Develop Reference

Regex to convert URL to string

REGEX ONLY
I exclusively need Javascript regex code to convert URLs like
https://hello.romeo-juliet.fr
https://hello.romeojuliet.co.uk
https://hello.romeo-jul-iet.fr
https://hello.romeo-juliet.com
into this string romeojuliet
Basically want to get the alphabetic domain name with removing all other characters and https://, com/co.uk/fr etc Top Level Domains
Would be helpful if done using JS replace.
I tried till here
let url="https://hello.romeo-juliet.fr";
const test=url.replace(/(^\w+:|^)\/\/(\w+.)/, '');
console.log(test);

A non regex solution:
Get the host of the URL (by parsing the string with the URL() constructor and getting its host property), split by a period and get the second item in the resulting array, then remove all occurences of -:
let url="https://hello.romeo-juliet.fr";
const test = new URL(url).host.split(".")[1].replaceAll("-", '');
console.log(test);

You can use it with no regex as the following:
let url="https://hello.romeo-juliet.fr";
url.substring(url.indexOf(".")+1, url.lastIndexOf("."));
// result: romeo-juliet
I hope this answers your question

Javascipt regex to get string between two characters except escaped without lookbehind

I am looking for a specific javascript regex without the new lookahead/lookbehind features of Javascript 2018 that allows me to select text between two asterisk signs but ignores escaped characters.
In the following example only the text "test" and the included escaped characters are supposed to be selected according the rules above:
\*jdjdjdfdf*test*dfsdf\*adfasdasdasd*test**test\**sd* (Selected: "test", "test", "test\*")
During my research I found this solution Regex, everything between two characters except escaped characters /(?<!\\)(%.*?(?<!\\)%)/ but it uses negative lookbehinds which is supported in javascript 2018 but I need to support IE11 as well, so this solution doesn't work for me.
Then i found another approach which is almost getting there for me here: Javascript: negative lookbehind equivalent?. I altered the answer of Kamil Szot to fit my needs: ((?!([\\])).|^)(\*.*?((?!([\\])).|^)\*) Unfortuantely it doesn't work when two asterisks ** are in a row.
I have already invested a lot of hours and can't seem to get it right, any help is appreciated!
An example with what i have so far is here: https://www.regexpal.com/?fam=117350
I need to use the regexp in a string.replace call (str.replace(regexp|substr, newSubStr|function); so that I can wrap the found strings with a span element of a specific class.

You can use this regular expression:
(?:\\.|[^*])*\*((?:\\.|[^*])*)\*
Your code should then only take the (only) capture group of each match.
Like this:
var str = "\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*";
var regex = /(?:\\.|[^*])*\*((?:\\.|[^*])*)\*/g
var match;
while (match = regex.exec(str)) {
console.log(match[1]);
}
If you need to replace the matches, for instance to wrap the matches in a span tag while also dropping the asterisks, then use two capture groups:
var str = "\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*";
var regex = /((?:\\.|[^*])*)\*((?:\\.|[^*])*)\*/g
var result = str.replace(regex, "$1<span>$2</span>");
console.log(result);
One thing to be careful with: when you use string literals in JavaScript tests, escape the backslash (with another backslash). If you don't do that, the string actually will not have a backslash! To really get the backslash in the in-memory string, you need to escape the backslash.

const testStr = `\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*`;
const m = testStr.match(/\*(\\.)*t(\\.)*e(\\.)*s(\\.)*t(\\.)*\*/g).map(m => m.substr(1, m.length-2));
console.log(m);
More generic code:
const prepareRegExp = (word, delimiter = '\\*') => {
const escaped = '(\\\\.)*';
return new RegExp([
delimiter,
escaped,
[...word].join(escaped),
escaped,
delimiter
].join``, 'g');
};
const testStr = `\\*jdjdjdfdf*test*dfsdf\\*adfasdasdasd*test**test\\**sd*`;
const m = testStr
.match(prepareRegExp('test'))
.map(m => m.substr(1, m.length-2));
console.log(m);
https://instacode.dev/#Y29uc3QgcHJlcGFyZVJlZ0V4cCA9ICh3b3JkLCBkZWxpbWl0ZXIgPSAnXFwqJykgPT4gewogIGNvbnN0IGVzY2FwZWQgPSAnKFxcXFwuKSonOwogIHJldHVybiBuZXcgUmVnRXhwKFsKICAgIGRlbGltaXRlciwKICAgIGVzY2FwZWQsCiAgICBbLi4ud29yZF0uam9pbihlc2NhcGVkKSwKICAgIGVzY2FwZWQsCiAgICBkZWxpbWl0ZXIKICBdLmpvaW5gYCwgJ2cnKTsKfTsKCmNvbnN0IHRlc3RTdHIgPSBgXFwqamRqZGpkZmRmKnRlc3QqZGZzZGZcXCphZGZhc2Rhc2Rhc2QqdGVzdCoqdGVzdFxcKipzZCpgOwpjb25zdCBtID0gdGVzdFN0cgoJLm1hdGNoKHByZXBhcmVSZWdFeHAoJ3Rlc3QnKSkKCS5tYXAobSA9PiBtLnN1YnN0cigxLCBtLmxlbmd0aC0yKSk7Cgpjb25zb2xlLmxvZyhtKTs=

Regex for filename at end of href attribute (replace filepath with filename)

I am trying to replace the filepath with just the filename using regex, and struggling.
I take a list of text like this:
xlink:href="file://C:\x\y & DRAWINGS\z - CONTROLLED\a \testsvg-A01.svg"
and i just want to output
xlink:href="testsvg-A01.svg"
I can get there with separate regex (javascript) with several bits of regex as such:
let inQuotes = hrefs[0].match(/"(.*?)"/gm);
inQuotes = inQuotes[0].match(/([^\\]+$)/gm);
inQuotes = inQuotes[0].replace(/"/g, "");
this will return just the filename, but i was wondering if there was a way to take this and replace the original text to the desired style.
EDIT:
i can get it for a single occurrance with this line of code:
let testHrefs = outText.match(/xlink:href="(.*?)"/gm)[0].match(/"(.*?)"/gm)[0].match(/([^\\]+$)/gm)[0].replace(/^/, 'xlink:href="');
but it looks awful and doesn't completely do what i want. Any advice?

