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How to match regular expression In Javascript

I have string [FBWS-1] comes first than [FBWS-2]
In this string, I want to find all occurance of [FBWS-NUMBER]
I tried this :
var term = "[FBWS-1] comes first than [FBWS-2]";
alert(/^([[A-Z]-[0-9]])$/.test(term));
I want to get all the NUMBERS where [FBWS-NUMBER] string is matched.
But no success. I m new to regular expressions.
Can anyone help me please.

Note that ^([[A-Z]-[0-9]])$ matches start of a string (^), a [ or an uppercase ASCII letter (with [[A-Z]), -, an ASCII digit and a ] char at the end of the string. So,basically, strings like [-2] or Z-3].
You may use
/\[[A-Z]+-[0-9]+]/g
See the regex demo.
NOTE If you need to "hardcode" FBWS (to only match values like FBWS-123 and not ABC-3456), use it instead of [A-Z]+ in the pattern, /\[FBWS-[0-9]+]/g.
Details
\[ - a [ char
[A-Z]+ - one or more (due to + quantifier) uppercase ASCII letters
- - a hyphen
[0-9]+ - one or more (due to + quantifier) ASCII digits
] - a ] char.
The /g modifier used with String#match() returns all found matches.
JS demo:
var term = "[FBWS-1] comes first than [FBWS-2]";
console.log(term.match(/\[[A-Z]+-[0-9]+]/g));

You can use:
[\w+-\d]
var term = "[FBWS-1] comes first than [FBWS-2]";
alert(/[\w+-\d]/.test(term));
There are several reasons why your existing regex doesn't work.
You trying to match the beginning and ending of your string when you
actually want everything in between, don't use ^$
Your only trying to match one alpha character [A-Z] you need to make this greedy using the +
You can shorten [A-Z] and [0-9] by using the shorthands \w and \d. The brackets are generally unnecessary.
Note your code only returns a true false value (your using test) ATM it's unclear if this is what you want. You may want to use match with a global modifier (//g) instead of test to get a collection.

Here is an example using string.match(reg) to get all matches strings:
var term = "[FBWS-1] comes first than [FBWS-2]";
var reg1 = /\[[A-Z]+-[0-9]\]/g;
var reg2 = /\[FBWS-[0-9]\]/g;
var arr1 = term.match(reg1);
var arr2 = term.match(reg2)
console.log(arr1);
console.log(arr2);

Your regular expression /^([[A-Z]-[0-9]])$/ is wrong.
Give this regex a try, /\[FBWS-\d\]/g
remove the g if you only want to find 1 match, as g will find all similar matches
Edit: Someone mentioned that you want ["any combination"-"number"], hence if that's what you're looking for then this should work /\[[A-Z]+-\d\]/

regex precceded by two or more special character

I am stuck with creating regex such that if the word is preceded or ended by special character more than one regex on each side regex 'exec' method should throw null. Only if word is wrap with exactly one bracket on each side 'exec' method should give result Below is the regular expression I have come up with.
If the string is like "(test)" or then only regex.exec should have values for other combination such as "((test))" OR "((test)" OR "(test))" it should be null. Below code is not throwing null which it should. Please suggest.
var w1 = "\(test\)";
alert(new RegExp('(^|[' + '\(\)' + '])(' + w1 + ')(?=[' + '\(\)' + ']|$)', 'g').exec("this is ((test))"))

If you have a list of words and want to filter them, you can do the following.
string.split(' ').filter(function(word) {
return !(/^[!##$%^&*()]{2,}.+/).test(word) || !(/[!##$%^&*()]{2,}$).test(word)
});
The split() function splits a string at a space character and returns an array of words, which we can then filter.
To keep the valid words, we will test two regex expressions to see if the word starts or ends with 2 or more special characters respectively.
RegEx Breakdown
^ - Expression starts with the following
[] - A single character in the block
!##$%^&*() - These are the special characters I used. Replace them with the ones you want.
{2,} - Matches 2 or more of the preceeding characters
.+ - Matches 1 or more of any character
$ - Expression ends with the following
To use the exec function this way do this
!(/^[!##$%^&*()]{2,}.+/).exec(string) || !(/[!##$%^&*()]{2,}$).exec(string)

If I understand correctly, you are looking for any string which contains (test), anywhere in it, and exactly that, right?
In that case, what you probably need is the following:
var regExp = /.*[^)]\(test\)[^)].*/;
alert(regExp.exec("this is ((test))")); // → null
alert(regExp.exec("this is (test))" )); // → null
alert(regExp.exec("this is ((test)" )); // → null
alert(regExp.exec("this is (test) ...")); // → ["this is (test) ..."]
Explanation:
.* matches any character (except newline) between zero and unlimited times, as many times as possible.
[^)] match a single character but not the literal character )
This makes sure there's your test string in the given string, but it is only ever wrapped with one brace in every side!

