You need to use the m flag for ^ and $ to work as expected:

[...] if m is used, ^ and $ change from matching at only the start or end of the entire string to the start or end of any line within the string.

Also put the * in the right place:

let iCalContent = "DATE:20091201T220000\r\nSUMMARY:Dad's birthday";
let result = /^SUMMARY\:(.*)$/gm.exec(iCalContent);
console.log(result && result[1]);

Each group defined by parenthesis () is captured during processing and each captured group content is pushed into result array in the same order as groups within pattern starts. See more on http://www.regular-expressions.info/brackets.html and http://www.regular-expressions.info/refcapture.html (choose right language to see supported features)

var source = "afskfsd33j"
var result = source.match(/a(.*)j/);

result: ["afskfsd33j", "fskfsd33"]

The reason why you received this exact result is following:

First value in array is the first found string which confirms the entire pattern. So it should definitely start with "a" followed by any number of any characters and ends with first "j".

Second value in the array is captured group defined by parenthesis. In your case group contains string consist of any character with length <0; infinite). Which is entire source without starting "a" and ending "j" which are defined outside parenthesis, so exactly "fskfsd33".

If you want to get rid of second value in the array you may define pattern like this:

/a(?:.*)j/

where "?:" means that group of chars which match the content in parenthesis will not be part of the resulting array.

Other options might be in this simple case to write pattern without any group because it is not necessary to use group at all:

/a.*j/

If you want to just check whether source text matches the pattern and does not care about found texts then you may try:

var result = /a.*j/.test(source);

The result should return then only true|false values. For more info see http://www.javascriptkit.com/javatutors/re3.shtml

I would recommend test page https://regex101.com/ where you can test the patterns with full explanation how the RegExp works.

You can try this:

let str = "abc (24 314 83 383)(-256)sa 0 (24 314) 1";
let replaced = str.replace(/(\s*\(|\))/g, '<REP>');
let arrFromStr = replaced.split('<REP>').filter(w => w.length != 0);

Variable "replaced" replaces all 1) 0 or more spaces + "(", and 2) all ")" symbols to "" string. arrFromStr creates an array from string and split it with "". Then we check is the element of array empty, or not.

The group (.+) will match as much as it can (it's "greedy"). To make it find a minimal match you can use (.+?).

you can use .split() to split the strings:

var str = "10M225S"
var m = str.split("M")[0];
var s = str.split("M")[1].split("S")[0];

console.log("m >>> " + m);
console.log("s >>> " + s);

no need for regular expressions in this case.

Regex for you will be like #(?<extract>[\s\S]*?)\$

Named group 'extract'will contain the values you want.

As Alex mentioned, if javascript doesnt support named group, we can use numbered group. So modified regex will be #([\s\S]*?)\$ and desired values will be in group number 1.

Use match instead, like this:

var regex = /@\{([^;]+);\s+([^@]+)/;

var matches = text.match(regex);

alert(matches[1] + ', ' + matches[2]);

http://jsfiddle.net/rooseve/bM2U6/

You can try this-

let myString = "This is my comment content. [~firstName.lastName] and the original name is [~John.Doe]";
const regex = /\[~([^\]]+)\.([^\]]+)\]/g;

let match = regex.exec(myString);
const names = [];

while(match !== null) {
  names.push({firstName: match[1], lastName: match[2]});
  match = regex.exec(myString);
}

// Remove the pattern from the original string.
myString = myString.replace(regex, '');

console.log(names, myString);

This code will find all the matches found for the pattern.