when i run a program like "ls" the output is to STDOUT so i can pipe it or
redirect it through something like sed or grep. but cvs doesn't do that. if
i run "cvs update > /dev/null" or "cvs update | sed -e 's//yyy/'"
it still prints the same thing it always does. how do i capture this
inform
gabriel wrote:
when i run a program like "ls" the output is to STDOUT so i can pipe it or
redirect it through something like sed or grep. but cvs doesn't do that. if
i run "cvs update > /dev/null" or "cvs update | sed -e 's//yyy/'"
it still prints the same thing it always does. how do i captu
gabriel wrote:
when i run a program like "ls" the output is to STDOUT so i can pipe it or
redirect it through something like sed or grep. but cvs doesn't do that. if
i run "cvs update > /dev/null" or "cvs update | sed -e 's//yyy/'"
it still prints the same thing it always does. how do i captu
> i run "cvs update > /dev/null" or "cvs update | sed -e 's//yyy/'"
> it still prints the same thing it always does. how do i capture this
> information?
cvs prints many of its messages to standard error (STDERR) instead of
standard output. In bourne-like shells (including bash), you use 2>
i
On January 28, 2004 02:17 pm, Diego Zamboni wrote:
> > i run "cvs update > /dev/null" or "cvs update | sed -e
> > 's//yyy/'" it still prints the same thing it always does.
> > how do i capture this information?
>
> cvs prints many of its messages to standard error (STDERR) instead of
> standard ou
