Develop Reference

Replace content present in the nested brackets

Input = ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))
Expected Output = ABCDEF,GHIJKLMN,OPQRSTUVW
Tried so far
Output = Input.replace(/ *\([^)]*\)*/g, "");

Using a regex here probably won't work, or scale, because you expect nested parentheses in your input string. Regex works well when there is a known and fixed structure to the input. Instead, I would recommend that you approach this using a parser. In the code below, I iterate over the input string, one character at at time, and I use a counter to keep track of how many open parentheses there are. If we are inside a parenthesis term, then we don't record those characters. I also have one simple replacement at the end to remove whitespace, which is an additional step which your output implies, but you never explicitly mentioned.
var pCount = 0;
var Input = "ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))";
var Output = "";
for (var i=0; i < Input.length; i++) {
if (Input[i] === '(') {
pCount++;
}
else if (Input[i] === ')') {
pCount--;
}
else if (pCount == 0) {
Output += Input[i];
}
}
Output = Output.replace(/ /g,'');
console.log(Output);

If you need to remove nested parentheses, you may use a trick from Remove Nested Patterns with One Line of JavaScript.
var Input = "ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))";
var Output = Input;
while (Output != (Output = Output.replace(/\s*\([^()]*\)/g, "")));
console.log(Output);
Or, you could use a recursive function:
function remove_nested_parens(s) {
let new_s = s.replace(/\s*\([^()]*\)/g, "");
return new_s == s ? s : remove_nested_parens(new_s);
}
console.log(remove_nested_parens("ABCDEF ((3) abcdef),GHIJKLMN ((4)(5) Value),OPQRSTUVW((4(5)) Value (3))"));
Here, \s*\([^()]*\) matches 0+ whitespaces, (, 0+ chars other than ( and ) and then a ), and the replace operation is repeated until the string does not change.

Javascript Remove strings in beginning and end

base on the following string
...here..
..there...
.their.here.
How can i remove the . on the beginning and end of string like the trim that removes all spaces, using javascript
the output should be
here
there
their.here

These are the reasons why the RegEx for this task is /(^\.+|\.+$)/mg:
Inside /()/ is where you write the pattern of the substring you want to find in the string:
/(ol)/ This will find the substring ol in the string.
var x = "colt".replace(/(ol)/, 'a'); will give you x == "cat";
The ^\.+|\.+$ in /()/ is separated into 2 parts by the symbol | [means or]
^\.+ and \.+$
^\.+ means to find as many . as possible at the start.
^ means at the start; \ is to escape the character; adding + behind a character means to match any string containing one or more that character
\.+$ means to find as many . as possible at the end.
$ means at the end.
The m behind /()/ is used to specify that if the string has newline or carriage return characters, the ^ and $ operators will now match against a newline boundary, instead of a string boundary.
The g behind /()/ is used to perform a global match: so it find all matches rather than stopping after the first match.
To learn more about RegEx you can check out this guide.

Try to use the following regex
var text = '...here..\n..there...\n.their.here.';
var replaced = text.replace(/(^\.+|\.+$)/mg, '');

Here is working Demo
Use Regex /(^\.+|\.+$)/mg
^ represent at start
\.+ one or many full stops
$ represents at end
so:
var text = '...here..\n..there...\n.their.here.';
alert(text.replace(/(^\.+|\.+$)/mg, ''));

Here is an non regular expression answer which utilizes String.prototype
String.prototype.strim = function(needle){
var first_pos = 0;
var last_pos = this.length-1;
//find first non needle char position
for(var i = 0; i<this.length;i++){
if(this.charAt(i) !== needle){
first_pos = (i == 0? 0:i);
break;
}
}
//find last non needle char position
for(var i = this.length-1; i>0;i--){
if(this.charAt(i) !== needle){
last_pos = (i == this.length? this.length:i+1);
break;
}
}
return this.substring(first_pos,last_pos);
}
alert("...here..".strim('.'));
alert("..there...".strim('.'))
alert(".their.here.".strim('.'))
alert("hereagain..".strim('.'))
and see it working here : http://jsfiddle.net/cettox/VQPbp/

