Dexter Filmore wrote: > > Well, shouldn't \ do th job as well? > Thing is, even if I pass the filename > foo\ bar > echo tells me the name is > foo bar
echo is a bad tool to use to determine where the shell is breaking the words. compare: a="moo cow" echo \""$a"\" echo \"$a\" They both provide the same results, but one is split and the other is not. This is why I offered you ls -b. You can also use test -f. sh-3.1$ touch "moo cow" sh-3.1$ a=moo\ cow sh-3.1$ test -f $a sh: test: moo: binary operator expected sh-3.1$ test -f "$a" $a got split, and "$a" remained as a single word. You can also play with set. sh-3.1$ a="moo cow"; set - $a; echo $#; set - "$a" ; echo $# 2 1 Stop trying to count the number of arguments with echo: it does not work. -john -- [email protected] http://www.kernel-panic.org/cgi-bin/mailman/listinfo/kplug-list
