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JavasScript Regex to exclude specific characters which are stored in a string variable, spaces, and change remaining characters to underscores

This is for a hangman-type guessing game. I already figured out how to use a Regex to display the letters as underscores on the page with appropriate spacing. Now I want use a Regex to do the following, all in one expression:
Check a string containing the correct answer this.answers[arraysIndex], against the string containing all of the user's correct guesses rightString
In the correct answer string: change only the letters that don't match the correct guesses string into underscores. This means I want to keep the spaces unchanged too.
I've tried this:
var regex = new RegExp("/(?![^"+rightString+"])/[\A-Za-z/])/","g");
newDisplay = (this.answers[arraysIndex]).replace(regex, "_");
...and this:
newDisplay = (this.answers[arraysIndex]).replace("/(?![^+rightString+])/[\A-Za-z/])/g", "_")
...and countless slight variations of each. I'm not married to the idea of using a string variable, I could use an array variable too, or maybe there's something that hasn't even occurred to me. I've researched exhaustively on here and many other resources (that's how I solved my first problem) but this one's got me beat. Any help is greatly appreciated.

Why would the replace with regex not work even though the regex does?

There may be a very simple answer to this, probably because of my familiarity (or possibly lack thereof) of the replace method and how it works with regex.
Let's say I have the following string: abcdefHellowxyz
I just want to strip the first six characters and the last four, to return Hello, using regex... Yes, I know there may be other ways, but I'm trying to explore the boundaries of what these methods are capable of doing...
Anyway, I've tinkered on http://regex101.com and got the following Regex worked out:
/^(.{6}).+(.{4})$/
Which seems to pass the string well and shows that abcdef is captured as group 1, and wxyz captured as group 2. But when I try to run the following:
"abcdefHellowxyz".replace(/^(.{6}).+(.{4})$/,"")
to replace those captured groups with "" I receive an empty string as my final output... Am I doing something wrong with this syntax? And if so, how does one correct it, keeping my original stance on wanting to use Regex in this manner...
Thanks so much everyone in advance...

The code below works well as you wish
"abcdefHellowxyz".replace(/^.{6}(.+).{4}$/,"$1")
I think that only use ()to capture the text you want, and in the second parameter of replace(), you can use $1 $2 ... to represent the group1 group2.
Also you can pass a function to the second parameter of replace,and transform the captured text to whatever you want in this function.
For more detail, as #Akxe recommend , you can find document on https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace.

You are replacing any substring that matches /^(.{6}).+(.{4})$/, with this line of code:
"abcdefHellowxyz".replace(/^(.{6}).+(.{4})$/,"")
The regex matches the whole string "abcdefHellowxyz"; thus, the whole string is replaced. Instead, if you are strictly stripping by the lengths of the extraneous substrings, you could simply use substring or substr.
Edit
The answer you're probably looking for is capturing the middle token, instead of the outer ones:
var str = "abcdefHellowxyz";
var matches = str.match(/^.{6}(.+).{4}$/);
str = matches[1]; // index 0 is entire match
console.log(str);

Grabbing the third fragment between square brackets

Still completely stuck with regex's and square brackets. Hopefully someone can help me out.
Say I have a string like this:
room_request[1][1][2011-08-21]
How would I grab the third fragment out of it?
I tried the following, but I'm not exactly sure what I'm doing so it's fairly hard to figure out where I'm going wrong.
.match(/\[(.*?)\]/);
But this returns the [1] fragment. (The first one, I guess).
So then, I asked here on SO and people told me to add a global flag:
.match(/\[(.*?)\]/g)[2];
In other cases that I've used this regex, this worked fine. However, in this case, I want the stuff INSIDE the square brackets. It returns:
[2011-08-21]
But I really want 2011-08-21.
How can I do this? Thanks a lot.
If anyone could recommend any decent resources about regular expressions, that'd be great aswell. I'm starting to understand the very basics but most of this stuff is far too confusing atm. Thanks.