Slightly more code-golfy, if not readable, non-regexp prototype extension:
String.prototype.strim = function(needle) {
var out = this;
while (0 === out.indexOf(needle))
out = out.substr(needle.length);
while (out.length === out.lastIndexOf(needle) + needle.length)
out = out.slice(0,out.length-needle.length);
return out;
}
var spam = "this is a string that ends with thisthis";
alert("#" + spam.strim("this") + "#");
Fiddle-ige

Use RegEx with javaScript Replace
var res = s.replace(/(^\.+|\.+$)/mg, '');

We can use replace() method to remove the unwanted string in a string
Example:
var str = '<pre>I'm big fan of Stackoverflow</pre>'
str.replace(/<pre>/g, '').replace(/<\/pre>/g, '')
console.log(str)
output:
Check rules on RULES blotter

Why is my RegExp ignoring start and end of strings?

I made this helper function to find single words, that are not part of bigger expressions
it works fine on any word that is NOT first or last in a sentence, why is that?
is there a way to add "" to regexp?
String.prototype.findWord = function(word) {
var startsWith = /[\[\]\.,-\/#!$%\^&\*;:{}=\-_~()\s]/ ;
var endsWith = /[^A-Za-z0-9]/ ;
var wordIndex = this.indexOf(word);
if (startsWith.test(this.charAt(wordIndex - 1)) &&
endsWith.test(this.charAt(wordIndex + word.length))) {
return wordIndex;
}
else {return -1;}
}
Also, any improvement suggestions for the function itself are welcome!
UPDATE: example: I want to find the word able in a string, I waht it to work in cases like [able] able, #able1 etc.. but not in cases that it is part of another word like disable, enable etc

A different version:
String.prototype.findWord = function(word) {
return this.search(new RegExp("\\b"+word+"\\b"));
}
Your if will only evaluate to true if endsWith matches after the word. But the last word of a sentence ends with a full stop, which won't match your alphanumeric expression.

Did you try word boundary -- \b?
There is also \w which match one word character ([a-zA-Z_]) -- this could help you too (depends on your word definition).
See RegExp docs for more details.

If you want your endsWith regexp also matches the empty string, you just need to append |^$ to it:
var endsWith = /[^A-Za-z0-9]|^$/ ;
Anyway, you can easily check if it is the beginning of the text with if (wordIndex == 0), and if it is the end with if (wordIndex + word.length == this.length).
It is also possible to eliminate this issue by operating on a copy of the input string, surrounded with non-alphanumerical characters. For example:
var s = "#" + this + "#";
var wordIndex = this.indexOf(word) - 1;
But I'm afraid there is another problems with your function:
it would never match "able" in a string like "disable able enable" since the call to indexOf would return 3, then startsWith.test(wordIndex) would return false and the function would exit with -1 without searching further.
So you could try:
String.prototype.findWord = function (word) {
var startsWith = "[\\[\\]\\.,-\\/#!$%\\^&\*;:{}=\\-_~()\\s]";
var endsWith = "[^A-Za-z0-9]";
var wordIndex = ("#"+this+"#").search(new RegExp(startsWith + word + endsWith)) - 1;
if (wordIndex == -1) { return -1; }
return wordIndex;
}

Javascript split only once and ignore the rest

I am parsing some key value pairs that are separated by colons. The problem I am having is that in the value section there are colons that I want to ignore but the split function is picking them up anyway.
sample:
Name: my name
description: this string is not escaped: i hate these colons
date: a date
On the individual lines I tried this line.split(/:/, 1) but it only matched the value part of the data. Next I tried line.split(/:/, 2) but that gave me ['description', 'this string is not escaped'] and I need the whole string.
Thanks for the help!

a = line.split(/:/);
key = a.shift();
val = a.join(':');

Use the greedy operator (?) to only split the first instance.
line.split(/: (.+)?/, 2);

If you prefer an alternative to regexp consider this:
var split = line.split(':');
var key = split[0];
var val = split.slice(1).join(":");
Reference: split, slice, join.

