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JavaScript Regex finding all substrings that matches with specific starting and ending pattern

I want a Javascript regex or with any possible solution,
For a given string finds all the substrings that start with a particular string and end with a particular character. The returned set of subStrings can be an Array.
this string can also have nested within parenthesis.
var str = "myfunc(1,2) and myfunc(3,4) or (myfunc(5,6) and func(7,8))";
starting char = "myfunc" ending char = ")" . here ending character should be first matching closing paranthesis.
output: function with arguments.
[myfunc(1,2),
myfunc(3,4),
myfunc(5,6),
func(7,8)]
I have tried with this. but, its returning null always.
var str = "myfunc(1,2) and myfunc(3,4) or (myfunc(5,6) and func(7,8))";
var re = /\myfunc.*?\)/ig
var match;
while ((match = re.exec(str)) != null){
console.log(match);
}
Can you help here?

I tested your regex and it seems to work fine:
let input = "myfunc(1,2) and myfunc(3,4) or (myfunc(5,6) and func(7,8))"
let pattern = /myfunc.*?\)/ig
// there is no need to use \m since it does nothing, and NO you dont need it even if you use 'm' at the beginning.
console.log(input.match(pattern))
//[ "myfunc(1,2)", "myfunc(3,4)", "myfunc(5,6)" ]
If you use (?:my|)func\(.+?\) you will be able to catch 'func(7,8)' too.
(?:my|)
( start of group
?: non capturing group
my| matches either 'my' or null, this will match either myfunc or func
) end of group
Test the regex here: https://regex101.com/r/3ujbdA/1

js Remove a part from a parameter which doesnt fit a pattern

const regex = /[1-9a-zA-Z]{3}-[1-9a-zA-Z]{3}-[1-9a-zA-Z]{3}/gm;
let m;
while ((m = regex.exec(tweet.text)) !== null) {
let newClass = tweet.text.replace(/[^1-9a-zA-Z]{3}-[^1-9a-zA-Z]{3}-[^1-9a-zA-Z]{3}/g, '');
console.log(`Found match: ${newClass}`);
};
when tweet.text = "123.qwe.456 test" I still get the same output but I want to remove anything which doesnt fit the pattern
/[1-9a-zA-Z]{3}-[1-9a-zA-Z]{3}-[1-9a-zA-Z]{3}/
any ideas?

You can use capture groups to extract exactly what gets matched in your string and then replace your original variable with this value. Something like
const regex = /([1-9a-zA-Z]{3}-[1-9a-zA-Z]{3}-[1-9a-zA-Z]{3})/
let match = tweet.text.match(regex)
tweet.text = match[1]

Instead of replace, you can get the match instead
\b[1-9a-zA-Z]{3}([-.])[1-9a-zA-Z]{3}\1[1-9a-zA-Z]{3}\b
Explanation
\b A word boundary
[1-9a-zA-Z]{3} Match 3 times any of the listed (Note that 1-9 does not match a 0)
([-.]) Capture in group 1 either an - or .
[1-9a-zA-Z]{3} Match 3 times any of the listed
\1 Back reference to group 1, match the same as captured in group 1
[1-9a-zA-Z]{3} Match 3 times any of the listed
\b A word boundary
Regex demo

const regex = /[1-9a-zA-Z]{3}-[1-9a-zA-Z]{3}-[1-9a-zA-Z]{3}/gm;
let m;
while ((m = regex.exec(tweet.text)) !== null) {
console.log(`Found match: ${m[0]}`);
figured the solution

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)

Javascript Regex: Capture between two asterisks with multiple asterisks in comma delimited string

I am trying to capture all characters between multiple instances of asterisks, which are comma delimited in a string. Here's an example of the string:
checkboxID0*,*checkboxID1*,&checkboxID2&,*checkboxID3*,!checkboxID4!,checkboxID5*
The caveat is that the phrase must start and end with an asterisk. I have been able to come close by using the following regex, however, it won't discard any matches when the captured string is missing the starting asterisk(*):
let str = "checkboxID0*,*checkboxID1*,&checkboxID2&,*checkboxID3*,!checkboxID4!,checkboxID5*"
const regex = /[^\,\*]+(?=\*)/gi;
var a = str.match(regex)
console.log(a) // answer should exclude checkboxID0 and checkboxID5
The answer returns the following, however, "checkboxID0 and checkboxID5" should be excluded as it doesn't start with an asterisk.
[
"checkboxID0",
"checkboxID1",
"checkboxID3",
"checkboxID5"
]
Thanks, in advance!

