <b>hello</b> <u>world</u> <b>foo <u>to the</u> bar</b>
Also, a few suggestions for cleaning up the RegEx:
var exptags = /\[(b|u|i|s|center|code)\](.+?)\[\/(\1)\]/ig;
{1} is assumed when no other count specifiers exist
{1,} can be shortened to +

Agree with Richard Szalay, but his regex didn't get quoted right:
var exptags = /\[(b|u|i|s|center|code)](.*)\[\/\1]/ig;
is cleaner. Note that I also change .+? to .*. There are two problems with .+?:
you won't match [u][/u], since there isn't at least one character between them (+)
a non-greedy match won't deal as nicely with the same tag nested inside itself (?)

Yes, you will have to loop. Alternatively since your tags looks so much like HTML ones you could replace [b] for <b> and [/b] for </b> separately. (.){1,}? is the same as (.*?) - that is, any symbols, least possible sequence length.
Updated: Thanks to MrP, (.){1,}? is (.)+?, my bad.

How about:
tagreg=/\[(.?)?(b|u|i|s|center|code)\]/gi;
"[b][i]helloworld[/i][/b]".replace(tagreg, "<$1$2>");
"[b]helloworld[/b]".replace(tagreg, "<$1$2>");
For me the above produces:
<b><i>helloworld</i></b>
<b>helloworld</b>
This appears to do what you want, and has the advantage of needing only a single pass.
Disclaimer: I don't code often in JS, so if I made any mistakes please feel free to point them out :-)

You are right about the inner pattern being troublesome.
((.){1,}?)
That is doing a captured match at least once and then the whole thing is captured. Every character inside your tag will be captured as a group.
You are also capturing your closing element name when you don't need it and are using {1} when that is implied. Below is a cleanup up version:
/\[(b|u|i|s|center|code)](.+?)\[\/\1]/ig
Not sure about the other problem.