javascript replace() not replacing text containing literal \r\n strings

Question

Using this bit of code trims out hidden characters like carriage returns and linefeeds with nothing using javascript just fine:

value = value.replace(/[\r\n]*/g, "");

but when the code actually contains \r\n text what do I do to trim it without affecting r's and n's in my content? I've tried this code:

value = value.replace(/[\\r\\n]+/g, "");

on this bit of text:

{"client":{"werdfasreasfsd":"asdfRasdfas\r\nMCwwDQYJKoZIhvcNAQEBBQADGw......

I end up with this:

{"cliet":{"wedfaseasfsd":"asdfRasdfasMCwwDQYJKoZIhvcNAQEBBQADGw......

Side note: It leaves the upper case versions of R and N alone because I didn't include the /i flag at the end and thats ok in this case.

What do I do to just remove \r\n text found in the string?

Answer 1

If you want to match literal \r and literal \n then you should use the following:

value = value.replace(/(?:\\[rn])+/g, "");

You might think that matching literal \r and \n with [\\r\\n] is the right way to do it and it is a bit confusing but it won't work and here is why:

Remember that in character classes, each single character represents a single letter or symbol, it doesn't represent a sequence of characters, it is just a set of characters.

So the character class [\\r\\n] actually matches the literal characters \, r and n as separate letters and not as sequences.

Edit: If you want to replace all carriage returns \r, newlines \n and also literal \r and '\n` then you could use:

value = value.replace(/(?:\\[rn]|[\r\n]+)+/g, "");

About (?:) it means a non-capturing group, because by default when you put something into a usual group () then it gets captured into a numbered variable that you can use elsewhere inside the regular expression itself, or latter in the matches array.

(?:) prevents capturing the value and causes less overhead than (), for more info see this article.