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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);

JavaScript: Replacing characters on both sides of a string

What I want to do is to match characters enclosed by ^^ and replace those ^^ while maintaining the string. In other words, turning this:
^^This is a test^^ this is ^^another test^^
into this:
<sc>This is a test</sc> this is <sc>another test</sc>
I got the regex to match them:
\^{2}[^^]+\^{2}
But I'm stuck there. I'm not sure what to do with the other .replace parameter:
.replace(/\^{2}[^^]+\^{2}/g, WHAT_TO_ADD_HERE?)
Any ideas?

You can use replace with regex and grouping like
var text = '^^This is a test^^ this is ^^another test^^'.replace(/\^\^(.*?)\^\^/g, '<sc>$1</sc>')

Here is a piece of code you can use:
var re = /(\^{2})([^^]+)(\^{2})/g;
var str = '^^This is a test^^ this is ^^another test^^\n\n<sc>This is a test</sc> this is <sc>another test</sc>';
var subst = '<sc>$2</sc>';
var result = str.replace(re, subst);
This is just an enhancement of your regex pattern where I added capturing groups. To improve performance and ensure you will be capturing all symbols between the ^^, you can use only one capturing group and . symbol with non-greedy quantificator:
var re = /\^{2}(.+?)\^{2}/g;
Have a look at the example.

In this case you need to use the group index to wrap the content.
var content = "^^This is a test^^ this is ^^another test^^";
content.replace(/\^{2}(.*?)\^{2}/g, '<sc>$1</sc>');
The (.*?) will help you to group the content and in your replace statement use $1 where 1 is the index of group.

non-capture group still showing in match

I know this topic has been thoroughly covered on StackOverflow, but I can't for the life of me get my regular expression to work. So without further repetitive ado ...
This is what I have.
String: <p model='cat'></p>
Regex: .match(/(?:model=')(.*)(?:')/g)
This is what my expression returns: model='cat'
This is what I want: cat
Why isn't my non capture group ignored? Is it that I don't understand what a non-capturing group does? Why isn't my Regex working?

The entire match will always be group 0, you need to access that specific group (group 1 in this case since the first group is non-capture), you can do it like this:
var str = "<p model='cat'></p>";
var regex = /(?:model=')(.*)(?:')/g
var match = regex.exec(str);
alert(match[1]); // cat
Fiddle
Also, I suppose you are probably wanting several matches within str, you could do that like this:
var str = "<p model='cat'></p><p model='dog'></p><p model='horse'></p>";
var regex = /(?:model=')([^']*)/g
var matches = [];
var match;
while (match = regex.exec(str)) {
matches.push(match[1]);
}
alert(matches); // cat,dog,horse
Fiddle

A non-capturing group is basically just a non-group ― a way to use parentheses without actually treating that part of the pattern as a group.
It looks like what you're actually looking for are the "match prefix but exclude" group (?<=) and the "match suffix but exclude" group (?=).
Note: This type of group does not seem to be supported in Internet Explorer.
If you use these, you get the desired result:
var str = "<p model='cat'></p><p model='dog'></p><p model='horse'></p>";
var regex = /(?<=model=')[^']*(?=')/g
var matches = str.match(regex);
console.log(matches);

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");

Regex to Match Phrase ending with a question mark

I'm trying to figure out a javascript regex that'll match an exact phrase that ends with a question mark, but isn't wrapped in quotes. So far I have this, which matches the phrase "some phrase", but I can't figure out how to match "some phrase?". Any help would be greatly appreciated.
(?<!"|')\some phrase\b(?!"|')

Lookbehinds don't exist in JavaScript. Use the following pattern:(?:[^"']|^)(some phrase\?)(?!["']). [^"']|^ means: any non-quote character or the beginning of a string.
Example:
var text = "....";
var pattern = /(?:[^"']|^)(some phrase\?)(?!["'])/;
var string = text.match(pattern);
var desiredString = string[1]; //Get the grouped text
var patternWithNQuoteGrouped = /([^"']|^)(some phrase\?)(?!["'])/;//Notice: No ?:
var replaceString = text.replace(patternWithNQuoteGrouped, '$1$2');
//$1 = non-quote character $2 = matched phrase
The parentheses around the phrase mark a referable group. (?: means: Create a group, but dereference it. To refer back to it, see the example code. Because lookbehinds don't exist in JavaScript, it's not possible to create a pattern which checks whether a prefix does not exist.

Try this:
var expr = /(^|(?!["']).)(some phrase\?)($|(?!["']).)/;
if (expr.test(searchText)) {
var matchingPhrase = RegExp.$2;
}
http://jsfiddle.net/gilly3/zCUsg/