Awk + Print All Line Content Except $1

Answer :

If you set $1 to "" you will leave the delimiting space. If you don't want to do that you have to iterate over the fields:

awk '{for (f=2; f<=NF; ++f) { if (f!=2) {printf("%s",OFS);} printf("%s",$f)}; printf "\n" }' 

Edit: fixed per Gilles' comment.

Another way to do the same thing:

awk '{d = ""; for (f=2; f<=NF; ++f) {printf("%s%s", d, $f); d = OFS}; printf("\n") }' 

Somehow I think this would be so much easier and more intuitive to do with the cut command:

echo /var/sysconfig/network/my_functions  alpha beta gama | cut -d' ' -f 2- 

The only problem is that cut doesn't support multiple different types of whitespace at once for delimiters. So if you have spaces or tabs, it won't work.