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regular expression replacement in JavaScript with some part remaining intact

I need to parse a string that comes like this:
-38419-indices-foo-7119-attributes-10073-bar
Where there are numbers followed by one or more words all joined by dashes. I need to get this:
[
0 => '38419-indices-foo',
1 => '7119-attributes',
2 => '10073-bar',
]
I had thought of attempting to replace only the dash before a number with a : and then using .split(':') - how would I do this? I don't want to replace the other dashes.

Imo, the pattern is straight-forward:
\d+\D+
To even get rid of the trailing -, you could go for
(\d+\D+)(?:-|$)
Or
\d+(?:(?!-\d|$).)+
You can see it here:
var myString = "-38419-indices-foo-7119-attributes-10073-bar";
var myRegexp = /(\d+\D+)(?:-|$)/g;
var result = [];
match = myRegexp.exec(myString);
while (match != null) {
// matched text: match[0]
// match start: match.index
// capturing group n: match[n]
result.push(match[1]);
match = myRegexp.exec(myString);
}
console.log(result);
// alternative 2
let alternative_results = myString.match(/\d+(?:(?!-\d|$).)+/g);
console.log(alternative_results);
Or a demo on regex101.com.

Logic
lazy matching using quantifier .*?
Regex
.*?((\d+)\D*)(?!-)
https://regex101.com/r/WeTzF0/1
Test string
-38419-indices-foo-7119-attributes-10073-bar-333333-dfdfdfdf-dfdfdfdf-dfdfdfdfdfdf-123232323-dfsdfsfsdfdf
Matches
Further steps
You need to split from the matches and insert into your desired array.

I need help getting the first n characters of a string up to when a number character starts

I'm working with a string where I need to extract the first n characters up to where numbers begin. What would be the best way to do this as sometimes the string starts with a number: 7EUSA8889er898 I would need to extract 7EUSA But other string examples would be SWFX74849948, I would need to extract SWFX from that string.
Not sure how to do this with regex my limited knowledge is blocking me at this point:
^(\w{4}) that just gets me the first four characters but I don't really have a stopping point as sometimes the string could be somelongstring292894830982 which would require me to get somelongstring

Using \w will match a word character which includes characters and digits and an underscore.
You could match an optional digit [0-9]? from the start of the string ^and then match 1+ times A-Za-z
^[0-9]?[A-Za-z]+
Regex demo
const regex = /^[0-9]?[A-Za-z]+/;
[
"7EUSA8889er898",
"somelongstring292894830982",
"SWFX74849948"
].forEach(s => console.log(s.match(regex)[0]));

Can use this regex code:
(^\d+?[a-zA-Z]+)|(^\d+|[a-zA-Z]+)
I try with exmaple and good worked:
1- somelongstring292894830982 -> somelongstring
2- 7sdfsdf5456 -> 7sdfsdf
3- 875werwer54556 -> 875werwer

If you want to create function where the RegExp is parametrized by n parameter, this would be
function getStr(str,n) {
var pattern = "\\d?\\w{0,"+n+"}";
var reg = new RegExp(pattern);
var result = reg.exec(str);
if(result[0]) return result[0].substr(0,n);
}

There are answers to this but here is another way to do it.
var string1 = '7EUSA8889er898';
var string2 = 'SWFX74849948';
var Extract = function (args) {
var C = args.split(''); // Split string in array
var NI = []; // Store indexes of all numbers
// Loop through list -> if char is a number add its index
C.map(function (I) { return /^\d+$/.test(I) === true ? NI.push(C.indexOf(I)) : ''; });
// Get the items between the first and second occurence of a number
return C.slice(NI[0] === 0 ? NI[0] + 1 : 0, NI[1]).join('');
};
console.log(Extract(string1));
console.log(Extract(string2));
Output
EUSA
SWFX7

Since it's hard to tell what you are trying to match, I'd go with a general regex
^\d?\D+(?=\d)

How to find all appearances of text after a specific word?

I get a string like:
str = “Test/hello/filename/12345678/first
Hddhkhd
Hdhal
filename/1212abcd/second”
I want to get an array of the all strings that comes after “filename//“ and I know that after the “/“ there is an 8 letter word that I want to get.
In this case, I want to get an array that will be:
strArr = [“12345678”, “1212abcd”]
How do I solve this problem?

A regex that captures the 8 characters that immediately follow a literal "filename//":
/filename\/\/(.{8})/

Try use this regex first:
filename\/\w{8}
and after it, slice from the result by this regex:
\w{8}$
First you will get:
filename/12345678
filename/1212abcd
Second you will get :
12345678
1212abcd

You might also capture in a group matching 8 times not a forward slash or a newline after matching /filename
\bfilename\/([^\/\n]{8})
Regex demo
If you want to match 8 or more times you could use {8,} instead or if you want to match 1 or more times you could use a +.
If you don't want to match whitespace characters you could change the \n to \s
const regex = /filename\/([^\/\n]{8})/g;
const str = `Test/hello/filename/12345678/first
Hddhkhd
Hdhal
filename/1212abcd/second`;
let m;
while ((m = regex.exec(str)) !== null) {
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
console.log(m[1]);
}

You can use the following code. It will match all characters after the filename/ until it encounters another /. After you get the matches in an array you can map it out and replace all the filename/ with '':
let a = /filename\/[^\/]+/g;
let b = 'Test/hello/filename/12345678/first Hddhkhd Hdhal filename/1212abcd/second';
let c = b.match(a).map(x=>x.replace('filename/',''));
console.log(c);

