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How to regex test a string for a pattern while excluding certain characters?

I'm getting nowhere with this...
I need to test a string if it contains %2 and at the same time does not contain /. I can't get it to work using regex. Here is what I have:
var re = new RegExp(/.([^\/]|(%2))*/g);
var s = "somePotentially%2encodedStringwhichMayContain/slashes";
console.log(re.test(s)) // true
Question:
How can I write a regex that checks a string if it contains %2 while not containing any / slashes?

While the link referred to by Sebastian S. is correct, there's an easier way to do this as you only need to check if a single character is not in the string.
/^[^\/]*%2[^\/]*$/
EDIT: Too late... Oh well :P

Try the following:
^(?!.*/).*%2

either use inverse matching as shown here: Regular expression to match a line that doesn't contain a word?
or use indexOf(char) in an if statement. indexOf returns the position of a string or char in a string. If not found, it will return -1:
var s = "test/";
if(s.indexOf("/")!=-1){
//contains "/"
}else {
//doesn't contain "/"
}

Finding image url via using Regex

Any working Regex to find image url ?
Example :
var reg = /^url\(|url\(".*"\)|\)$/;
var string = 'url("http://domain.com/randompath/random4509324041123213.jpg")';
var string2 = 'url(http://domain.com/randompath/random4509324041123213.jpg)';
console.log(string.match(reg));
console.log(string2.match(reg));
I tied but fail with this reg
pattern will look like this, I just want image url between url(" ") or url( )
I just want to get output like http://domain.com/randompath/random4509324041123213.jpg
http://jsbin.com/ahewaq/1/edit

I'd simply use this expression:
/url.*\("?([^")]+)/
This returns an array, where the first index (0) contains the entire match, the second will be the url itself, like so:
'url("http://domain.com/randompath/random4509324041123213.jpg")'.match(/url.*\("?([^")]+)/)[1];
//returns "http://domain.com/randompath/random4509324041123213.jpg"
//or without the quotes, same return, same expression
'url(http://domain.com/randompath/random4509324041123213.jpg)'.match(/url.*\("?([^")]+)/)[1];
If there is a change that single and double quotes are used, you can simply replace all " by either '" or ['"], in this case:
/url.*\(["']?([^"')]+)/

Try this regexp:
var regex = /\burl\(\"?(.*?)\"?\)/;
var match = regex.exec(string);
console.log(match[1]);
The URL is captured in the first subgroup.

If the string will always be consistent, one option would be simply to remove the first 4 characters url(" and the last two "):
var string = 'url("http://domain.com/randompath/random4509324041123213.jpg")';
// Remove last two characters
string = string.substr(0, string.length - 2);
// Remove first five characters
string = string.substr(5, string.length);
Here's a working fiddle.
Benefit of this approach: You can edit it yourself, without asking StackOverflow to do it for you. RegEx is great, but if you don't know it, peppering your code with it makes for a frustrating refactor.

JavaScript Regex to check that the searched string is preceded by either nothing, or spaces

I have the following code:
var requestData = {};
var byPattern = /by=(\w+)/;
var value = byPattern.exec(stringToSearch);
if (value && value.length === 2)
requestData.by = value[1];
The first problem with this regex, was that if stringToSearch is for example "standby=foo", it matched and returned "foo". I want it to fail there. I want the regex to match only if there is nothing before "by", or spaces.
So I replaced by /^\s*by=(\w+)/
That's better, but I want the regex to match if stringToSearch is for example "city=paris by=foo". It should match and return "foo". Not the case here :/
Can someone help me fix the regex? Thanks a lot!

the magic word is \b:
var byPattern = /\bby=(\w+)/;
console.log(byPattern.exec("by=3"));
console.log(byPattern.exec(" by=3"));
console.log(byPattern.exec("standby=3"));
Only the first two are matched, which is (I think) what you want.

If you want spaces or nothing before the capturing group, that would be ( +|^):
/( |^)by=(\w+)/
Technically, I'm matching space or nothing, but the effect is the same.

How can I get a substring located between 2 quotes?

I have a string that looks like this: "the word you need is 'hello' ".
What's the best way to put 'hello' (but without the quotes) into a javascript variable? I imagine that the way to do this is with regex (which I know very little about) ?
Any help appreciated!

Use match():
> var s = "the word you need is 'hello' ";
> s.match(/'([^']+)'/)[1];
"hello"
This will match a starting ', followed by anything except ', and then the closing ', storing everything in between in the first captured group.

http://jsfiddle.net/Bbh6P/
var mystring = "the word you need is 'hello'"
var matches = mystring.match(/\'(.*?)\'/); //returns array
​alert(matches[1]);​

If you want to avoid regular expressions then you can use .split("'") to split the string at single quotes , then use jquery.map() to return just the odd indexed substrings, ie. an array of all single-quoted substrings.
var str = "the word you need is 'hello'";
var singleQuoted = $.map(str.split("'"), function(substr, i) {
return (i % 2) ? substr : null;
});
DEMO
CAUTION
This and other methods will get it wrong if one or more apostrophes (same as single quote) appear in the original string.

Extracting numbers from a string using regular expressions

I am clueless about regular expressions, but I know that they're the right tool for what I'm trying to do here: I'm trying to extract a numerical value from a string like this one:
approval=not requested^assignment_group=12345678901234567890123456789012^category=Test^contact_type=phone^
Ideally, I'd extract the following from it: 12345678901234567890123456789012 None of the regexes I've tried have worked. How can I get the value I want from this string?

This will get all the numbers:
var myValue = /\d+/.exec(myString)

mystr.match(/assignment_group=([^\^]+)/)[1]; //=> "12345678901234567890123456789012"
This will find everything from the end of "assignment_group=" up to the next caret ^ symbol.

Try something like this:
/\^assignment_group=(\d*)\^/
This will get the number for assignment_group.
var str = 'approval=not requested^assignment_group=12345678901234567890123456789012^category=Test^contact_type=phone^',
regex = /\^assignment_group=(\d*)\^/,
matches = str.match(regex),
id = matches !== null ? matches[1] : '';
console.log(id);

If there is no chance of there being numbers anywhere but when you need them, you could just do:
\d+
the \d matches digits, and the + says "match any number of whatever this follows"