awk how to print the rest

Question

my file contains lines like this

any1 aaa bbb ccc

The delimiter is space. the number of words in the line is unknown

I want to put the first word into a var1 . It's simple with

awk '{print $1}'

Now I want to put the rest of the line into a var2 with awk.

How I can print the rest of the line with awk ?

Answer 1

Better to use read here:

s="any1 aaa bbb ccc"
read var1 var2 <<< "$s"
echo "$var1"
any1
echo "$var2"
aaa bbb ccc

For awk only solution use:

echo "$s" | awk '{print $1; print substr($0, index($0, " ")+1)}'
any1
aaa bbb ccc

Answer 2

$ var=$(awk '{sub(/^[^[:space:]]+[[:space:]]+/,"")}1' file)
$ echo "$var"
aaa bbb ccc

or in general to skip some number of fields use a RE interval:

$ awk '{sub(/^[[:space:]]*([^[:space:]]+[[:space:]]+){1}/,"")}1' file
aaa bbb ccc
$ awk '{sub(/^[[:space:]]*([^[:space:]]+[[:space:]]+){2}/,"")}1' file
bbb ccc
$ awk '{sub(/^[[:space:]]*([^[:space:]]+[[:space:]]+){3}/,"")}1' file
ccc

Note that doing this gets much more complicated if you have a FS that's more than a single char, and the above is just for the default FS since it additionally skips any leading blanks if present (remove the first [[:space:]]* if you have a non-default but still single-char FS).

Answer 3

awk解决方案：

awk '{$1 = ""; print $0;}'`