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Regex: How do I remove the character BEFORE the matched string?

I am intercepting messages which contain the following characters:
*_-
However, whenever any one of these characters comes through, it will always be preceded by a \. The \ is just for formatting though and I want to remove it before sending it off to my server. I know how to easily create a regex which would remove this backslash from a single letter:
'omg\_bbq\_everywhere'.replace(/\\_/g, '')
And I recognize I could just do this operation 3 times: once for each character I want to remove the preceding backslash for. But how can I create a single regex which would detect all three characters and remove the preceding backslash in all 3 cases?

You can use a character class like [*_-].
To remove only the backslash before these characters:
document.body.innerHTML =
"omg\\-bbq\\*everywhere\\-".replace(/\\([*_-])/g, '$1');
When you place a subpattern into a capturing group ((...)), you capture that subtext into a numbered buffer, and then you can reference it with a $1 backreference (1 because there is only one (...) in the pattern.)

This is a good time to use atomic matching. Specifically you want to check for the slash and then positive lookahead for any of those characters.
Ignoring the code, the raw regex you want is:
\\(?=[*_-])
A literal backslash, with one of these characters in front of it: *_-
So now you are matching the slash. The atomic match is a 0 length match, so it doesn't match anything, but sets a requirement that "for this to be a valid match, it needs to be followed by [*_-]"
Atomic groups: http://www.regular-expressions.info/atomic.html
Lookaround statements: http://www.regular-expressions.info/lookaround.html
Positive and negative lookahead and lookbehind matches are available.

JavaScript regular expressions to match no digits, whitespace and selected symbols

Thanks for taking a look.
My goal is to come up with a regexp that will match input that contains no digits, whitespace or the symbols !#£$%^&*()+= or any other symbol I may choose.
I am however struggling to grasp precisely how regular expressions work.
I started out with the simple pattern /\D/, which from my understanding will match the first non-digit character it can find. This would match the string 'James' which is correct but also 'James1' which I don't want.
So, my understanding is that if I want to ensure that a pattern is not found anywhere in a given string, I use the ^ and $ characters, as in /^\D$/. Now because this will only match a single character that is not a digit, I needed to use + to specify that 1 or more digits should not be founds in the entire string, giving me the expression /^\D+$/. Brilliant, it no longer matches 'James1'.
Question 1
Is my reasoning up to this point correct?
The next requirement was to ensure no whitespace is in the given string. \s will match a single whitespace and [^\s] will match the first non-whitespace character. So, from my understanding I just had to add this to what I have already to match strings that contain no digits and no whitespace. Again, because [^\s] will only match a single non-white space character, I used + to match one or more whitespace characters, giving the new regexp of /^\D+[^\s]+$/.
This is where I got lost, as the expression now matches 'James1' or even 'James Smith25'. What? Massively confused at this point.
Question 2
Why is /^\D+[^\s]+$/ matching strings that contain spaces?
Question 3
How would I go about writing the regular expression I'm trying to solve?
While I am keen to solve the problem I am more interested in figuring where my understanding of regular expressions is lacking, so any explanations would be helpful.

Not quite; ^ and $ are actually "anchors" - they mean "start" and "end", it's actually a little more complicated, but you can consider them to mean the start and end of a line for now - look up the various modifiers on regular expressions if you're interested in learning more about this. Unfortunately ^ has an overloaded meaning; if used inside square brackets it means "not", which is the meaning you are already acquainted with. It's very important that you understand the difference between these two meanings and that the definition in your head actually applies only to character range matching!
Contributing further to your confusion is that \d means "a numerical digit" and \D means "not a numerical digit". Similarly \s means "a whitespace (space/tab/newline/etc.) character" and \S means "not a whitespace character."
It's worth noting that \d is effectively a shortcut for [0-9] (note that - has a special meaning inside square brackets), and \D is a shortcut for [^0-9].
The reason it's matching strings that contain spaces is that you've asked for "1+ non-numerical digits followed by 1+ non-space characters" - so it'll match lots of strings! I think that perhaps you don't understand that regular expressions match bits of strings, you're not adding constraints as you go, but rather building up bots of matchers that will match bits of corresponding strings.
/^[^\d\s!#£$%^&*()+=]+$/ is the answer you're looking for - I'd look at it like this:
i. [] - match a range of characters
ii. []+ - match one or more of that range of characters
iii. [^\d\s]+ - match one or more characters that do not match \d (numerical digit) or \s (whitespace)
iv. [^\d\s!#£$%^&*()+=]+ - here's a bunch of other characters I don't want you to match
v. ^[^\d\s!#£$%^&*()+=]+$ - now there are anchors applied, so this matcher has to apply to the whole line otherwise it fails to match
A useful website to explore regexs is http://regexr.com/3b9h7 - which I supply with my suggested solution as an example. Edit: Pruthvi Raj's link to debuggerx is awesome!

