Regex Removing Hash from Beginning

Question

I am running up against a weird problem here. When I run:

"#/a/b/c/d".replace("#\/","")

I get what I would expect: a/b/c/d.

But When I precede this regex with a start of string character ^, To get:

"#/a/b/c/d".replace("^#\/","")

This returns the original string "#a/b/c/d".

Could anybody explain why the hash isn't removed and possibly suggest an alternative that would remove the hash only if it appears at the beginning of the string?

Regexes have a special syntax with forward slash delimiters, not quotes. — trincot, Apr 07 '16 at 12:23
Thats awesome, could explain why that works? what is the purpose of the double slash? If you make this an answer I'll accept it as the correct answer :) — user2662833, Apr 07 '16 at 12:25
@user2662833: I see some one decided to answer a duplicate - please consider accepting/upvoting my answer as you promised. +1. — Wiktor Stribiżew, Apr 07 '16 at 20:11

score 1 · Answer 1 · answered Apr 07 '16 at 13:31

1

The first argument to replace can be either a string or a regular expression.

You are passing a string, so it is looking for an exact match for that string.

Pass a regular expression instead.

"#/a/b/c/d".replace(/^#\//,"")

answered Apr 07 '16 at 13:31

Quentin

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score 0 · Answer 2 · answered Apr 07 '16 at 13:31

String.replace takes as its first argument either a string or a regexp. If you give it a string, it is not interpreted as or converted to a regexp; it is treated literally. Therefore, when you add ^ to your string, it replaces nothing, since there is no ^ in the input.

As explained in the comments, you can either pass a regexp literal (/.../) as the first argument, or you could construct a regexp with new RegExp (but why would you do that?).

score 0 · Accepted Answer · answered Apr 07 '16 at 20:10

When you pass a string as the first argument to .replace() method, the string is used as a literal and only 1 replacement can be made.

See the MDN replace reference:

substr (pattern)
A String that is to be replaced by newSubStr. It is treated as a verbatim string and is not interpreted as a regular expression.

What you need is to pass the regex object as the first argument:

"#/a/b/c/d".replace(/^#\//,"")
                    ^^^^^^

There is no point in using a constructor notation (RegExp("^#/") since the pattern is not built dynamically from variables.

There are no double slashes in the pattern! The outer /.../ are called regex delimiters. Then, what is insde is a pattern. The pattern is ^#/, but since the / are used as regex delimiters, the / in the pattern must be escaped to match a literal / (only in the regex literal notation).

Another way too: `name = (name.charAt(0) === '#') ? name.slice(1) : name;` — JREAM, Apr 20 '19 at 19:52

Regex Removing Hash from Beginning

3 Answers3