Routine to remove extra newline chars and carriage return

[vflyer]

I there a quick routine that removes trailing newline chars and
carriage return?

vFlyer Labs
http://www.vflyerlabs.com

[noagbodjivictor]

You can write one:

String.prototype.rtrim = function() {
return this.replace(/\n\r/,"");
}

[scripts.contact]

On May 29, 8:11 pm, [e-mail] wrote:

You can write one:

String.prototype.rtrim = function() {
return this.replace(/\n\r/,"");}

String.prototype.rtrim = function() {
return this.replace(/[\n\r]/,"");
}

[Bart Van der Donck]

scripts.contact wrote:

On May 29, 8:11 pm, [e-mail] wrote:

You can write one:

String.prototype.rtrim = function() {
return this.replace(/\n\r/,"");}

String.prototype.rtrim = function() {
return this.replace(/[\n\r]/,"");}

No. The three most common EOLs are:

\n (Linefeed)
\r (Carriage return)
\r\n (Carriage return followed by linefeed)

But you also have:

NEL = Next Line (\u0085)
FF = Form Feed (\u000C)
LS = Line Separator (\u2028)
PS = Paragraph Separator (\u2029)

A fully backwards compatible, Unicode-compliant regexp would be:

/(\r\n|\r|\n|\u0085|\u000C|\u2028|\u2029)/g

--
Bart

[scripts.contact]

On May 31, 2:29 pm, Bart Van der Donck <[e-mail]:

String.prototype.rtrim = function() {
return this.replace(/[\n\r]/,"");}

forgot the $ :
return this.replace(/[\n\r]$/,"");}

A fully backwards compatible, Unicode-compliant regexp would be:

/(\r\n|\r|\n|\u0085|\u000C|\u2028|\u2029)/g

/(\r\n|\r|\n|\u0085|\u000C|\u2028|\u2029)$/

[scripts.contact]

On May 31, 2:29 pm, Bart Van der Donck <[e-mail]:

String.prototype.rtrim = function() {
return this.replace(/[\n\r]/,"");}

forgot the $ :
return this.replace(/[\n\r]$/,"");}

A fully backwards compatible, Unicode-compliant regexp would be:

/(\r\n|\r|\n|\u0085|\u000C|\u2028|\u2029)/g

/(\r\n|\r|\n|\u0085|\u000C|\u2028|\u2029)$/