Develop Reference

regex for ignoring character if inside () parenthesis?

I was doing some regex, but I get this bug:
I have this string for example "+1/(1/10)+(1/30)+1/50" and I used this regex /\+.[^\+]*/g
and it working fine since it gives me ['+1/(1/10)', '+(1/30)', '+1/50']
BUT the real problem is when the + is inside the parenthesis ()
like this: "+1/(1+10)+(1/30)+1/50"
because it will give ['+1/(1', '+10)', '+(1/30)', '+1/50']
which isn't what I want :(... the thing I want is ['+1/(1+10)', '+(1/30)', '+1/50']
so the regex if it see \(.*\) skip it like it wasn't there...
how to ignore in regex?
my code (js):
const tests = {
correct: "1/(1/10)+(1/30)+1/50",
wrong : "1/(1+10)+(1/30)+1/50"
}
function getAdditionArray(string) {
const REGEX = /\+.[^\+]*/g; // change this to ignore the () even if they have the + sign
const firstChar = string[0];
if (firstChar !== "-") string = "+" + string;
return string.match(REGEX);
}
console.log(
getAdditionArray(test.correct),
getAdditionArray(test.wrong),
)

You can exclude matching parenthesis, and then optionally match (...)
\+[^+()]*(?:\([^()]*\))?
The pattern matches:
\+ Match a +
[^+()]* Match optional chars other than + ( )
(?: Non capture group to match as a whole part
\([^()]*\) Match from (...)
)? Close the non capture group and make it optional
See a regex101 demo.
Another option could be to be more specific about the digits and the + and / and use a character class to list the allowed characters.
\+(?:\d+[+/])?(?:\(\d+[/+]\d+\)|\d+)
See another regex101 demo.

How can I replace all duplicated paths of a url with a JS Regex

For the following URL:
https://www.google.es/test/test/hello/world
I want to replace all the occurrences of "/test/", and its important that it "test" starts with and ends with a "/".
I tried with:
let url = "https://www.google.es/test/test/hello/world"
url.replace(/\/test\//g, "/");
But it returns:
'https://www.google.es/test/hello/world'
It doesn't replace the second "/test/"
Any clues on how I could do this with a regex?
I basically want to replace the content that lives inside the dashes, but not the dashes themselves.

Something like this would work:
/(\/[^\/]+)(?:\1)+\//g
( - open capture group #1
\/ - find a literal slash
[^\/]+ - capture at least one non-slash char
) - close capture group #1
(?:\1)+ - repeat capture group #1 one or more times
\/ - ensure a closing slash
/g - global modifier
https://regex101.com/r/NgJA3X/1
var regex = /(\/[^\/]+)(?:\1)+\//g;
var str = `https://www.google.es/test/test/hello/world
https://www.google.es/test/test/test/test/hello/world
https://www.google.es/test/test/hello/test/hello/hello/world
https://www.google.es/test/hello/world/world
https://www.google.es/test/hello/helloworld/world`;
var subst = ``;
// The substituted value will be contained in the result variable
var result = str.replace(regex, subst);
console.log(result);

You can do this with a regular expression, but it sounds like your intent is to replace only individual parts of the pathname component of a URL.
A URL has other components (such as the fragment identifier) which could contain the pattern that you describe, and handling that distinction with a regular expression is more challenging.
The URL class is designed to help solve problems just like this, and you can replace just the path parts using a functional technique like this:
function replacePathParts (url, targetStr, replaceStr = '') {
url = new URL(url);
const updatedPathname = url.pathname
.split('/')
.map(str => str === targetStr ? replaceStr : str)
.filter(Boolean)
.join('/');
url.pathname = updatedPathname;
return url.href;
}
const inputUrl = 'https://www.google.es/test/test/hello/world';
const result1 = replacePathParts(inputUrl, 'test');
console.log(result1); // "https://www.google.es/hello/world"
const result2 = replacePathParts(inputUrl, 'test', 'message');
console.log(result2); // "https://www.google.es/message/message/hello/world"

Based on conversation in comment, you can use this solution:
let url = "https://www.google.es/test/test/hello/world"
console.log( url.replace(/(?:\/test)+(?=\/)/g, "/WORKS") );
//=> https://www.google.es/WORKS/hello/world
RegEx Breakdown:
(?:\/test)+: Match 1+ repeats of text /test
(?=\/): Lookahead to assert that we have a / ahead

Regex to extract city and city code from a dynamic URL in React

I am storing the URL in location variale ,and which will be dynamic for ex.
1- https://abc.go.com/United_States,US/
2- https://abc.go.com/US/
3- https://abc.go.com/Uganda,UG/
4- https://abc.go.com/United_States,US
5- https://abc.go.com/
the URL's are totally dynamic ,
I was able to extract the state Code with following regex :-
const cityCode =location.split(/(?=[A-Z][A-Z])/)[1].split('/')[0];
//this will get two consecutive Uppercase characters and remove "/" in the end if present
Is there any possible regex to extract the country and State Code in a single variable if present in the url.