Is my reasoning up to this point correct?
Almost. /\D/ matches any character other than a digit, but not just the first one (if you use g option).
and [^\s] will match the first non-whitespace character
Almost, [^\s] will match any non-whitespace character, not just the first one (if you use g option).
/^\D+[^\s]+$/ matching strings that contain spaces?
Yes, it does, because \D matches a space (space is not a digit).
Why is /^\D+[^\s]+$/ matching strings that contain spaces?
Because \D+ in /^\D+[^\s]+$/can match spaces.
Conclusion:
Use
^[^\d\s!#£$%^&*()+=]+$
It will match strings that have no digits and spaces, and the symbols you do not allow.
Mind that to match a literal -, ] or [ with a character class, you either need to escape them, or use at the start or end of the expression. To play it safe, escape them.

Just insert every character you don't want to include in a negated character class as follows:
^[^\s\d!#£$%^&*()+=]*$
DEMO
Debuggex Demo
^ - start of the string
[^...] - matches one character that is not in `...`
\s - matches a whitespace (space, newline,tab)
\d - matches a digit from 0 to 9
* - a quantifier that repeats immediately preceeding element by 0 or more times
so the regex matches any string that has
1. string that has a beginning
2. containing 0 or more number of characters that is not whitesapce, digit, and all the symbols included in the character class ( In this example !#£$%^&*()+=) i.e., characters that are not included in the character class `[...]`
3.that has ending
NOTE:
If the symbols you don't want it to have also includes - , a hyphen, don't put it in between some other characters because it is a metacharacter in character class, put it at last of character class

RegEx works in tester not on site using JavaScript

I'm trying to write a RegEx that returns true if the string starts with / or http: and only allows alpha numeric characters, the dash and underscore. Any white space and any other special characters should fire a false response when tested.
Below works fine (except that it allows special characters, I have not figured out how to do that yet) when tested at https://www.regex101.com/#javascript. Unfortunately returns false when I implement it in my site and test it with /products/homedecor/tablecloths. What am I doing wrong and is there a better regEx to use that would accomplish my goals?
^(\\/|(?:http:))\S+[a-zA-Z0-9-_]+$

Keep unescaped hyphen at first or at last position in character class:
^(\/|(?:http:))[/.a-zA-Z0-9_-]+$
Or even simpler:
^(\/|http:)[/\w.-]+$
Since \w is same as [a-zA-Z0-9_]
To match URL you may need to match DOT and forward slash as well.

Just remove the \S+ from your regex and put the hyphen inside the character class at the first or at the last. Note that \S+ matches any non-space characters (including non-word characters).
^(\/|http:)[a-zA-Z0-9_-]+$

unable to parse - in Regular expression in Javascript

I am a bit new to the regular expressions in Javascript.
I am trying to write a function called parseRegExpression()
which parses the attributes passed and generates a key/value pairs
It works fine with the input:
"iconType:plus;iconPosition:bottom;"
But it is not able to parse the input:
"type:'date';locale:'en-US';"
Basically the - sign is being ignored. The code is at:
http://jsfiddle.net/visibleinvisibly/ZSS5G/
The Regular Expression key value pair is as below
/[a-z|A-Z|-]*\s*:\s*[a-z|A-Z|'|"|:|-|_|\/|\.|0-9]*\s*;|[a-z|A-Z|-]*\s*:\s*[a-z|A-Z|'|"|:|-|_|\/|\.|0-9]*\s*$/gi;

There are a few problems:
A | inside a character class means a literal | character, not an alternation.
A . inside a character class means a literal . character, so there's no need to escape it.
A - as the first or last character inside a character class means a literal - character, otherwise it means a character range.
There's no need to use [a-zA-Z] when you use the case-insensitive modifier (i); [a-z] is enough.
The only difference between your alterations is the last bit; this can be simplified significantly by just limiting your alternation to that part which is different.
This should be equivalent to your original pattern:
/[a-z-]*\s*:\s*[a-z0-9'":_\/.-]*\s*(?:;|$)/gi

