Regex remove string in url

Look at this code that is using regex

var url = "https://randomsitename-dd555959b114a0.mydomain.com";
var res = url.replace(/ *\-[^.]*\. */g, ".");

http://jsfiddle.net/VYw9Y/

It's usually best to code for all possible cases and since hyphens are allowed within any part of domain names, you'll more than likely want to use a more specific RexExp such as:

^                  # start of string
(                  # start first capture group
    [a-z]+         # one or more letters
)                  # end first capture group
://                # literal separator
(                  # start second capture group
    [^.-]+         # one or more chars except dot or hyphen
)                  # end second capture group
(?:                # start optional non-capture group
    -              # literal hyphen
    [^.]+          # one or more chars except dot
)?                 # end optional non-capture group
(                  # start third capture group
    .+             # one or more chars
)                  # end third capture group
$                  # end of string

Or without comments:

^([a-z]+)://([^.-])(?:-[^.]+)?(.+)$

(Remember to escape slashes if you use the literal form for RegExps rather than creating them as objects, i.e. /literal\/form/ vs. new RegExp('object/form'))

Used in a string replacement, the second argument should then be: $1://$2$3

Previous answers will fail for URLs like http://foo.bar-baz.com or http://foo-bar.baz-blarg.com.

You could try this regex,

(.*)(-[^\.]*)(.*$)

Your code should be,

var url = "https://randomsitename-dd555959b114a0.mydomain.com";
var res = url.replace(/(.*)(-[^\.]*)(.*$)/, "$1$3");
//=>https://randomsitename.mydomain.com

Explanation:

• (.*) matches any character 0 or more times and it was stored into group 1 because we enclose those characters within paranthesis. Whenever the regex engine finds -, it stops storing it into group1.

• (-[^\.]*) From - upto a literal . are stored into group2. It stops storing when it finds a literal dot.

• (.*$) From the literal dot upto the last character are stored into group3.

• $1$3 at the replacement part prints only the stored group1 and 3.

OR

(.*)(?:-[^\.]*)(.*$)

If you use this regex, in the replacement part you need to put only $1 and $2.

DEMO