如何切割第一个 n 列和最后一个 n 列？

you can use awk, for example, cut off 1st,2nd and last 3 columns

awk '{for(i=3;i<=NF-3;i++} print $i}' file

if you have a programing language such as Ruby (1.9+)

$ ruby -F"\t" -ane 'print $F[2..-3].join("\t")' file

Cut can take several ranges in -f:

Columns up to 4 and from 7 onwards:

cut -f -4,7-

or for fields 1,2,5,6 and from 10 onwards:

cut -f 1,2,5,6,10-

etc

To use AWK to cut off the first and last fields:

awk '{$1 = ""; $NF = ""; print}' inputfile

Unfortunately, that leaves the field separators, so

aaa bbb ccc

becomes

[space]bbb[space]

To do this using kurumi's answer which won't leave extra spaces, but in a way that's specific to your requirements:

awk '{delim = ""; for (i=2;i<=NF-1;i++) {printf delim "%s", $i; delim = OFS}; printf "\n"}' inputfile

This also fixes a couple of problems in that answer.

To generalize that:

awk -v skipstart=1 -v skipend=1 '{delim = ""; for (i=skipstart+1;i<=NF-skipend;i++) {printf delim "%s", $i; delim = OFS}; printf "\n"}' inputfile

Then you can change the number of fields to skip at the beginning or end by changing the variable assignments at the beginning of the command.

Try the following:

echo a#b#c | awk -F"#" '{$1 = ""; $NF = ""; print}' OFS=""

You can use Bash for that:

while read -a cols; do echo ${cols[@]:0:1} ${cols[@]:1,-1}; done < file.txt

You can cut using following ,
-d: delimiter ,-f for fields
\t used for tab separated fields

cut -d$'\t' -f 1-3,7-

The first part of your question is easy. As already pointed out, cut accepts omission of either the starting or the ending index of a column range, interpreting this as meaning either “from the start to column n (inclusive)” or “from column n (inclusive) to the end,” respectively:

$ printf 'this:is:a:test' | cut -d: -f-2
this:is
$ printf 'this:is:a:test' | cut -d: -f3-
a:test

It also supports combining ranges. If you want, e.g., the first 3 and the last 2 columns in a row of 7 columns:

$ printf 'foo:bar:baz:qux:quz:quux:quuz' | cut -d: -f-3,6-
foo:bar:baz:quux:quuz

However, the second part of your question can be a bit trickier depending on what kind of input you’re expecting. If by “last n columns” you mean “last n columns (regardless of their indices in the overall row)” (i.e. because you don’t necessarily know how many columns you’re going to find in advance) then sadly this is not possible to accomplish using cut alone. In order to effectively use cut to pull out “the last n columns” in each line, the total number of columns present in each line must be known beforehand, and each line must be consistent in the number of columns it contains.

If you do not know how many “columns” may be present in each line (e.g. because you’re working with input that is not strictly tabular), then you’ll have to use something like awk instead. E.g., to use awk to pull out the last 2 “columns” (awk calls them fields, the number of which can vary per line) from each line of input:

$ printf '/a\n/a/b\n/a/b/c\n/a/b/c/d\n' | awk -F/ '{print $(NF-1) FS $(NF)}'
/a
a/b
b/c
c/d

Use

cut -b COLUMN_N_BEGINS-COLUMN_N_UNTIL INPUT.TXT > OUTPUT.TXT

-f doesn't work if you have "tabs" in the text file.