You can avoid the regex:
var test1 = "iconType:plus;iconPosition:bottom;";
var test2 = "type:'date';locale:'en-US';";
function toto(str) {
var result = new Array();
var temp = str.split(';');
for (i=0; i<temp.length-1; i++) {
result[i] = temp[i].split(':',1);
}
return result;
}
console.log(toto(test1));
console.log(toto(test2));

Inside a character set atom [...] the pipe char | is just a regular char and doesn't mean "or".
A character set atom lists characters or ranges you want to accept (or exclude if the character set starts with ^) and "or" is implicit.
You can use a backslash in a character set if you need to include/exclude a close bracket ], the ^ sign, the dash - that is used for ranges, the backslash \ itself, an unprintable character or if you want to use a non-ASCII unicode char specifying the code instead of literally.
Regular expression syntax however also lets you to avoid backslash-escaping in a character set atom by placing the character in a position where it cannot have the special meaning... for example a dash - as first or last in the set (it cannot mean a range there).
Note also that if you need to be able to match as values quoted strings, including backslash escaping, the regular expression is more complex, for example
'(?:[^'\\]|\\.)*'|"(?:[^"\\]|\\.)*"
matches a single-quoted or double-quoted string including backslash escaping, the meaning being:
A single quote '
Zero or more of either:
Any char except the single quote ' or the backslash \
A pair composed of a backslash \ followed by any char
A single quote '
or the same with double quotes " instead.
Note that the groups have been delimited with (?:...) instead of plain (...) to avoid capture

It doesn't match hyphens because it interpreting |-| as a range that starts at | and ends at |. (I would have expected that to be treated as a syntax error, but there you have it. It works the same in every regex flavor I've tried, too.)
Have a look at this regex:
/(?:^|;)([a-z-]*)\s*:\s*([a-z'":_\/.0-9-]*)\s*(?=;|$)/ig
As suggested by the other responders, I collapsed it to one alternative, removed the unneeded pipes, and escaped the hyphen by moving it to the end. I also anchored it at the beginning as well as the end. Or anchored it as well as I can, anyway. I used a lookahead to match the trailing semicolon so it will still be there when the next match starts. It's far from foolproof, but it should work okay as long as the input is well formed.

Replace regular expressions in your code as follow:
regExpKeyValuePair = /[-a-z]*\s*:\s*[-a-z'":_\/.0-9]*\s*;|[-a-z]*\s*:\s*[-a-z'":-_\/.0-9]*\s*$/gi;
regExpKey = /[-a-z]*/gi;
regExpValue = /[-a-z:_\/.0-9]*/gi;
You don't need escape . inside [].
No need to put | between elements [].
Because you are using /i flag, [A-Z] is not needed.
- should be at the beginning or at the end.

Why does my regexp not work when the strings end with spaces?

I am using this regexp - [^\s\da-zA-ZåäöÅÄÖ]+$ to filter out anything but A-Z, 0-9 plus the Swedish characters ÅÄÖ. It works as expected as long as the string isn't ending with whitespace and I am a bit confused on what I need correct to make it accept strings even if they end with whitespace. The \s is there but is apparently not enough.
What is wrong in my regexp?
"something #¤%&/()=?".replace(/[^\s\da-zA-ZåäöÅÄÖ]+$/, '') # => a string
"something ending with whitespace #¤%&/()=? ".replace(/[^\s\da-zA-ZåäöÅÄÖ]+$/, '')# => a string ending with space #¤%&/()=?

You're using a negated character class ("anything that is not a space, a digit, a letter etc."), therefore your regex fails to match.
Drop the \s from it, and also the $ (which ties the match to the end of the string), and it should work.
If you do want to keep spaces inside the string and only remove them at the end, use
"something with whitespace #¤%&/()=? ".replace(/[^\s\da-zA-ZåäöÅÄÖ]+|\s+$/g, '')
Result:
something with whitespace

Your regex says: "match one or more instances of the characters not in the following range, followed by end-of-string". This essentially means that your regex will match only sequences of not-allowed characters appearing at the end of the string. Since your test string ends with a whitespace, which is allowed by your logic, there's no 'sequence of not-allowed characters appearing at the end of the string' and so the regex doesn't match anything.
You can achieve your desired filtering if you remove the $ from the end of the regex and instead use the g flag to make it globally replace anything not in the specified character range with the empty string.
If you additionally want to trim trailing whitespace, it'd be better to do so using another regex, or a simpler trimRight call.