Using this bit of code trims out hidden characters like carriage returns and linefeeds with nothing using javascript just fine:
value = value.replace(/[\\r\\
To just remove them, this seems to work for me:
value = value.replace(/[\r\n]/g, "");
You don't need the * after the character set because the g flag solves that for you.
Note, this will remove all \r or \n chars whether they are in this exact sequence or not.
Working demo of this option: http://jsfiddle.net/jfriend00/57GtJ/
If you want to remove these characters only when in this exact sequence (e.g. only when a \r is directly followed by a \n, you could use this:
value = value.replace(/\r\n/g, "");
Working demo of this option: http://jsfiddle.net/jfriend00/Ta3sn/
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.
If you have text with a lot of \r\n and want to save all of them try this one
value.replace(/(?:\\[rn]|[\r\n])/g,"<br>")
http://jsfiddle.net/57GtJ/63/