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How to match all words starting with dollar sign but not slash dollar

I want to match all words which are starting with dollar sign but not slash and dollar sign.
I already try few regex.
(?:(?!\\)\$\w+)
\\(\\?\$\w+)\b
String
$10<i class="">$i01d</i>\$id
Expected result
*$10*
*$i01d*
but not this
*$id*
After find all expected matching word i want to replace this my object.

One option is to eliminate escape sequences first, and then match the cleaned-up string:
s = String.raw`$10<i class="">$i01d</i>\$id`
found = s.replace(/\\./g, '').match(/\$\w+/g)
console.log(found)

The big problem here is that you need a negative lookbehind, however, JavaScript does not support it. It's possible to emulate it crudely, but I will offer an alternative which, while not great, will work:
var input = '$10<i class="">$i01d</i>\\$id';
var regex = /\b\w+\b\$(?!\\)/g;
//sample implementation of a string reversal function. There are better implementations out there
function reverseString(string) {
return string.split("").reverse().join("");
}
var reverseInput = reverseString(input);
var matches = reverseInput
.match(regex)
.map(reverseString);
console.log(matches);
It is not elegant but it will do the job. Here is how it works:
JavaScript does support a lookahead expression ((?>)) and a negative lookahead ((?!)). Since this is the reverse of of a negative lookbehind, you can reverse the string and reverse the regex, which will match exactly what you want. Since all the matches are going to be in reverse, you need to also reverse them back to the original.
It is not elegant, as I said, since it does a lot of string manipulations but it does produce exactly what you want.
See this in action on Regex101
Regex explanation Normally, the "match x long as it's not preceded by y" will be expressed as (?<!y)x, so in your case, the regex will be
/(?<!\\)\$\b\w+\b/g
demonstration (not JavaScript)
where
(?<!\\) //do not match a preceding "\"
\$ //match literal "$"
\b //word boundary
\w+ //one or more word characters
\b //second word boundary, hence making the match a word
When the input is reversed, so do all the tokens in order to match. Furthermore, the negative lookbehind gets inverted into a negative lookahead of the form x(?!y) so the new regular expression is
/\b\w+\b\$(?!\\)/g;

This is more difficult than it appears at first blush. How like Regular Expressions!
If you have look-behind available, you can try:
/(?<!\\)\$\w+/g
This is NOT available in JS. Alternatively, you could specify a boundary that you know exists and use a capture group like:
/\s(\$\w+)/g
Unfortunately, you cannot rely on word boundaries via /b because there's no such boundary before '\'.
Also, this is a cool site for testing your regex expressions. And this explains the word boundary anchor.

If you're using a language that supports negative lookback assertions you can use something like this.
(?<!\\)\$\w+
I think this is the cleanest approach, but unfortunately it's not supported by all languages.
This is a hackier implementation that may work as well.
(?:(^\$\w+)|[^\\](\$\w+))
This matches either
A literal $ at the beginning of a line followed by multiple word characters. Or...
A literal $ this is preceded by any character except a backslash.
Here is a working example.

Constructor Regex in JavaScript

My code is as follows:
var regex = new RegExp ('(.*/*)');
console.log(regex);
I think the result is:
/(.*/*)/
But the actual result is:
/(.*\/*)/
Could someone please explain this for me?

The leading and trailing forward slashes are just how JavaScript represents a regex in string form and as a literal. The mean the same thing:
var regex = new RegExp ('(.*/*)');
is the same as
var regex = /(.*\/*)/;
It is important to escape the middle / otherwise it would interpret it as the end of the literal.

If you expect the last asterisk * to be literal you need to escape it, otherwise it's a quantifier in regex meaning "match between zero and unlimited times". It's also recommened to escape the forward slash /, even if it might work without doing so.
(.*\/\*)
If you want to match any string /* any string then use:
(.*\/\*.*)
https://regex101.com/r/aJ4eA4/1

Regex-Groups in Javascript

I have a problem using a Javascript-Regexp.
This is a very simplified regexp, which demonstrates my Problem:
(?:\s(\+\d\w*))|(\w+)
This regex should only match strings, that doesn't contain forbidden characters (everything that is no word-character).
The only exception is the Symbol +
A match is allowed to start with this symbol, if [0-9] is trailing.
And a + must not appear within words (44+44 is not a valid match, but +4ad is)
In order to allow the + only at the beginning, I said that there must be a whitespace preceding. However, I don't want the whitespace to be part of the match.
I tested my regex with this tool: http://regex101.com/#javascript and the resultig matches look fine.
There are 2 Issues with that regexp:
If I use it in my JS-Code, the space is always part of the match
If +42 appears at the beginning of a line, it won't be matched
My Questions:
How should the regex look like?
Why does this regex add the space to the matches?
Here's my JS-Code:
var input = "+5ad6 +5ad6 sd asd+as +we";
var regexp = /(?:\s(\+\d\w*))|(\w+)/g;
var tokens = input.match(regexp);
console.log(tokens);

