Advertisement
Advertisement


How can I remove a character from a string using Javascript?


Question

I am so close to getting this, but it just isn't right. All I would like to do is remove the character r from a string. The problem is, there is more than one instance of r in the string. However, it is always the character at index 4 (so the 5th character).

example string: crt/r2002_2

What I want: crt/2002_2

This replace function removes both r

mystring.replace(/r/g, '')

Produces: ct/2002_2

I tried this function:

String.prototype.replaceAt = function (index, char) {
    return this.substr(0, index) + char + this.substr(index + char.length);
}
mystring.replaceAt(4, '')

It only works if I replace it with another character. It will not simply remove it.

Any thoughts?

2019/09/21
1
378
9/21/2019 8:29:25 PM


A simple functional javascript way would be

mystring = mystring.split('/r').join('/')

simple, fast, it replace globally and no need for functions or prototypes

2014/10/29

There's always the string functions, if you know you're always going to remove the fourth character:

str.slice(0, 4) + str.slice(5, str.length))
2016/07/24

Your first func is almost right. Just remove the 'g' flag which stands for 'global' (edit) and give it some context to spot the second 'r'.

Edit: didn't see it was the second 'r' before so added the '/'. Needs \/ to escape the '/' when using a regEx arg. Thanks for the upvotes but I was wrong so I'll fix and add more detail for people interested in understanding the basics of regEx better but this would work:

mystring.replace(/\/r/, '/')

Now for the excessive explanation:

When reading/writing a regEx pattern think in terms of: <a character or set of charcters> followed by <a character or set of charcters> followed by <...

In regEx <a character or set of charcters> could be one at a time:

/each char in this pattern/

So read as e, followed by a, followed by c, etc...

Or a single <a character or set of charcters> could be characters described by a character class:

/[123!y]/
//any one of these
/[^123!y]/
//anything but one of the chars following '^' (very useful/performance enhancing btw)

Or expanded on to match a quantity of characters (but still best to think of as a single element in terms of the sequential pattern):

/a{2}/
//precisely two 'a' chars - matches identically as /aa/ would

/[aA]{1,3}/
//1-3 matches of 'a' or 'A'

/[a-zA-Z]+/
//one or more matches of any letter in the alphabet upper and lower
//'-' denotes a sequence in a character class

/[0-9]*/
//0 to any number of matches of any decimal character (/\d*/ would also work)

So smoosh a bunch together:

   var rePattern = /[aA]{4,8}(Eat at Joes|Joes all you can eat)[0-5]+/g
   var joesStr = 'aaaAAAaaEat at Joes123454321 or maybe aAaAJoes all you can   eat098765';

   joesStr.match(rePattern);

   //returns ["aaaAAAaaEat at Joes123454321", "aAaAJoes all you can eat0"]
   //without the 'g' after the closing '/' it would just stop at the first   match and return:
   //["aaaAAAaaEat at Joes123454321"]

And of course I've over-elaborated but my point was simply that this:

/cat/

is a series of 3 pattern elements (a thing followed by a thing followed by a thing).

And so is this:

/[aA]{4,8}(Eat at Joes|Joes all you can eat)[0-5]+/

As wacky as regEx starts to look, it all breaks down to series of things (potentially multi-character things) following each other sequentially. Kind of a basic point but one that took me a while to get past so I've gone overboard explaining it here as I think it's one that would help the OP and others new to regEx understand what's going on. The key to reading/writing regEx is breaking it down into those pieces.

2017/01/29

For global replacement of '/r', this code worked for me.

mystring = mystring.replace(/\/r/g,'');
2013/09/12

Just fix your replaceAt:

String.prototype.replaceAt = function(index, charcount) {
  return this.substr(0, index) + this.substr(index + charcount);
}

mystring.replaceAt(4, 1);

I'd call it removeAt instead. :)

2012/03/29

Source: https://stackoverflow.com/questions/9932957
Licensed under: CC-BY-SA with attribution
Not affiliated with: Stack Overflow
Email: [email protected]