You need to use asterisks on both ends of the pattern and capture all 1 or more chars other than commas and asterisks in between:
/\*([^,*]+)\*/g
See the regex demo
Pattern details
\* - an asterisk
([^,*]+) - Capturing group 1: one or more chars other than , and *
\* - an asterisk
JS demo:
var regex = /\*([^,*]+)\*/g;
var str = "checkboxID0*,*checkboxID1*,&checkboxID2&,*checkboxID3*,!checkboxID4!,checkboxID5*";
var m, res = [];
while (m = regex.exec(str)) {
res.push(m[1]);
}
console.log(res);

Regex to grab strings between square brackets

I have the following string: pass[1][2011-08-21][total_passes]
How would I extract the items between the square brackets into an array? I tried
match(/\[(.*?)\]/);
var s = 'pass[1][2011-08-21][total_passes]';
var result = s.match(/\[(.*?)\]/);
console.log(result);
but this only returns [1].
Not sure how to do this.. Thanks in advance.

You are almost there, you just need a global match (note the /g flag):
match(/\[(.*?)\]/g);
Example: http://jsfiddle.net/kobi/Rbdj4/
If you want something that only captures the group (from MDN):
var s = "pass[1][2011-08-21][total_passes]";
var matches = [];
var pattern = /\[(.*?)\]/g;
var match;
while ((match = pattern.exec(s)) != null)
{
matches.push(match[1]);
}
Example: http://jsfiddle.net/kobi/6a7XN/
Another option (which I usually prefer), is abusing the replace callback:
var matches = [];
s.replace(/\[(.*?)\]/g, function(g0,g1){matches.push(g1);})
Example: http://jsfiddle.net/kobi/6CEzP/

var s = 'pass[1][2011-08-21][total_passes]';
r = s.match(/\[([^\]]*)\]/g);
r ; //# => [ '[1]', '[2011-08-21]', '[total_passes]' ]
example proving the edge case of unbalanced [];
var s = 'pass[1]]][2011-08-21][total_passes]';
r = s.match(/\[([^\]]*)\]/g);
r; //# => [ '[1]', '[2011-08-21]', '[total_passes]' ]

add the global flag to your regex , and iterate the array returned .
match(/\[(.*?)\]/g)

I'm not sure if you can get this directly into an array. But the following code should work to find all occurences and then process them:
var string = "pass[1][2011-08-21][total_passes]";
var regex = /\[([^\]]*)\]/g;
while (match = regex.exec(string)) {
alert(match[1]);
}
Please note: i really think you need the character class [^\]] here. Otherwise in my test the expression would match the hole string because ] is also matches by .*.

'pass[1][2011-08-21][total_passes]'.match(/\[.+?\]/g); // ["[1]","[2011-08-21]","[total_passes]"]
Explanation
\[ # match the opening [
Note: \ before [ tells that do NOT consider as a grouping symbol.
.+? # Accept one or more character but NOT greedy
\] # match the closing ] and again do NOT consider as a grouping symbol
/g # do NOT stop after the first match. Do it for the whole input string.
You can play with other combinations of the regular expression
https://regex101.com/r/IYDkNi/1

[C#]
string str1 = " pass[1][2011-08-21][total_passes]";
string matching = #"\[(.*?)\]";
Regex reg = new Regex(matching);
MatchCollection matches = reg.Matches(str1);
you can use foreach for matched strings.