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JavaScript - strip everything before and including a character

I am relatively new to RegEx and am trying to achieve something which I think may be quite simple for someone more experienced than I.
I would like to construct a snippet in JavaScript which will take an input and strip anything before and including a specific character - in this case, an underscore.
Thus 0_test, 1_anotherTest, 2_someOtherTest would become test, anotherTest and someOtherTest, respectively.
Thanks in advance!

You can use the following regex (which can only be great if your special character is not known, see Alex's solution for just _):
^[^_]*_
Explanation:
^ - Beginning of a string
[^_]* - Any number of characters other than _
_ - Underscore
And replace with empty string.
var re = /^[^_]*_/;
var str = '1_anotherTest';
var subst = '';
document.getElementById("res").innerHTML = result = str.replace(re, subst);
<div id="res"/>
If you have to match before a digit, and you do not know which digit it can be, then the regex way is better (with the /^[^0-9]*[0-9]/ or /^\D*\d/ regex).

Simply read from its position to the end:
var str = "2_someOtherTest";
var res = str.substr(str.indexOf('_') + 1);

Regex trying to match characters before and after symbol

I'm trying to match characters before and after a symbol, in a string.
string: budgets-closed
To match the characters before the sign -, I do: ^[a-z]+
And to match the other characters, I try: \-(\w+) but, the problem is that my result is: -closed instead of closed.
Any ideas, how to fix it?
Update
This is the piece of code, where I was trying to apply the regex http://jsfiddle.net/trDFh/1/
I repeat: It's not that I don't want to use split; it's just I was really curious, and wanted to see, how can it be done the regex way. Hacking into things spirit
Update2
Well, using substring is a solution as well: http://jsfiddle.net/trDFh/2/ and is the one I chosed to use, since the if in question, is actually an else if in a more complex if syntax, and the chosen solutions seems to be the most fitted for now.

Use exec():
var result=/([^-]+)-([^-]+)/.exec(string);
result is an array, with result[1] being the first captured string and result[2] being the second captured string.
Live demo: http://jsfiddle.net/Pqntk/

I think you'll have to match that. You can use grouping to get what you need, though.
var str = 'budgets-closed';
var matches = str.match( /([a-z]+)-([a-z]+)/ );
var before = matches[1];
var after = matches[2];
For that specific string, you could also use
var str = 'budgets-closed';
var before = str.match( /^\b[a-z]+/ )[0];
var after = str.match( /\b[a-z]+$/ )[0];
I'm sure there are better ways, but the above methods do work.

If the symbol is specifically -, then this should work:
\b([^-]+)-([^-]+)\b
You match a boundry, any "not -" characters, a - and then more "not -" characters until the next word boundry.
Also, there is no need to escape a hyphen, it only holds special properties when between two other characters inside a character class.
edit: And here is a jsfiddle that demonstrates it does work.

How do I make a regular expression that matches everything on a line after a given character?

If I have a String in JavaScript
key=value
How do I make a RegEx that matches key excluding =?
In other words:
var regex = //Regular Expression goes here
regex.exec("key=value")[0]//Should be "key"
How do I make a RegEx that matches value excluding =?
I am using this code to define a language for the Prism syntax highlighter so I do not control the JavaScript code doing the Regular Expression matching nor can I use split.

Well, you could do this:
/^[^=]*/ // anything not containing = at the start of a line
/[^=]*$/ // anything not containing = at the end of a line
It might be better to look into Prism's lookbehind property, and use something like this:
{
'pattern': /(=).*$/,
'lookbehind': true
}
According to the documentation this would cause the = character not to be part of the token this pattern matches.

use this regex (^.+?)=(.+?$)
group 1 contain key
group 2 contain value
but split is better solution

.*=(.*)
This will match anything after =
(.*)=.*
This will match anything before =
Look into greedy vs ungreedy quantifiers if you expect more than one = character.
Edit: as OP has clarified they're using javascript:
var str = "key=value";
var n=str.match(/(.*)=/i)[1]; // before =
var n=str.match(/=(.*)/i)[1]; // after =

var regex = /^[^=]*/;
regex.exec("key=value");

Can regex matches in javascript match any word after an equal operator?

I am trying to target ?state=wildcard in this statement :
?state=uncompleted&dancing=yes
I would like to target the entire line ?state=uncomplete, but also allow it to find whatever word would be after the = operator. So uncomplete could also be completed, unscheduled, or what have you.
A caveat I am having is granted I could target the wildcard before the ampersand, but what if there is no ampersand and the param state is by itself?

Try this regular expression:
var regex = /\?state=([^&]+)/;
var match = '?state=uncompleted&dancing=yes'.match(regex);
match; // => ["?state=uncompleted", "uncompleted"]
It will match every character after the string "\?state=" except an ampersand, all the way to the end of the string, if necessary.

Alternative regex: /\?state=(.+?)(?:&|$)/
It will match everything up to the first & char or the end of the string

IMHO, you don't need regex here. As we all know, regexes tend to be slow, especially when using look aheads. Why not do something like this:
var URI = '?state=done&user=ME'.split('&');
var passedVals = [];
This gives us ['?state=done','user=ME'], now just do a for loop:
for (var i=0;i<URI.length;i++)
{
passedVals.push(URI[i].split('=')[1]);
}
Passed Vals wil contain whatever you need. The added benefit of this is that you can parse a request into an Object:
var URI = 'state=done&user=ME'.split('&');
var urlObjects ={};
for (var i=0;i<URI.length;i++)
{
urlObjects[URI[i].split('=')[0]] = URI[i].split('=')[1];
}
I left out the '?' at the start of the string, because a simple .replace('?','') can fix that easily...

You can match as many characters that are not a &. If there aren't any &s at all, that will of course also work:
/(\?state=[^&]+)/.exec("?state=uncompleted");
/(\?state=[^&]+)/.exec("?state=uncompleted&a=1");
// both: ["?state=uncompleted", "?state=uncompleted"]

RegExp.test not working?

I am trying to validate year using Regex.test in javascript, but no able to figure out why its returning false.
var regEx = new RegExp("^(19|20)[\d]{2,2}$");
regEx.test(inputValue) returns false for input value 1981, 2007
Thanks

As you're creating a RegExp object using a string expression, you need to double the backslashes so they escape properly. Also [\d]{2,2} can be simplified to \d\d:
var regEx = new RegExp("^(19|20)\\d\\d$");
Or better yet use a regex literal to avoid doubling backslashes:
var regEx = /^(19|20)\d\d$/;

Found the REAL issue:
Change your declaration to remove quotes:
var regEx = new RegExp(/^(19|20)[\d]{2,2}$/);

Do you mean
var inputValue = "1981, 2007";
If so, this will fail because the pattern is not matched due to the start string (^) and end string ($) characters.
If you want to capture both years, remove these characters from your pattern and do a global match (with /g)
var regEx = new RegExp(/(?:19|20)\d{2}/g);
var inputValue = "1981, 2007";
var matches = inputValue.match(regEx);
matches will be an array containing all matches.

I've noticed, for reasons I can't explain, sometimes you have to have two \\ in front of the d.
so try [\\d] and see if that helps.