You can use the following regex:
(^|[^(])(\(test\))(?!\))
See regex demo here, replace with $1<span style="new">$2</span>.
The regex features an alternation group (^|[^(]) that matches either start of string ^ or any character other than (. This alternation is a kind of a workaround since JS regex engine does not support look-behinds.
Then, (\(test\)) matches and captures (test). Note the round brackets are escaped. If they were not, they would be treated as a capturing group delimiters.
The (?!\)) is a look-ahead that makes sure there is no literal ) right after test). Look-aheads are supported fully by JS regex engine.
A JS snippet:
var re = /(^|[^(])(\(test\))(?!\))/gi;
var str = 'this is (test)\nthis is ((test))\nthis is ((test)\nthis is (test))\nthis is ((test\nthis is test))';
var subst = '$1<span style="new">$2</span>';
var result = str.replace(re, subst);
alert(result);

Javascript reg exp not right

Here is a string str = '.js("aaa").js("bbb").js("ccc")', I want to write a regular expression to return an Array like this:
[aaa, bbb, ccc];
My regular expression is:
var jsReg = /.js\(['"](.*)['"]\)/g;
var jsAssets = [];
var js;
while ((js = jsReg.exec(find)) !== null) {
jsAssets.push(js[1]);
}
But the jsAssets result is
[""aaa").js("bbb").js("ccc""]
What's wrong with this regular expression?

Use the lazy version of .*:
/\.js\(['"](.*?)['"]\)/g
^
And it would be better if you escape the first dot.
This will match the least number of characters until the next quote.
jsfiddle demo
If you want to allow escaped quotes, use something like this:
/\.js\(['"]((?:\\['"]|[^"])+)['"]\)/g
regex101 demo

I believe it can be done in one-liner with replace and match method calls:
var str = '.js("aaa").js("bbb").js("ccc")';
str.replace(/[^(]*\("([^"]*)"\)[^(]*/g, '$1,').match(/[^,]+/g);
//=> ["aaa", "bbb", "ccc"]

The problem is that you are using .*. That will match any character. You'll have to be a bit more specific with what you are trying to capture.
If it will only ever be word characters you could use \w which matches any word character. This includes [a-zA-Z0-9_]: uppercase, lowercase, numbers and an underscore.
So your regex would look something like this :
var jsReg = /js\(['"](\w*)['"]\)/g;

In
/.js\(['"](.*)['"]\)/g
matches as much as possible, and does not capture group 1, so it matches
"aaa").js("bbb").js("ccc"
but given your example input.
Try
/\.js\(('(?:[^\\']|\\.)*'|"(?:[\\"]|\\.)*"))\)/
To break this down,
\. matches a literal dot
\.js\( matches the literal string ".js("
( starts to capture the string.
[^\\']|\\. matches a character other than quote or backslash or an escaped non-line terminator.
(?:[\\']|\\.)* matches the body of a string
'(?:[\\']|\\.)*' matches a single quoted string
(...|...) captures a single quoted or double quoted string
)\) closes the capturing group and matches a literal close parenthesis
The second major problem is your loop.
You're doing a global match repeatedly which makes no sense.
Get rid of the g modifier, and then things should work better.

Try this one - http://jsfiddle.net/UDYAq/
var str = new String('.js("aaa").js("bbb").js("ccc")');
var regex = /\.js\(\"(.*?)\"\){1,}/gi;
var result = [];
result = str.match (regex);
for (i in result) {
result[i] = result[i].match(/\"(.*?)\"/i)[1];
}
console.log (result);

To be sure that matched characters are surrounded by the same quotes:
/\.js\((['"])(.*?)\1\)/g

Why isn't this regex matching the expected way?

