Develop Reference

Finding ++ in Regular Expression

I want to find ++ or -- or // or ** sign in in string can anyone help me?
var str = document.getElementById('screen').innerHTML;
var res = str.substring(0, str.length);
var patt1 = ++,--,//,**;
var result = str.match(patt1);
if (result)
{
alert("you cant do this :l");
document.getElementById('screen').innerHTML='';
}

This finds doubles of the characters by a backreference:
/([+\/*-])\1/g
[from q. comments]: i know this but when i type var patt1 = /[++]/i; code find + and ++
[++] means one arbitrary of the characters. Normally + is the qantifier "1 or more" and needs to be escaped by a leading backslash when it should be a literal, except in brackets where it does not have any special meaning.
Characters that do need to be escaped in character classes are e.g. the escape character itself (backslash), the expression delimimiter (slash), the closing bracket and the range operator (dash/minus), the latter except at the end of the character class as in my code example.
A character class [] matches one character. A quantifier, e.g. [abc]{2} would match "aa", "bb", but "ab" as well.
You can use a backreference to a match in parentheses:
/(abc)\1
Here the \1 refers to the first parentheses (abc). The entire expression would match "abcabc".
To clarify again: We could use a quantifier on the backreference:
/([+\/*-])\1{9}/g
This matches exactly 10 equal characters out of the class, the subpattern itself and 9 backreferences more.
/.../g finds all occurrences due to the modifier global (g).
test-case on regextester.com

Define your pattern like this:
var patt1 = /\+\+|--|\/\/|\*\*/;
Now it should do what you want.
More info about regular expressions: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions

You can use:
/\+\+|--|\/\/|\*\*/
as your expression.
Here I have escaped the special characters by using a backslash before each (\).
I've also used .test(str) on the regular expression as all you need is a boolean (true/false) result.
See working example below:
var str = document.getElementById('screen').innerHTML;
var res = str.substring(0, str.length);
var patt1 = /\+\+|--|\/\/|\*\*/;
var result = patt1.test(res);
if (result) {
alert("you cant do this :l");
document.getElementById('screen').innerHTML = '';
}
<div id="screen">
This is some++ text
</div>

Try this:-
As
n+:- Matches any string that contains at least one n
n* Matches any string that contains zero or more occurrences of n
We need to use backslash before this special characters.
var str = document.getElementById('screen').innerHTML;
var res = str.substring(0, str.length);
var patt1 = /\+\+|--|\/\/|\*\*/;
var result = str.match(patt1);
if (result)
{
alert("you cant do this :l");
document.getElementById('screen').innerHTML='';
}
<div id="screen">2121++</div>

Need help finding a plus sign using javascript regex

I am using the code below to find a match for a plus sign but it keeps returning false. I am not sure what I am doing wrong. Any help will be really appreciated it. Thanks!
var str = '+2443';
var result = /d\+1/.test(str);
console.log(result); // true

var str = '+2443';
var result = /\+/.test(str);
console.log(result); // true

Your /d\+1/ regex matches the first occurrence of a d+1 substring in any string.
To check if a string contains a +, you do not need a regex. Use indexOf:
var str = '+2443';
if (~str.indexOf("+")) {
console.log("Found a `+`");
} else {
console.log("A `+` is not found");
}
A regex will be more appropriate when you need to match a + in some context. For example, to check if the string starts with a plus, and then only contains digits, you would use
var str = '+2443';
var rx = /^\+\d+$/;
console.log(rx.test(str));
where ^ assets the position at the end of the string, \+ matches a literal +, \d+ matches 1+ digits and the $ anchor asserts the position at the end of the string.

Getting each 'word' after every underscore in a string in Javascript using regex

I'm wanting to extract each block of alphanumeric characters that come after underscores in a Javascript string. I currently have it working using a combination of string methods and regex like so:
var string = "ignore_firstMatch_match2_thirdMatch";
var firstValGone = string.substr(string.indexOf('_'));
// returns "_firstMatch_match2_thirdMatch"
var noUnderscore = firstValGone.match(/[^_]+/g);
// returns ["firstMatch", "match2" , "thirdMatch"]
I'm wondering if there's a way to do it purely using regex? Best I've managed is:
var string = "ignore_firstMatch_match2_thirdMatch";
var matchTry = string.match(/_[^_]+/g);
// returns ["_firstMatch", "_match2", "_thirdMatch"]
but that returns the preceding underscore too. Given you can't use lookbehinds in JS I don't know how to match the characters after, but exclude the underscore itself. Is this possible?

You can use a capture group (_([^_]+)) and use RegExp#exec in a loop while pushing the captured values into an array:
var re = /_([^_]+)/g;
var str = 'ignore_firstMatch_match2_thirdMatch';
var res = [];
while ((m = re.exec(str)) !== null) {
res.push(m[1]);
}
document.body.innerHTML = "<pre>" + JSON.stringify(res, 0, 4) + "</pre>";
Note that using a string#match() with a regex defined with a global modifier /g will lose all the captured texts, that's why you cannot just use str.match(/_([^_]+)/g).

Since lookbehind is not supported in JS the only way I can think of is using a group like this.
Regex: _([^_]+) and capture group using \1 or $1.
Regex101 Demo
var myString = "ignore_firstMatch_match2_thirdMatch";
var myRegexp = /_([^_]+)/g;
match = myRegexp.exec(myString);
while (match != null) {
document.getElementById("match").innerHTML += "<br>" + match[0];
match = myRegexp.exec(myString);
}
<div id="match">
</div>
An alternate way using lookahead would be something like this.
But it takes long in JS. Killed my page thrice. Would make a good ReDoS exploit
Regex: (?=_([A-Za-z0-9]+)) and capture groups using \1 or $1.
Regex101 Demo

Why do you assume you need regex? a simple split will do the job:
string str = "ignore_firstMatch_match2_thirdMatch";
IEnumerable<string> matches = str.Split('_').Skip(1);

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

how to config RegExp when string contains parentheses

I'm sure this is an easy one, but I can't find it on the net.
This code:
var new_html = "foo and bar(arg)";
var bad_string = "bar(arg)";
var regex = new RegExp(bad_string, "igm");
var bad_start = new_html.search(regex);
sets bad_start to -1 (not found). If I remove the (arg), it runs as expected (bad_start == 8). Is there something I can do to make the (very handy) "new Regexp" syntax work, or do I have to find another way? This example is trivial, but in the real app it would be doing global search and replace, so I need the regex and the "g". Or do I?
TIA

Escape the brackets by double back slashes \\. Try this.
var new_html = "foo and bar(arg)";
var bad_string = "bar\\(arg\\)";
var regex = new RegExp(bad_string, "igm");
var bad_start = new_html.search(regex);
Demo

Your RegEx definition string should be:
var bad_string = "bar\\(arg\\)";
Special characters need to be escaped when using RegEx, and because you are building the RegEx in a string you need to escape your escape character :P
http://www.regular-expressions.info/characters.html

You need to escape the special characters contained in string you are creating your Regex from. For example, define this function:
function escapeRegex(string) {
return string.replace(/[/\-\\^$*+?.()|[\]{}]/g, '\\$&');
}
And use it to assign the result to your bad_string variable:
let bad_string = "bar(arg)"
bad_string = escapeRegex(bad_string)
// You can now use the string to create the Regex :v: