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How do I check for brackets in a specific place in the string?

I have this code and it needs to returns true or false based on the string you give it.
This is the only example on which it doesn't work. How can I check if brackets exist in a specific index of the string?
function telephoneCheck(str) {
var newStr = str.replace(/-/g,'').replace(/ /g,'').replace(/\(|\)/g,'');
var valid = true;
var re = /\([^()]*\)/g;
while (str.match(re))
str = str.replace(re, '');
if (str.match(/[()]/)){
valid = false;
}
if(newStr.length === 10 && valid === true && str.indexOf()){
return true;
}else if(newStr.length === 11 && str[0] != "-" && newStr[0] == 1 && valid === true){
return true;
}else{
return false;
}
}
telephoneCheck("(6505552368)");

Based on your code I think you might be looking for something like this:
'(6505552368)'.replace(/^\((\d+)\)$/, '$1');
The ^ and $ in the RegExp will match the start and the end of the string. The \d+ will match one or more numbers. The extra parentheses form a capture group that is then used in the replacement as $1.
You might be better off doing more work using RegExps rather than doing all that replacing but without knowing the exact requirements it's difficult to be more precise. I highly suggest learning more about RegExp syntax.
If you literally just want to know whether 'brackets exist in a specific index' then you can just use:
str.charAt(index) === '(';

To check if there are brackets at a specific index in the string:
/[()]/.test(str[index])
To check if there are any brackets in the string at all:
/[()]/.test(str)
If you want to test for a specific bracket type (e.g. opening but not closing) remove the other one (e.g. closing) from the regex.

Some parts of my regular expression not working

I'm trying to use a regular expression to validate the input on a textbox
The expression should allow only numbers, maxmium two decimals, max one comma (,) and one minus symbol in front of the number (optional).
Valid:
0,25
10,2
-7000
-175,33
15555555555555,99
invalid:
9,999
15.03
77,77,77
etc
I'm using ^[-+]?[\d ]+(,\d{0,2})?$
The regex is used in a Jquery code to prevent the user from entering invalid numbers (event.preventDefault()):
$("input[name*='TB_mytbx']").on('keypress', function (event) {
var regex = new RegExp("^[-+]?[\d ]+(,\d{0,2})?$", "g");
var key = String.fromCharCode(!event.charCode ? event.which : event.charCode);
if (!regex.test(key)) {
event.preventDefault();
return false;
}
});
Only a part of the regular expression seems to work.
It works with numbers (It does not allow me to enter letters) but it also won't allow commas (,) and the minus (-).
What am I doing wrong?
Edit
Before I used:
if (focused.val().indexOf(',') != -1) {
var number = (focused.val().split(','));
if (number[1] && number[1].length >= 2) {
event.preventDefault();
return;
}
But this gives annoying behavior. As soon as you enter a number with two digits you can't make edits anymore. For example: you can't change 200,50 to 300,50 or 100 300,50. (You get the point). I hoped that a regex could change that somehow.

I think you're massively over-complicating the regex. This should be plenty:
^-?\d+(,\d\d)?$
^ Start of line,
-? Optional minus sign,
\d+ Followed by a bunch of digits,
(,\d\d)? Followed by a comma and 2 digits, which are all 3 optional.
(alternative: (,\d{2})?)
$ End of line.
var regex = /^-?\d+(,\d\d)?$/;
console.log(regex.test('0,25'));
console.log(regex.test('-175,33'));
console.log(regex.test('15555555555555,99'));
console.log(regex.test('9,999'));
console.log(regex.test('15.03'));
console.log(regex.test('77,77,77'));
There you have a regex to validate the input value.
Now, that block of code can be replaced with this:
$("input[name*='TB_mytbx']").on('keypress', function (event) {
var regex = /^-?\d+(,\d\d)?$/;
var value = $(this).val(); // Use the field's value, instead of the pressed key.
if (!regex.test(value)) {
event.preventDefault();
return false;
}
});

For those of you who wanna know, I solved it using this code
$("input[name*='mythingy']").on('keypress', function (event) {
var theEvent = event || window.event;
var key = theEvent.keyCode || theEvent.which;
key = String.fromCharCode(key);
var value = this.value;
var value = value.replace(value.substring(theEvent.currentTarget.selectionStart, theEvent.currentTarget.selectionEnd), "");
value = [value.slice(0, theEvent.currentTarget.selectionStart), key, value.slice(theEvent.currentTarget.selectionStart)].join('');
var regex = /^[-+]?([\d ]+(,\d{0,2})?)?$/;
if (!regex.test(value)) {
theEvent.returnValue = false;
if (theEvent.preventDefault) theEvent.preventDefault();
}
});

REGEX in javascript not working for dots (.) in JavaScript

I am using a regex to validate an email address in JavaScript.
The regex is pretty simple. It checks three things: 1)'#' , 2)'.' ('dot' as in something#gmail.com), and 3) 'a-z' in an email address. If all three return true, email address is valid (according to my validation, atleast)
Here is the code:
function checkemail(){
var e = document.getElementById("email").value;
if((e.match(/#/g)==null)||(e.match(/[a-z]/ig)==null)||(e.match(/./g)==null)){
//display error message
}
}
My question is:
(e.match(/./g)==null); //returns false even though there are no dots in the string e
returns false even when there are no dots in string.
For example:
("thisIsMyEmail".match(/./ig))==null //returns false
Why does it return false when it should be true?

