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Write a regex for usernames

I want a Regex for my mongoose schema to test if a username contains only letters, numbers and underscore, dash or dot. What I got so far is
/[a-zA-Z0-9-_.]/
but somehow it lets pass everything.

Your regex is set to match a string if it contains ANY of the contained characters, but it doesn't make sure that the string is composed entirely of those characters.
For example, /[a-zA-Z0-9-_.]/.test("a&") returns true, because the string contains the letter a, regardless of the fact that it also includes &.
To make sure all characters are one of your desired characters, use a regex that matches the beginning of the string ^, then your desired characters followed by a quantifier + (a plus means one or more of the previous set, a * would mean zero or more), then end of string $. So:
const reg = /^[a-zA-Z0-9-_.]+$/
console.log(reg.test("")) // false
console.log(reg.test("I-am_valid.")) // true
console.log(reg.test("I-am_not&")) // false

Try like this with start(^) and end($),
^[a-zA-Z0-9-_.]+$
See demo : https://regex101.com/r/6v0nNT/3

/^([a-zA-Z0-9]|[-_\.])*$/
This regex should work.
^ matches at the beginning of the string. $ matches at the end of the string. This means it checks for the entire string.
The * allows it to match any number of characters or sequences of characters. This is required to match the entire password.
Now the parentheses are required for this as there is a | (or) used here. The first stretch was something you already included, and it is for capital/lowercase letters, and numbers. The second area of brackets are used for the other characters. The . must be escaped with a backslash, as it is a reserved character in regex, used for denoting that something can be any character.

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.

Javascript test method not working ( as expected )

Regular expression
[A-Za-z_-]+
should match strings that only have upper and lower case letters, underscores, and a dash
but when I run in chrome console
/[A-Za-z_-]+/.test("johmSmith12")
Why it returns true

Because you didn't anchor the expression. You need to add ^ and $, which match beginning and end of string.
For example:
^[A-Za-z_-]+$
Just the [A-Za-z_-]+ will match johnSmith in your example, leaving out the 12 (as David Starkey pointed out).

It is due to your regex looking for any sequence of characters within the test string that matches the regex. In your example, "johnSmith" matches your regex criteria, and so test returns true.
If you instead put ^ (start of string) and $ (end of string) at the ends of your regex, then you would assert that the entire string must match your regex:
/^[A-Za-z_-]+$/.test("johnSmith12");
This will return false.

can someone help to explain this regular expression in javascript?

This code is used to get rid of mime type from rawdata.but I can not understand how it works
content.replace(/^[^,]*,/ , '')
it seems quite different from java.... any help will be appreciated.

Your mime-type probably is seperated by a comma , and at the beginning of your raw data.
This regex says take everything from the beginning (^) that is NOT a comma ([^,]*) (the star makes it as many characters until there is a comma) and take the comma itself (,). Then replace it by nothing ('').
This one only gets the first appearence because it is marked by the beginning ^ that it must be at the beginning of the string.

The first thing you need to know is that there are regex literals in JavaScript, constructed by pairs of slashes. So like "..." is a string, /.../ is a regex. That's actually the only difference your code shows as compared to a Java regex.
Then, [abc] within a regex is called a character class, meaning "one character out of a, b or c". Conversely, [^abc] is a negated character class, meaning "one character except a, b or c".
So your sample means:
/ # Start of regex literal
^ # Start the match at the start of the string
[^,]* # Match any number of characters except commas
, # Match a comma
/ # End of regex literal

The regular expression is the text between the two forward slashes, the first carat (^) means at the begining of the string, the brackets mean a character class, the carat inside the brackets means any character except a comma, then asterisk after the closing bracket means match zero or more of the character defined by the character class (which again is any character except the comma), and then finally the last comma means match the comma after all this. Then its used in a replace function so the matching result will be replaced with the second parameter, in your case: an empty string.
Basically it matches the first characters up to and including the first comma in the 'content' variable and then replaces it with an empty string.

Regular Expression - Match any character except +, empty string should also be matched

I am having a bit of trouble with one part of a regular expression that will be used in JavaScript. I need a way to match any character other than the + character, an empty string should also match.
[^+] is almost what I want except it does not match an empty string. I have tried [^+]* thinking: "any character other than +, zero or more times", but this matches everything including +.

Add a {0,1} to it so that it will only match zero or one times, no more no less:
[^+]{0,1}
Or, as FailedDev pointed out, ? works too:
[^+]?
As expected, testing with Chrome's JavaScript console shows no match for "+" but does match other characters:
x = "+"
y = "A"
x.match(/[^+]{0,1}/)
[""]
y.match(/[^+]{0,1}/)
["A"]
x.match(/[^+]?/)
[""]
y.match(/[^+]?/)
["A"]

[^+] means "match any single character that is not a +"
[^+]* means "match any number of characters that are not a +" - which almost seems like what I think you want, except that it will match zero characters if the first character (or even all of the characters) are +.
use anchors to make sure that the expression validates the ENTIRE STRING:
^[^+]*$
means:
^ # assert at the beginning of the string
[^+]* # any character that is not '+', zero or more times
$ # assert at the end of the string

If you're just testing the string to see if it doesn't contain a +, then you should use:
^[^+]*$
This will match only if the ENTIRE string has no +.