For explanation check this REGEX
var arr = "Test/hello/filename/12345678/first Hddhkhd Hdhal filename/1212abcd/second".match(/(?<=filename\/)(.*?)(?=\/)/g);
console.log(arr)
OR
For unsupported Lookbehinds browser use Array#map after regex
var arr = "Test/hello/filename/12345678/first Hddhkhd Hdhal filename/1212abcd/second".match(/filename\/(.*?)\//g).map(i=> i.split('/')[1]);
console.log(arr)

Excluding matcher [duplicate]

After coming to the shocking realization that regular expressions in JavaScript are somewhat different from the ones in PCE, I am stuck with the following.
In php I extract a number after x:
(?x)[0-9]+
In JavaScript the same regex doesn't work, due to invalid group resulting from the capturing parenthesis difference.
So I am trying to achieve the same trivial functionality, but I keep getting both the x and the number:
(?:x)([0-9]+)
How do I capture the number after x without including x?

This works too:
/(?:x)([0-9]+)/.test('YOUR_STRING');
Then, the value you want is:
RegExp.$1 // group 1

You can try the following regex: (?!x)[0-9]+
fiddle here: https://jsfiddle.net/xy6x938e/1/
This is assuming that you are now looking for an x followed by a number, it uses a capture group to capture just the numbers section.
var myString = "x12345";
var myRegexp = /x([0-9]+)/g;
var match = myRegexp.exec(myString);
var myString2 = "z12345";
var match2 = myRegexp.exec(myString2);
if(match != null && match.length > 1){
alert('match1:' + match[1]);
}
else{
alert('no match 1');
}
if(match2 != null && match2.length > 1){
alert('match2:' + match2[1]);
}
else{
alert('no match 2');
}

(\d+) try this!
i have tested on this tool with x12345
http://www.regular-expressions.info/javascriptexample.html

How do I capture the number after x without including x?
In fact, you just want to extract a sequence of digits after a fixed string/known pattern.
Your PCRE (PHP) regex, (?x)[0-9]+, is wrong becaue (?x) is an inline version of a PCRE_EXTENDED VERBOSE/COMMENTS flag (see "Pattern Modifiers"). It does not do anything meaningful in this case, (?x)[0-9]+ is equal to [0-9]+ or \d+.
You can use
console.log("x15 x25".match(/(?<=x)\d+/g));
You can also use a capturing group and then extract Group 1 value after a match is obtained:
const match = /x(\d+)/.exec("x15");
if (match) {
console.log(match[1]); // Getting the first match
}
// All matches
const matches = Array.from("x15,x25".matchAll(/x(\d+)/g), x=>x[1]);
console.log(matches);

You still can use exclusive pattern (?!...)
So, for your example it will be /(?!x)[0-9]+/. Give a try to the following:
/(?!x)\d+/.exec('x123')
// => ["123"]

JavaScript regex get number after string

After coming to the shocking realization that regular expressions in JavaScript are somewhat different from the ones in PCE, I am stuck with the following.
In php I extract a number after x:
(?x)[0-9]+
In JavaScript the same regex doesn't work, due to invalid group resulting from the capturing parenthesis difference.
So I am trying to achieve the same trivial functionality, but I keep getting both the x and the number:
(?:x)([0-9]+)
How do I capture the number after x without including x?

This works too:
/(?:x)([0-9]+)/.test('YOUR_STRING');
Then, the value you want is:
RegExp.$1 // group 1

You can try the following regex: (?!x)[0-9]+
fiddle here: https://jsfiddle.net/xy6x938e/1/
This is assuming that you are now looking for an x followed by a number, it uses a capture group to capture just the numbers section.
var myString = "x12345";
var myRegexp = /x([0-9]+)/g;
var match = myRegexp.exec(myString);
var myString2 = "z12345";
var match2 = myRegexp.exec(myString2);
if(match != null && match.length > 1){
alert('match1:' + match[1]);
}
else{
alert('no match 1');
}
if(match2 != null && match2.length > 1){
alert('match2:' + match2[1]);
}
else{
alert('no match 2');
}

(\d+) try this!
i have tested on this tool with x12345
http://www.regular-expressions.info/javascriptexample.html

How do I capture the number after x without including x?
In fact, you just want to extract a sequence of digits after a fixed string/known pattern.
Your PCRE (PHP) regex, (?x)[0-9]+, is wrong becaue (?x) is an inline version of a PCRE_EXTENDED VERBOSE/COMMENTS flag (see "Pattern Modifiers"). It does not do anything meaningful in this case, (?x)[0-9]+ is equal to [0-9]+ or \d+.
You can use
console.log("x15 x25".match(/(?<=x)\d+/g));
You can also use a capturing group and then extract Group 1 value after a match is obtained:
const match = /x(\d+)/.exec("x15");
if (match) {
console.log(match[1]); // Getting the first match
}
// All matches
const matches = Array.from("x15,x25".matchAll(/x(\d+)/g), x=>x[1]);
console.log(matches);

You still can use exclusive pattern (?!...)
So, for your example it will be /(?!x)[0-9]+/. Give a try to the following:
/(?!x)\d+/.exec('x123')
// => ["123"]