Develop Reference

Use only one of the characters in regular expression javascript

I guess that should be smth very easy, but I'm stuck with that for at least 2 hours and I think it's better to ask the question here.
So, I've got a reg expression /&t=(\d*)$/g and it works fine while it is not ?t instead of &t in url. I've tried different combinations like /\?|&t=(\d*)$/g ; /\?t=(\d*)$|/&t=(\d*)$/g ; /(&|\?)t=(\d*)$/g and various others. But haven't got the expected result which is /\?t=(\d*)$/g or /&t=(\d*)$/g url part (whatever is placed to input).
Thx for response. I think need to put some details here. I'm actually working on this peace of code
var formValue = $.trim($("#v").val());
var formValueTime = /&t=(\d*)$/g.exec(formValue);
if (formValueTime && formValueTime.length > 1) {
formValueTime = parseInt(formValueTime[1], 10);
formValue = formValue.replace(/&t=\d*$/g, "");
}
and I want to get the t value whether reference passed with &t or ?t in references like youtu.be/hTWKbfoikeg?t=82 or similar one youtu.be/hTWKbfoikeg&t=82

To replace, you may use
var formValue = "some?some=more&t=1234"; // $.trim($("#v").val());
var formValueTime;
formValue = formValue.replace(/[&?]t=(\d*)$/g, function($0,$1) {
formValueTime = parseInt($1,10);
return '';
});
console.log(formValueTime, formValue);
To grab the value, you may use
/[?&]t=(\d*)$/g.exec(formValue);
Pattern details
[?&] - a character class matching ? or &
t= - t= substring
(\d*) - Group 1 matching zero or more digits
$ - end of string

/\?t=(\d*)|\&t=(\d*)$/g
you inverted the escape character for the second RegEx.
http://regexr.com/3gcnu

I want to thank you all guys for trying to help. Special thanks to #Wiktor Stribiżew who gave the closest answer.
Now the piece of code I needed looks exactly like this:
/[?&]t=(\d*)$/g.exec(formValue);
So that's the [?&] part that solved the problem.
I use array later, so /\?t=(\d*)|\&t=(\d*)$/g doesn't help because I get an array like [t&=50,,50] when reference is & type and the correct answer [t?=50,50] when reference is ? type just because of the order of statements in RegExp.
Now, if you're looking for a piece of RegExp that picks either character in one place while the rest of RegExp remains the same you may use smth like this [?&] for the example where wanted characters are ? and &.

JS / RegEx to remove characters grouped within square braces

I hope I can explain myself clearly here and that this is not too much of a specific issue.
I am working on some javascript that needs to take a string, find instances of chars between square brackets, store any returned results and then remove them from the original string.
My code so far is as follows:
parseLine : function(raw)
{
var arr = [];
var regex = /\[(.*?)]/g;
var arr;
while((arr = regex.exec(raw)) !== null)
{
console.log(" ", arr);
arr.push(arr[1]);
raw = raw.replace(/\[(.*?)]/, "");
console.log(" ", raw);
}
return {results:arr, text:raw};
}
This seems to work in most cases. If I pass in the string [id1]It [someChar]found [a#]an [id2]excellent [aa]match then it returns all the chars from within the square brackets and the original string with the bracketed groups removed.
The problem arises when I use the string [id1]It [someChar]found [a#]a [aa]match.
It seems to fail when only a single letter (and space?) follows a bracketed group and starts missing groups as you can see in the log if you try it out. It also freaks out if i use groups back to back like [a][b] which I will need to do.
I'm guessing this is my RegEx - begged and borrowed from various posts here as I know nothing about it really - but I've had no luck fixing it and could use some help if anyone has any to offer. A fix would be great but more than that an explanation of what is actually going on behind the scenes would be awesome.
Thanks in advance all.

You could use the replace method with a function to simplify the code and run the regexp only once:
function parseLine(raw) {
var results = [];
var parsed = raw.replace(/\[(.*?)\]/g, function(match,capture) {
results.push(capture);
return '';
});
return { results : results, text : parsed };
}

The problem is due to the lastIndex property of the regex /\[(.*?)]/g; not resetting, since the regex is declared as global. When the regex has global flag g on, lastIndex property of RegExp is used to mark the position to start the next attempt to search for a match, and it is expected that the same string is fed to the RegExp.exec() function (explicitly, or implicitly via RegExp.test() for example) until no more match can be found. Either that, or you reset the lastIndex to 0 before feeding in a new input.
Since your code is reassigning the variable raw on every loop, you are using the wrong lastIndex to attempt the next match.
The problem will be solved when you remove g flag from your regex. Or you could use the solution proposed by Tibos where you supply a function to String.replace() function to do replacement and extract the capturing group at the same time.