Two possible methods. To grab the third bracketed expression:
.match(/\[.*?\]\[.*?\]\[(.*?)\]/);
Or, if you know that the expression you want is always at the end of the string:
.match(/\[(.*?)\]$/);

var str = "room_request[1][1][2011-08-21]"
var val = str.match(/\[[^\]]*\]\[[^\]]*\]\[([^\]]*)\]/);
alert(val[1]);

This is a little less messy I think:
var r = "room_request[1][1][2011-08-21]";
var match = r.match(/(?:\[([^\]]+)\]){3}/);
console.log(match[1]);
Basically, it picks out the third match of the square brackets containing something. You get the match result back with two matches - the whole [1][1][2011-08-21] (for whatever reason) and the matched date: 2011-08-21
My regex is a little rusty, but this certainly works.

Strip brackets and the word "The" from a string

I have been trying to figure out regex myself and to no avail can I get the last bracket to disappear from a string.
For example:
[The Day the Earth Stood Still]
I can only get:
Day the Earth Stood Still]
with the following RegEx code:
/(\[|\](^The ))\2/
I'm aiming for just:
Day the Earth Stood Still
Any help would be greatly appreciated. I've spent 3 hours trying to figure it out on my own... This is me giving in. :3

You can try:
\[The\s(.*)]
If you need this to work to strip out brackets even when 'The' is not present you can try:
(?:\[The\s|\[)(.*)]
If you think you will run into a case where you may have 'the' or 'The' you can try:
(?:\[[Tt]he\s|\[)(.*)]
Here is some code to implement capturing text without the brackets and 'The':
var title = new Array();
title[0] = "[The Day the Earth Stood Still]";
title[1] = "[Independence Day]";
title[2] = "[the Day the Earth Stood Still]";
alert(title[0].match(/(?:\[[Tt]he\s|\[)(.*)]/)[1]);
alert(title[1].match(/(?:\[[Tt]he\s|\[)(.*)]/)[1]);
alert(title[2].match(/(?:\[[Tt]he\s|\[)(.*)]/)[1]);
Try it out here: http://jsfiddle.net/aSwYz/

Maybe you should take the approach of thinking of what you want to keep instead of what you want to get rid of
(?<=\[The\s*)[\w\s]*(?=\])
Otherwise, you are trying to do a match in separate parts of the string and should be done with 2 different matches

Check that the user is entering a time format? eg 13:00

Basically, I'd like to have an input that when blur'd, will check the input to make sure it's in the format...
[24hour] : [minutes]
So for example 13:00, or 15:30.
So I guess I have to split it up into three parts, check the first bit is between 0 and 24, then check it has the semi-colon, then check it has a number between 0 and 60.
Going more complicated than that, it'd be fantastic to have it so if the user enters 19 it'll complete it as 19:00 for example.
I am using jQuery fwiw, but regular code is fine, for example I'm using this little piece of code so far which works fine, to convert . inputs to :
tempval = $(this).val().replace(".", ":");
$(this).val(tempval);
Not sure where to start with this, if anyone could recommend me some reading that'd be fantastic, thank you!

([0-1 ]?[0-9]|2[0-3]):([0-5][0-9])
I think that's the regex you're looking for (not specifically for javascript though).
http://www.regular-expressions.info/
This site has an excellent amount of info for language-specific regular expressions! Cheers!

I suggest using masked input That way the wrong input will be prevented in the first place.
Disclaimer: I haven't used that plugin myself, just found it by keywords "masked input"

There are a bunch of widgets that already deal with time validation - try googling for "jQuery time widget" - the first result doesn't look bad.

var re = /^(\d+)(:\d+)?$/;
var match = re.match(yourstring);
Now if the match has succeeded match is an array with the matched pieces: match[0] is the whole of yourstring (you don't care about that), match[1] has the digits before the colon (if any colon, else just digits), match[2] if it exists has the colon followed by the digits after it. So now you just need to perform your numeric tests on match[1], and possibly match[2] minus the leading colon, to ensure the numbers are correct.