You can use a capture group with an optional part for the first part and the comma
https?:\/\/\S+?\/((?:\w+,)?[A-Z][A-Z])
https?:\/\/ Match the protocol with optional s
\S+?\/ Match 1+ non whitespace chars as least as possible
( Capture group 1 (Which is accessed by m[1] in the example code)
(?:\w+,)? Optionally match 1+ word characters and a comma
[A-Z][A-Z] Match 2 uppercase chars
) Close group 1
Regex demo
const pattern = /https?:\/\/\S+?\/((?:\w+,)?[A-Z][A-Z])/;
[
"https://abc.go.com/United_States,US/",
"https://abc.go.com/US/",
"https://abc.go.com/Uganda,UG/",
"https://abc.go.com/United_States,US",
"https://abc.go.com/"
].forEach(s => {
const m = s.match(pattern);
console.log(m ? m[1] : "no match");
});
Or if it is the part after the first forward slash
https?:\/\/[^\s\/]+\/((?:\w+,)?[A-Z][A-Z])
Regex demo

Reference: String.prototype.match()
urls = ['https://abc.go.com/United_States,US/',
'https://abc.go.com/US/',
'https://abc.go.com/Uganda,UG/',
'https://abc.go.com/United_States,US',
'https://abc.go.com/'
];
urls.forEach(url => {
console.log(url.match(/.*:\/\/.*?\/(.*)/m)[1].replace(/\/$/, ''))
});

var pathName = location.pathname;
var countryCode = (pathName.split('/')).length!= 0 ? pathName.split('/')[1].split(',') : [];
console.log(countryCode);
// if the Window URL is - "https://test.org/United_States,US" then result is - ["United_States", "US"]
// if the Window URL is - "https://test.org/United_States" then result is - ["United_States"]

Manipulate a string containing a file's path to get only the file name

I got a file path as
falsefile:///var/mobile/Containers/Data/Application/D4B6F6CD-5E5C-4459-90CC-0C649B3B31B8/Documents/ExponentExperienceData/%2540hherax%252Fiia-mas-app-new//IIAMASATTCHMENTS/BD6FE729-70F1-48B0-83EB-8E7D956E599E.MOV
as file extension will change as file type
file path will also change
how could I manipulate string to get file name as
BD6FE729-70F1-48B0-83EB-8E7D956E599E"
is in given example
2nd example of path and file type change
falsefile:///var/mobile/Containers/Data/Application/D4B6F6CD-5E5C-4459-90CC-0C649B3B31B8/Documents/ExponentExperienceData/%2540ppphrx%252Fiia-mas-app-new//IIAMASATTCHMENTS/DD6FE729-60F2-58B0-8M8B-8E759R6E547K.jpeg

you can do simply
let str="falsefile:///var/mobile/Containers/Data/Application/D4B6F6CD-5E5C-4459-90CC-0C649B3B31B8/Documents/ExponentExperienceData/%2540hherax%252Fiia-mas-app-new//IIAMASATTCHMENTS/BD6FE729-70F1-48B0-83EB-8E7D956E599E.MOV"
console.log( str.split(".")[0].split("/").pop()
)
just remember split split pop

Some variation of slice/split would work
const str = 'falsefile:///var/mobile/Containers/Data/Application/D4B6F6CD-5E5C-4459-90CC-0C649B3B31B8/Documents/ExponentExperienceData/%2540hherax%252Fiia-mas-app-new//IIAMASATTCHMENTS/BD6FE729-70F1-48B0-83EB-8E7D956E599E.MOV'
console.log(
str.slice(str.lastIndexOf("/")+1).split(".")[0]
)
// or
console.log(
str.split("/").pop().split(".")[0]
)

You can use regular expression for example.
The first thing comes in my mind is:
const filepath = 'falsefile:///var/mobile/Containers/Data/Application/D4B6F6CD-5E5C-4459-90CC-0C649B3B31B8/Documents/ExponentExperienceData/%2540hherax%252Fiia-mas-app-new//IIAMASATTCHMENTS/BD6FE729-70F1-48B0-83EB-8E7D956E599E.MOV'
const filenameWithoutExtension = filepath.match(/IIAMASATTCHMENTS\/(.*)\./)[1] // "BD6FE729-70F1-48B0-83EB-8E7D956E599E"
console.log(filenameWithoutExtension)