You need to escape the last bracket: \[(.*?)\].

Changing character between two strings

How can i change the character after "#overlay/" and before "/" after that first one?
var x = "www.foo.com/#overlay/2/";
x.replace(/#overlay\/([^]*)\//, "1"); // i'm expecting: www.foo.com/#overlay/1/
I'm using this code, but no success. I don't understand that much from regex.
I've searched some questions but without success.

I would not use a regular expression here. You can just use .split().
var url, newUrl, peices;
url = 'www.foo.com/#overlay/2/';
// Split the string apart by /
peices = url.split('/');
// Changing the 3 element in the array to 1, it was originally 2.
peices[2] = 1;
// Let's put it back together...
newUrl = peices.join('/');

You're making 3 mistakes :
you're replacing too much
you don't use the returned value. replace doesn't change the passed string (strings are immutable) but returns a new one
you forgot to precise in your capturing group when to stop (in fact it doesn't even have to be a capturing group)
You can do this :
x = x.replace(/(#overlay\/)[^\/]*\//, "$11/");
$1 here refers to the first captured group, so that you don't have to type it in the replacement string.
For example it changes
"www.foo.com/#overlay/2/rw/we/2345"
into
"www.foo.com/#overlay/1/rw/we/2345"

regex - get numbers after certain character string

I have a text string that can be any number of characters that I would like to attach an order number to the end. Then I can pluck off the order number when I need to use it again. Since there's a possibility that the number is variable length, I would like to do a regular expression that catch's everything after the = sign in the string ?order_num=
So the whole string would be
"aijfoi aodsifj adofija afdoiajd?order_num=3216545"
I've tried to use the online regular expression generator but with no luck. Can someone please help me with extracting the number on the end and putting them into a variable and something to put what comes before the ?order_num=203823 into its own variable.
I'll post some attempts of my own, but I foresee failure and confusion.

var s = "aijfoi aodsifj adofija afdoiajd?order_num=3216545";
var m = s.match(/([^\?]*)\?order_num=(\d*)/);
var num = m[2], rest = m[1];
But remember that regular expressions are slow. Use indexOf and substring/slice when you can. For example:
var p = s.indexOf("?");
var num = s.substring(p + "?order_num=".length), rest = s.substring(0, p);

I see no need for regex for this:
var str="aijfoi aodsifj adofija afdoiajd?order_num=3216545";
var n=str.split("?");
n will then be an array, where index 0 is before the ? and index 1 is after.
Another example:
var str="aijfoi aodsifj adofija afdoiajd?order_num=3216545";
var n=str.split("?order_num=");
Will give you the result:
n[0] = aijfoi aodsifj adofija afdoiajd and
n[1] = 3216545

You can substring from the first instance of ? onward, and then regex to get rid of most of the complexities in the expression, and improve performance (which is probably negligible anyway and not something to worry about unless you are doing this over thousands of iterations). in addition, this will match order_num= at any point within the querystring, not necessarily just at the very end of the querystring.
var match = s.substr(s.indexOf('?')).match(/order_num=(\d+)/);
if (match) {
alert(match[1]);
}

Javascript regex match for string "game_1"

I just can't get this thing to work in javascript. So, I have a text "game_1" without the quotes and now i want to get that number out of it and I tried this:
var idText = "game_1";
re = /game_(.*?)/;
found = idText.match(re);
var ajdi = found[1];
alert( ajdi );
But it doesn't work - please point out where am I going wrong.

If you're only matching a number, you may want to try
/game_([0-9]+)/
as your regular expression. That will match at least one number, which seems to be what you need. You entered a regexp that allows for 0 characters (*) and let it select the shortest possible result (?), which may be a problem (and match you 0 characters), depending on the regex engine.

If this is the complete text, then there is no need for regular expressions:
var id = +str.split('_')[1];
or
var id = +str.replace('game_', '');
(unary + is to convert the string to a number)
If you insist on regular expression, you have to anchor the expression:
/^game_(.*?)$/
or make the * greedy by omitting the ?:
/game_(.*)/
Better is to make the expression more restrictive as #Naltharial suggested.

Simple string manipulation:
var idText = "game_1",
adji = parseInt(idText.substring(5), 10);

* means zero or more occurrences. It seems that combining it with a greediness controller ? results in zero match.
You could replace * with + (which means one or more occurrences), but as #Felix Kling notes, it would only match one digit.
Better to ditch the ? completely.
http://jsfiddle.net/G8Qt7/2/

Try "game_1".replace(/^(game_)/, '')
this will return the number

You can simply use this re /\d+/ to get any number inside your string