How should the regex look like?
You've got multiple choices to reach your goal:
It's fine as you have it. You might allow the string beginning in place of the whitespace as well, though. Just get the capturing groups (tokens[1], tokens[2]) out of it, which will not include the whitespace.
If you didn't use JavaScript, a lookbehind could help. Unfortunately it's not supported.
Require a non-word-boundary before the +, which would make every \w character before the + prevent the match:
/\B\+\d\w+|\w+/
Why does this regex add the space to the matches?
Because the regex does match the whitespace. It does not add the \s(\+\d\w+) to the captured groups, though.

How do I ignore $1 replace backreferencing in javascript

I have a string that a user can edit at any time, and a regex that is being conducted on the string, to add it to an xml and then save it but they can add '$1' to the string. I just want the text '$1' to be saved but I have to perform a regular expression on the same string that $1 is in. It replaces the $1 with a character from the regex every time.
How do I find, and replace, the $1 in this string?
Example of what is happening:
string1 = '<item id="1">i have $100</item>'
regexp = new RegExp('<item id="1"([^<]|<[^\/]|<\/[^i]|<\/i[^t]|<\/it[^e]|<\/ite[^m]|<\/item[^>])*<\/item>');
data = '<data><item id="1">i have no money</item><item id="2">i have no money</item></data>'
data = data.replace(regexp, string1);
Results
<data><item id="1">i have >00</item><item id="2">i have no money</item></data>

If you have a variable string that you want to put in your replace() call which might possibly have $N's in it, you can prevent the $N from being treated as a backreference by replacing $ with $$. Apparently, unlike other special characters in JS regex, the $ character cannot be escaped with a \ - it must be escaped with a preceding $ (go figure).
In your example, you could do the following to fix the issue:
data = data.replace(regexp, string1.replace('$', '$$$'));
This should turn any $'s into $$ in string1, preventing them from being treated as backreferences.
(Note: I found this little nugget here)

This should only happen if you have a capturing group in the regex.
If you don't want your groups to capture, then place ?: inside the start of the group.
/foo(?:bar)/

You can escape the $. Eg:
var replacement = '<item id="1">i have \\$100</item>';
Useful when you have capturing groups and need to write a $.

Javascript string replace with regex to strip off illegal characters

Need a function to strip off a set of illegal character in javascript: |&;$%#"<>()+,
This is a classic problem to be solved with regexes, which means now I have 2 problems.
This is what I've got so far:
var cleanString = dirtyString.replace(/\|&;\$%#"<>\(\)\+,/g, "");
I am escaping the regex special chars with a backslash but I am having a hard time trying to understand what's going on.
If I try with single literals in isolation most of them seem to work, but once I put them together in the same regex depending on the order the replace is broken.
i.e. this won't work --> dirtyString.replace(/\|<>/g, ""):
Help appreciated!

What you need are character classes. In that, you've only to worry about the ], \ and - characters (and ^ if you're placing it straight after the beginning of the character class "[" ).
Syntax: [characters] where characters is a list with characters.
Example:
var cleanString = dirtyString.replace(/[|&;$%#"<>()+,]/g, "");

I tend to look at it from the inverse perspective which may be what you intended:
What characters do I want to allow?
This is because there could be lots of characters that make in into a string somehow that blow stuff up that you wouldn't expect.
For example this one only allows for letters and numbers removing groups of invalid characters replacing them with a hypen:
"This¢£«±Ÿ÷could&*()\/<>be!##$%^bad".replace(/([^a-z0-9]+)/gi, '-');
//Result: "This-could-be-bad"

You need to wrap them all in a character class. The current version means replace this sequence of characters with an empty string. When wrapped in square brackets it means replace any of these characters with an empty string.
var cleanString = dirtyString.replace(/[\|&;\$%#"<>\(\)\+,]/g, "");

Put them in brackets []:
var cleanString = dirtyString.replace(/[\|&;\$%#"<>\(\)\+,]/g, "");