awk + print all line content except $1

Question

I write the follwoing awk script:

% echo /var/sysconfig/network/my_functions  alpha beta gama  | \
      awk -v word=alpha '$2  == word { print $0 }'

how to tell awk that I want to print all line except $1 (/var/sysconfig/network/my_functions PATH ) so I will get the following:

alpha beta gama

instead of

/var/sysconfig/network/my_functions alpha beta gama

remark: line content can be anything and not limit by strings/word quantity

Answer 1

I think in awk there's no way but to remove the first field manually. (There are other ways if you're willing to normalize the inter-field space.)

awk '$2 == word {match($0, "("FS")+"); print substr($0, RSTART+RLENGTH);}'

Answer 2

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") }'

Answer 3

 % echo /var/sysconfig/network/my_functions  alpha beta gama | \
      awk -v word=alpha | \
             '$2  == word { $1=""; print $0 }'

Answer 4

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.

Answer 5

Everyone always makes it harder than it is

$ echo /var/sysconfig/network/my_functions alpha beta gama |
awk -v word=alpha '$2==word{print substr($0,index($0,$2))}'
alpha beta gama