Develop Reference

JavaScript - more concise way to grab string between two characters in URL?

Is there a more concise or more standard way to grab a string between the last slash and query question mark of a URL than this?
const recordId= window.location.href.split("item/")[1].split("?")[0]
In this case I'm using item/ because my URLs are always:
mysite.com/item/recordIdIwantToGrab?foo=bar&life=42

A regular expression can do the trick - match a /, followed by word characters, up until a ?.
const str = 'mysite.com/item/recordIdIwantToGrab?foo=bar&life=42';
const result = str.match(/\/(\w+)\?/)[1];
console.log(result);
\/ - match a literal /
(\w+) - capturing group, match word characters
\ - match a literal ?
[1] - extract the value matched by the capturing group

We can achieve with URL class in javascript.
let url = 'http://example.com/item/recordIdIwantToGrab?foo=bar&life=42';
url = new URL(url);
let result = url.pathname.split('/').at('-1');
console.log(result);

You can achieve the result using lastIndexOf and indexOf
const str = `mysite.com/item/recordIdIwantToGrab?foo=bar&life=42`;
const result = str.slice(str.lastIndexOf("/") + 1, str.indexOf("?"));
console.log(result);

URL regular expression pattern

I would like to parse URLs with Regular Expressions and find a pattern that matches with https://*.global.
Here is my URL test string on regex101.
Ideally, the regex would return https://app8.global instead of cover other https string.
const URL = `https://temp/"https://app8.global"https://utility.localhost/`;
const regex = /https:\/\/(.+?)\.global(\/|'|"|`)/gm;
const found = URL.match(regex);
console.log(found);
How would I manipulate the regex so it will return the https://*.global?

First of all, you need to exclude slashes from the starting part, otherwise it'll match things from the previous url:
const regex = /https:\/\/([^\/]+?)\.global(\/|'|"|`)/gm;
Now, you can convert the weird 4 character or with a character group:
const regex = /https:\/\/([^\/]+?)\.global[\/'"`]/gm;
And now you can get the matches and trim off that last character:
const matches = URL.match(regex).map(v => v.slice(0, -1));
Then, matches would evaluate to ["https://app8.global"].

Using Group RegExp.$1
const URL = `https://temp/"https://app8.global"https://utility.localhost/`;
const regex = /(https:\/\/([^\/]+?)\.global[\/'"`])/;
const found = URL.match(regex);
console.log(RegExp.$1);

Replace after char '-' or '/' match

I'm trying to execute regex replace after match char, example 3674802/3 or 637884-ORG
The id can become one of them, in that case, how can I use regex replace to match to remove after the match?
Input var id = 3674802/3 or 637884-ORG;
Expected Output 3674802 or 637884

You could use sbustring method to take part of string only till '/' OR '-':
var input = "3674802/3";
var output = input.substr(0, input.indexOf('/'));
var input = "637884-ORG";
var output = input.substr(0, input.indexOf('-'));
var input = "3674802/3";
if (input.indexOf('/') > -1)
{
input = input.substr(0, input.indexOf('/'));
}
console.log(input);
var input = "637884-ORG";
if (input.indexOf('-') > -1)
{
input = input.substr(0, input.indexOf('-'));
}
console.log(input);

You can use a regex with a lookahead assertion
/(\d+)(?=[/-])/g
var id = "3674802/3"
console.log((id.match(/(\d+)(?=[/-])/g) || []).pop())
id = "637884-ORG"
console.log((id.match(/(\d+)(?=[/-])/g) || []).pop())

You don't need Regex for this. Regex is far more powerful than what you need.
You get away with the String's substring and indexOf methods.
indexOf takes in a character/substring and returns an integer. The integer represents what character position the character/substring starts at.
substring takes in a starting position and ending position, and returns the new string from the start to the end.
If are having trouble getting these to work; then, feel free to ask for more clarification.

You can use the following script:
var str = '3674802/3 or 637884-ORG';
var id = str.replace(/(\d+)[-\/](?:\d+|[A-Z]+)/g, '$1');
Details concerning the regex:
(\d+) - A seuence of digits, the 1st capturing group.
[-\/] - Either a minus or a slash. Because / are regex delimiters,
it must be escaped with a backslash.
(?: - Start of a non-capturing group, a "container" for alternatives.
\d+ - First alternative - a sequence of digits.
| - Alternative separator.
[A-Z]+ - Second alternative - a sequence of letters.
) - End of the non-capturing group.
g - global option.
The expression to replace with: $1 - replace the whole finding with
the first capturing group.