I'm trying to get rid of the slash character in case it exists at the end of my string. I used the following expression, intending to match any character not being slash at the end of the line.
var str = "http://hazaa.com/blopp/";
str.match("[^/$]+", "g");
For some reason (surely logical and explainable but not graspabled to me on my own), I get the split into three string looking as follows.
["http:", "hazaa.com", "blopp"]
What am I assuming wrongly?
How to resolve it?

In str.match("[^/$]+", "g");, why put dollar sign inside bracket? It's supposed to be outside, namely, str.match("[^/]+$", "g");.
To remove all the trailing slash, you can use str.replace(/\/+$/, ""). (If you'd like to remove the last trailing slash ONLY, remove the + in the replace's regex)
Update:
One more way that doesn't use replace:
function stripEndingSlashes(str) {
var matched = str.match("(.*[^/]+)/*$");
return matched ? matched[1] : "";
}

The regexp is choosing "everything except slash". That is why match() returns the parts of the string between slashes.
You can resolve it with the replace() function:
var str = "http://hazaa.com/blopp/";
//replace the last slash with an empty string to remove it
str = str.replace(/\/$/,'');
The regexp literal should always be surrounded between / characters. So here the regexp is:
\/ : this means a single slash character. In order to prevent Javascript from interpreting your slash as the end of regexp, it needs to be 'escaped' with a backslash.
$ : this means the end of the string

Your current regex will match the portion of string until the first / or $ is encountered. The second parameter is ignored; there is no second parameter for String.match.
To remove the trailing slash, use the String.replace function:
var str = "http://hazaa.com/blopp/";
str = str.replace(/\/$/, "");
console.log(str);
// "http://hazaa.com/blopp"
If you need to check whether a string ends with a slash, use the String.match method like this:
var str = "http://hazaa.com/blopp/";
var match = str.match(/\/$/);
console.log(match);
// null if string does not end with /
// ["/"] if string ends with a /
If you need to grab every thing except the last character(s) being /, use this:
var r = /(.+?)\/*$/;
console.log("http://hazaa.com/blopp//".match(r)); // ["http://hazaa.com/blopp//", "http://hazaa.com/blopp"]
console.log("http://hazaa.com/blopp/".match(r)); // ["http://hazaa.com/blopp/", "http://hazaa.com/blopp"]
console.log("http://hazaa.com/bloppA".match(r)); // ["http://hazaa.com/bloppA", "http://hazaa.com/bloppA"]
The 2nd index in the returned array contains the desired portion of the URL. The regex works as follows:
(.+?) un-greedy match (and capture) any character
\/*$ matches optional trailing slash(es)
The first portion regex is intentionally changed to un-greedy. If it was greedy, it would attempt to find the biggest match as long the the whole regex matches (consuming the trailing / in the process). When ungreedy, it will find the smallest match as long as the whole regex matches.

Why use regex? Just check if the last symbol of the string is a slash and then slice. Like this:
if (str.slice(-1) === '/') {
str = str.slice(0, -1);
}

How to convert a string with regex

How can i convert a string with regex, so that it contains just alphabetical (a-z) or a hyphen.
It should get rid " ' ! ? . and so on. Even if they appear multiple times.
// if i have e.g.
var test = '"test!!!"';
// how can i get the value "test"?
Can sombody help. RegEx is totaly new to me.

Simply replace the characters you don't want:
'"test!!!"'.replace(/[^a-z-]/gi, '')
[^a-z-] matches all characters but a-z and the hyphen. The /g flag makes the regexp apply multiple times. The /i flag (optional) makes it match case-insensitively, i.e. not replace upper-case characters.

That's quite simple: You build a character class that matches everything except those chars you want and remove them by replacing each occurence (global flag) with the empty string:
return str.replace(/[^a-z-]/g, "");

str = "hello!! my + name $ is slim-shady";
console.log(str.replace(/[^a-z-]+/g, ''));
$ node src/java/regex/alphanum.js
hellomynameisslim-shady

Use the replace method for any string variable and specify the characters you wish to remove.
Here's an example:
var sampleString = ("Hello World!!"); //Sample of what you have.
var holdData = sampleString.replace(/!!/gi, '');
window.alert(holdData);

var str = "test!!!";
str = str.replace(/[^A-Za-z\-]/g,"");