awk '{ for(i=3; i<=NF; ++i) printf $i""FS; print "" }'

Lauhub在这里提出了这个正确、简单、快速的解决方案

打印所有列:

awk '{print $0}' somefile

打印除第一列以外的所有内容:

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

打印除前两列以外的所有内容:

awk '{$1=$2=""; print $0}' somefile

我想把建议的答案扩展到字段可能由几个空格分隔的情况——我想这就是OP不使用cut的原因。

我知道OP问了关于awk的问题，但sed方法在这里可以工作(例如从第5列打印到最后一列):

pure sed approach sed -r 's/^\s*(\S+\s+){4}//' somefile Explanation: s/// is the standard command to perform substitution ^\s* matches any consecutive whitespace at the beginning of the line \S+\s+ means a column of data (non-whitespace chars followed by whitespace chars) (){4} means the pattern is repeated 4 times. sed and cut sed -r 's/^\s+//; s/\s+/\t/g' somefile | cut -f5- by just replacing consecutive whitespaces by a single tab; tr and cut: tr can also be used to squeeze consecutive characters with the -s option. tr -s [:blank:] <somefile | cut -d' ' -f5-

打印从#2开始的列(输出在开始时没有尾随空格):

ls -l | awk '{sub(/[^ ]+ /, ""); print $0}'

以下是我在所有推荐中更喜欢的:

从第六列到最后一列打印。

ls -lthr | awk '{out=$6; for(i=7;i<=NF;i++){out=out" "$i}; print out}'

or

ls -lthr | awk '{ORS=" "; for(i=6;i<=NF;i++) print $i;print "\n"}'

Perl:

@m=`ls -ltr dir | grep ^d | awk '{print \$6,\$7,\$8,\$9}'`;
foreach $i (@m)
{
        print "$i\n";

}