If you know the format of the value you want to capture, you might get a more exact match using a regex and capture your value in the first capturing group.
You might use the /i flag to make the match case insensitive.
([A-Z0-9]+(?:-[A-Z0-9]+){4})\.\w+$
That will match:
( Capturing group
[A-Z0-9]+ Match 1+ times what is listed in the character class
(?:-[A-Z0-9]+){4} Repeat 4 times matching a hyphen and 1+ times what is listed in the character class
) Close capturing group
\.\w+$ Match a dot, 1+ times a word char and assert the end of the string
Regex demo
let strs = [
`falsefile:///var/mobile/Containers/Data/Application/D4B6F6CD-5E5C-4459-90CC-0C649B3B31B8/Documents/ExponentExperienceData/%2540hherax%252Fiia-mas-app-new//IIAMASATTCHMENTS/BD6FE729-70F1-48B0-83EB-8E7D956E599E.MOV`,
`falsefile:///var/mobile/Containers/Data/Application/D4B6F6CD-5E5C-4459-90CC-0C649B3B31B8/Documents/ExponentExperienceData/%2540ppphrx%252Fiia-mas-app-new//IIAMASATTCHMENTS/DD6FE729-60F2-58B0-8M8B-8E759R6E547K.jpeg`
];
let pattern = /([A-Z0-9]+(?:-[A-Z0-9]+){4})\.\w+$/i;
strs.forEach(str => console.log(str.match(pattern)[1]));

You could use regular expressions like here:
function get_filename(str) {
const regex = /\/([A-Z0-9\-_]+)\.[\w\d]+/gm;
let m = regex.exec(str);
return m[1];
}
console.log(
get_filename(`falsefile:///var/mobile/Containers/Data/Application/D4B6F6CD-5E5C-4459-90CC-0C649B3B31B8/Documents/ExponentExperienceData/%2540ppphrx%252Fiia-mas-app-new//IIAMASATTCHMENTS/DD6FE729-60F2-58B0-8M8B-8E759R6E547K.jpeg`)
)

var filpath = "falsefile:///var/mobile/Containers/Data/Application/D4B6F6CD-5E5C-4459-90CC-0C649B3B31B8/Documents/ExponentExperienceData/%2540hherax%252Fiia-mas-app-new//IIAMASATTCHMENTS/BD6FE729-70F1-48B0-83EB-8E7D956E599E.MOV"
console.log(
filpath.substring(filpath.lastIndexOf('/') + 1, filpath.length).substring(1, filpath.substring(filpath.lastIndexOf('/') + 1, filpath.length).lastIndexOf('.'))
)

var str = "falsefile:///var/mobile/Containers/Data/Application/D4B6F6CD-5E5C-4459-90CC-0C649B3B31B8/Documents/ExponentExperienceData/%2540hherax%252Fiia-mas-app-new//IIAMASATTCHMENTS/BD6FE729-70F1-48B0-83EB-8E7D956E599E.MOV",
re = /[\w|-]*\.\w*/
stringNameWithExt = str.match(re)
stringNameWithoutExt = str.match(re)[0].split(".")[0]
console.log(stringNameWithoutExt)

regex to extract numbers starting from second symbol

Sorry for one more to the tons of regexp questions but I can't find anything similar to my needs. I want to output the string which can contain number or letter 'A' as the first symbol and numbers only on other positions. Input is any string, for example:
---INPUT--- -OUTPUT-
A123asdf456 -> A123456
0qw#$56-398 -> 056398
B12376B6f90 -> 12376690
12A12345BCt -> 1212345
What I tried is replace(/[^A\d]/g, '') (I use JS), which almost does the job except the case when there's A in the middle of the string. I tried to use ^ anchor but then the pattern doesn't match other numbers in the string. Not sure what is easier - extract matching characters or remove unmatching.

I think you can do it like this using a negative lookahead and then replace with an empty string.
In an non capturing group (?:, use a negative lookahad (?! to assert that what follows is not the beginning of the string followed by ^A or a digit \d. If that is the case, match any character .
(?:(?!^A|\d).)+
var pattern = /(?:(?!^A|\d).)+/g;
var strings = [
"A123asdf456",
"0qw#$56-398",
"B12376B6f90",
"12A12345BCt"
];
for (var i = 0; i < strings.length; i++) {
console.log(strings[i] + " ==> " + strings[i].replace(pattern, ""));
}

You can match and capture desired and undesired characters within two different sides of an alternation, then replace those undesired with nothing:
^(A)|\D
JS code:
var inputStrings = [
"A-123asdf456",
"A123asdf456",
"0qw#$56-398",
"B12376B6f90",
"12A12345BCt"
];
console.log(
inputStrings.map(v => v.replace(/^(A)|\D/g, "$1"))
);

You can use the following regex : /(^A)?\d+/g
var arr = ['A123asdf456','0qw#$56-398','B12376B6f90','12A12345BCt', 'A-123asdf456'],
result = arr.map(s => s.match(/(^A|\d)/g).join(''));
console.log(result);