You could use a regex to remove the text you don't want to keep (i.e. everything between the href=" and the filename:
let href = 'xlink:href="file://C:\\x\\y & DRAWINGS\\z - CONTROLLED\\a \\testsvg-A01.svg"';
console.log(href);
console.log(href.replace(/href=".*?([^\\\/]+)$/, 'href="$1'));
Note I've used [\\\/] to allow for Unix and Windows style filenames, if you only want Windows style filenames that can simply be [\\].

First, keep in mind that backslashes in strings need to be escaped, as the backslash itself is an escape character. You'll likely want to escape these as \\.
After doing this, you can make use of a positive look-behind on the \ with the regex (?<=\\)[^\\]*:
var xlink = "file://C:\\x\\y & DRAWINGS\\z - CONTROLLED\\a \\testsvg-A01.svg";
var regex = /(?<=\\)[^\\]*/gm;
console.log(xlink.match(regex));
This splits the string into each of the folders (which may be useful within itself), though if you exclusively want the filename, you can use xlink.match(regex)[4] (assuming the length is consistent), or xlink.match(regex)[xlink.match(regex).length - 1] if it isn't.
var xlink = "file://C:\\x\\y & DRAWINGS\\z - CONTROLLED\\a \\testsvg-A01.svg";
var regex = /(?<=\\)[^\\]*/gm;
console.log(xlink.match(regex)[xlink.match(regex).length - 1]);

How would I write a Regular Expression to capture the value between Last Slash and Query String?

Problem:
Extract image file name from CDN address similar to the following:
https://cdnstorage.api.com/v0/b/my-app.com/o/photo%2FB%_2.jpeg?alt=media&token=4e32-a1a2-c48e6c91a2ba
Two-stage Solution:
I am using two regular expressions to retrieve the file name:
var postLastSlashRegEx = /[^\/]+$/,
preQueryRegEx = /^([^?]+)/;
var fileFromURL = urlString.match(postLastSlashRegEx)[0].match(preQueryRegEx)[0];
// fileFromURL = "photo%2FB%_2.jpeg"
Question:
Is there a way I can combine both regular expressions?
I've tried using capture groups, but haven't been able to produce a working solution.

From my comment
You can use a lookahead to find the "?" and use [^/] to match any non-slash characters.
/[^/]+(?=\?)/
To remove the dependency on the URL needing a "?", you can make the lookahead match a question mark or the end of line indicator (represented by $), but make sure the first glob is non-greedy.
/[^/]+?(?=\?|$)/

You don't have to use regex, you can just use split and substr.
var str = "https://cdnstorage.api.com/v0/b/my-app.com/o/photo%2FB%_2.jpeg?alt=media&token=4e32-a1a2-c48e6c91a2ba".split("?")[0];
var fileName = temp.substr(temp.lastIndexOf('/')+1);
but if regex is important to you, then:
str.match(/[^?]*\/([^?]+)/)[1]

The code using the substring method would look like the following -
var fileFromURL = urlString.substring(urlString.lastIndexOf('/') + 1, urlString.lastIndexOf('?'))

What RegEx would clean up this set of inputs?

I'm trying to figure out a RegEx that would match the following:
.../string-with-no-spaces -> string-with-no-spaces
or
string-with-no-spaces:... -> string-with-no-spaces
or
.../string-with-no-spaces:... -> string-with-no-spaces
where ... can be anything in these example strings:
example.com:8080/string-with-no-spaces:latest
string-with-no-spaces:latest
example.com:8080/string-with-no-spaces
string-with-no-spaces
and a bonus would be
http://example.com:8080/string-with-no-spaces:latest
and all would match string-with-no-spaces.
Is it possible for a single RegEx to cover all those cases?
So far I've gotten as far as /\/.+(?=:)/ but that not only includes the slash, but only works for case 3. Any ideas?
Edit: Also I should mention that I'm using Node.js, so ideally the solution should pass all of these: https://jsfiddle.net/ys0znLef/

How about:
(?:.*/)?([^/:\s]+)(?::.*|$)

Consider the following solution using specific regex pattern and String.match function:
var re = /(?:[/]|^)([^/:.]+?)(?:[:][^/]|$)/,
// (?:[/]|^) - passive group, checks if the needed string is preceded by '/' or is at start of the text
// (?:[:][^/]|$) - passive group, checks if the needed string is followed by ':' or is at the end of the text
searchString = function(str){
var result = str.match(re);
return result[1];
};
console.log(searchString("example.com:8080/string-with-no-spaces"));
console.log(searchString("string-with-no-spaces:latest"));
console.log(searchString("string-with-no-spaces"));
console.log(searchString("http://example.com:8080/string-with-no-spaces:latest"));
The output for all the cases above will be string-with-no-spaces

Here's the expression I've got... just trying to tweak to use the slash but not include it.
Updated result works in JS
\S([a-zA-Z0-9.:/\-]+)\S
//works on regexr, regex storm, & regex101 - tested with a local html file to confirm JS matches strings
var re = /\S([a-zA-Z0-9.:/\-]+)\S/;