Thanks To everyone who responded to my question, was really helpful to resolve my issue.
Here is My answer that I built:
var str = ['8484683*ORG','7488575/2','647658-ORG'];
for(i=0;i<str.length;i++){
var regRep = /((\/\/[^\/]+)?\/.*)|(\-.*)|(\*.*)/;
var txt = str[i].replace(regRep,"");
console.log(txt);
}

Regex check both side of match but not include in match string

I want get match with checking both side expropriation of main match.
var str = 1234 word !!! 5678 another *** 000more))) get word and another
console.log(str.match(/(?!\d+\s?)\w+(?=\s?\W+)/g))
>> (3) ["word", "another", "more"]
it check both side but not include in the main match sets.
But in html it not working [not working]
var str = ''; get url, url2 and url3
console.log(str.match(/(?!href=")[^"]+?(?=")/g))
>> (6) ["<a href=", "url3"]
I try to Negative lookarounds using (?!href=") and Positive lookarounds using (?=") to match only the value of its attribute but it return more attributes.
Is there any way to so like this here, Thanks

What you could do for your example data is capture what is between double quotes href="([^"]+) in an captured group and loop through the result:
var str = '';
var pattern = /href="([^"]+)/g;
var match = pattern.exec(str);
while (match != null) {
console.log(match[1]);
match = pattern.exec(str);
}

In other flavors of regex you could have used e.g. positive lookbehind
((?<=href="), but unfortunately Javascript regex does not support
lookbehinds.
A reasonable solution is:
Match href=" as "ordinary" content, to be ignored.
Match the attribute value as a capturing group ((\w+)),
to be "consumed".
Set the boundary of the above group with a *positive lookup"
((?=")), just as you did.
So the whole regex can be:
href="(\w+)(?=")
and read "your" value from group 1.

You can't parse HTML with regex. Because HTML can't be parsed by regex.
Have you tried using the DOM parser that's right at your fingertips?
var str = '';
var div = document.createElement('div');
div.innerHTML = str; // parsing magic!
var links = Array.from(div.getElementsByTagName("a"));
var urls = links.map(function(a) {return a.href;});
// above returns fully-resolved absolute URLs.
// for the literal attribute value, try a.getAttribute("href")
console.log(urls);

How can I remove all characters up to and including the 3rd slash in a string?

I'm having trouble with removing all characters up to and including the 3 third slash in JavaScript. This is my string:
http://blablab/test
The result should be:
test
Does anybody know the correct solution?

To get the last item in a path, you can split the string on / and then pop():
var url = "http://blablab/test";
alert(url.split("/").pop());
//-> "test"
To specify an individual part of a path, split on / and use bracket notation to access the item:
var url = "http://blablab/test/page.php";
alert(url.split("/")[3]);
//-> "test"
Or, if you want everything after the third slash, split(), slice() and join():
var url = "http://blablab/test/page.php";
alert(url.split("/").slice(3).join("/"));
//-> "test/page.php"

var string = 'http://blablab/test'
string = string.replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'')
alert(string)
This is a regular expression. I will explain below
The regex is /[\s\S]*\//
/ is the start of the regex
Where [\s\S] means whitespace or non whitespace (anything), not to be confused with . which does not match line breaks (. is the same as [^\r\n]).
* means that we match anywhere from zero to unlimited number of [\s\S]
\/ Means match a slash character
The last / is the end of the regex

var str = "http://blablab/test";
var index = 0;
for(var i = 0; i < 3; i++){
index = str.indexOf("/",index)+1;
}
str = str.substr(index);
To make it a one liner you could make the following:
str = str.substr(str.indexOf("/",str.indexOf("/",str.indexOf("/")+1)+1)+1);

You can use split to split the string in parts and use slice to return all parts after the third slice.
var str = "http://blablab/test",
arr = str.split("/");
arr = arr.slice(3);
console.log(arr.join("/")); // "test"
// A longer string:
var str = "http://blablab/test/test"; // "test/test";

You could use a regular expression like this one:
'http://blablab/test'.match(/^(?:[^/]*\/){3}(.*)$/);
// -> ['http://blablab/test', 'test]
A string’s match method gives you either an array (of the whole match, in this case the whole input, and of any capture groups (and we want the first capture group)), or null. So, for general use you need to pull out the 1th element of the array, or null if a match wasn’t found:
var input = 'http://blablab/test',
re = /^(?:[^/]*\/){3}(.*)$/,
match = input.match(re),
result = match && match[1]; // With this input, result contains "test"

let str = "http://blablab/test";
let data = new URL(str).pathname.split("/").pop();
console.log(data);