Slightly more elegant:
a = line.match(/(.*?):(.*)/);
key = a[1];
val = a[2];

May be this approach will be the best for such purpose:
var a = line.match(/([^:\s]+)\s*:\s*(.*)/);
var key = a[1];
var val = a[2];
So, you can use tabulations in your config/data files of such structure and also not worry about spaces before or after your name-value delimiter ':'.
Or you can use primitive and fast string functions indexOf and substr to reach your goal in, I think, the fastest way (by CPU and RAM)
for ( ... line ... ) {
var delimPos = line.indexOf(':');
if (delimPos <= 0) {
continue; // Something wrong with this "line"
}
var key = line.substr(0, delimPos).trim();
var val = line.substr(delimPos + 1).trim();
// Do all you need with this key: val
}

Split string in two at first occurrence
To split a string with multiple i.e. columns : only at the first column occurrence
use Positive Lookbehind (?<=)
const a = "Description: this: is: nice";
const b = "Name: My Name";
console.log(a.split(/(?<=^[^:]*):/)); // ["Description", " this: is: nice"]
console.log(b.split(/(?<=^[^:]*):/)); // ["Name", " My Name"]
it basically consumes from Start of string ^ everything that is not a column [^:] zero or more times *. Once the positive lookbehind is done, finally matches the column :.
If you additionally want to remove one or more whitespaces following the column,
use /(?<=^[^:]*): */
Explanation on Regex101.com

function splitOnce(str, sep) {
const idx = str.indexOf(sep);
return [str.slice(0, idx), str.slice(idx+1)];
}
splitOnce("description: this string is not escaped: i hate these colons", ":")

Javascript regular expression: remove first and last slash

I have these strings in javascript:
/banking/bonifici/italia
/banking/bonifici/italia/
and I would like to remove the first and last slash if it's exists.
I tried ^\/(.+)\/?$ but it doesn't work.
Reading some post in stackoverflow I found that php has trim function and I could use his javascript translation (http://phpjs.org/functions/trim:566) but I would prefer a "simple" regular expression.

return theString.replace(/^\/|\/$/g, '');
"Replace all (/.../g) leading slash (^\/) or (|) trailing slash (\/$) with an empty string."

There's no real reason to use a regex here, string functions will work fine:
var string = "/banking/bonifici/italia/";
if (string.charAt(0) == "/") string = string.substr(1);
if (string.charAt(string.length - 1) == "/") string = string.substr(0, string.length - 1);
// string => "banking/bonifici/italia"
See this in action on jsFiddle.
References:
String.substr
String.charAt

In case if using RegExp is not an option, or you have to handle corner cases while working with URLs (such as double/triple slashes or empty lines without complex replacements), or utilizing additional processing, here's a less obvious, but more functional-style solution:
const urls = [
'//some/link///to/the/resource/',
'/root',
'/something/else',
];
const trimmedUrls = urls.map(url => url.split('/').filter(x => x).join('/'));
console.log(trimmedUrls);
In this snippet filter() function can implement more complex logic than just filtering empty strings (which is default behavior).
Word of warning - this is not as fast as other snippets here.

One liner, no regex, handles multiple occurences
const trimSlashes = str => str.split('/').filter(v => v !== '').join('/');
console.log(trimSlashes('/some/path/foo/bar///')); // "some/path/foo/bar"

Just in case that someone needs a premature optimization here...
http://jsperf.com/remove-leading-and-trailing-slashes/5
var path = '///foo/is/not/equal/to/bar///'
var count = path.length - 1
var index = 0
while (path.charCodeAt(index) === 47 && ++index);
while (path.charCodeAt(count) === 47 && --count);
path = path.slice(index, count + 1)

you can check with str.startsWith and str.endsWith
then substr if exist
var str= "/aaa/bbb/";
var str= str.startsWith('/') ? str.substr(1) : str;
var str= str.endsWith('/') ? str.substr(0,str.length - 1) : str;
or you can write custom function
trimSlashes('/aaa/bbb/');
function trimSlashes(str){
str= str.startsWith('/') ? str.substr(1) : str;
str= str.endsWith('/') ? str.substr(0,str.length - 1) : str;
return str;
}