/./g (or /./ig) will match any string that as at least one character in it. . is special in regular expressions, it means "any character here."
For an actual dot, escape it with a backslash: /\./g.

First off, you don't need to check if the string is null. Simply use this:
var email = "Godisgood#gmail.com";
if (email.match(/^\S+\#\S+\.\S+$/i)){
alert("Email address passed validation.");
}

you have to escape the .
The unescaped period means matches any character.
Meaning having a string "abc" using your expression would result in an array containing the characters 'a', 'b', and 'c'.
In your snippet the correct answer is
(e.match(/\./g)==null);
This should result to what you're expecting

Try this
(e.match(/\./g)==null);
. matches any character so needs escaping /\./g

I know you have already got the answer.
But I just want to give an advice.
My advice is - don't use the javascript code to validate any email address; because as per your code, #domain., #domain.com these all are also valid email, but everybody knows these are not a valid email address.
So use the below code:
let email = $(this).val();
var positionOfAt = email.indexOf("#");
var positionOfDot = email.lastIndexOf(".");
if(email.search("#") == -1 || //if '#' is not present
email.search(" ") >= 1 || //if blank space is present
email.search(".") == -1 || //if "." is not present
positionOfAt < 1 || //if there is no character before "#", at least one character should be present before "#"
positionOfDot - positionOfAt <= 2 || //between '#' and '.', if there is not at least two character
email.length - positionOfDot <= 2) //if after '.' there is not at least two character)
{
console.log("Invalid email id")
}
else
{
console.log("Valid email id")
}

RegEx which allows only english text and no special charactes

I want to validate a text field (first name) using javascript. such that it should only contain text. NO special characters and No numbers.and since it is just the first name it should only contain one word. (no spaces)
Allowed:
John
john
Not Allowed
john kennedy.
John kennedy.
john123.
123john.
I tried this but its not working.
if( !validateName($fname))
{
alert("name invalid");
}
function validateName($name) {
var nameReg = /^A-Za-z*/;
if( !nameReg.test( $name ) ) {
return false;
} else {
return true;
}
}
EDIT:
I tried
var nameReg = /^[A-Za-z]*/;
but it still doesn't show the alert box when I enter john123 or 123john.

nameReg needs to be /^[a-z]+$/i (or some varient). The ^ matches the start of the string and $ matches the end. This is "one or more a-z characters from the start to the end of the string, case-insensitive." You can change + to *, but then the string could be empty.
http://jsfiddle.net/ExplosionPIlls/pwYV3/1/

Use a character class:
var nameReg = /^[A-Za-z]*/;
Without the containing [] (making it a character class), you're specifying a literal A-Za-z.
UPDATE:
Add a $ to the end of the Regex.
var nameReg = /^[A-Za-z]*$/;
Otherwise, john123 returns valid as the Regex is matching john and can ignore the 123 portion of the string.
Working demo: http://jsfiddle.net/GNVck/

Remove special character from the starting of a string and search # symbol.in javascript

I want to remove special characters from the starting of the string only.
i.e, if my string is like {abc#xyz.com then I want to remove the { from the starting. The string shoould look like abc#xyz.com
But if my string is like abc{#xyz.com then I want to retain the same string as it is ie., abc{#xyz.com.
Also I want to check that if my string has # symbol present or not. If it is present then OK else show a message.

The following demonstrates what you specified (or it's close):
var pat = /^[^a-z0-9]*([a-z0-9].*?#.*?$)/i; //pattern for optional non-alphabetic start followed by alphabetic, followed by '#' somewhere
var testString = "{abc#xyz.com"; //Try with {abcxyz.com for alert
arr = pat.exec(testString);
var adjustedString;
if (arr != null) { adjustedString = arr[1]; } //The potentially adjustedString (chopped off non-alphabetic start) will be in capture group 1
else { adjustedString = ""; alert(testString + " does not conform to pattern"); }
adjustedString;

I have used two separate regex objects to achieve what you require .It checks for both the conditions in the string.I know its not very efficient but it will serve your purpose.
var regex = new RegExp(/(^{)/);
var regex1 = new RegExp(/(^[^#]*$)/);
var str = "abc#gmail.com";
if(!regex1.test(str)){
if(regex.test(str))
alert("Bracket found at the beginning")
else
alert("Bracket not found at the beginning")
}
else{
alert("doesnt contain #");
}
Hope this helps