使用awk打印选定行的选定部分

Question

Assume I have a file called 'file.csv' that contains the following data

A:qwe
B:ert
C:rty

How would I be able to print out everything before the ':' in rows 2 and 3. So I want my output to be just B and C. Have tried

cat file.csv | awk -F ':' '{print $1}'

but that prints out A as well and not too sure how to implement 'NR==2,NR==3' to this

Answer 1

Awk seems like ridiculous overkill here. How about this:

tail -n 2 | cut -d ":" -f 1

Answer 2

For a range of lines from n to m, you can use sed|cut combination

sed -n '2,3p' file | cut -d: -f1

or just awk

awk -F: 'NR>=2 && NR<=3 {print $1}' file

Answer 3

sed is easier than awk for this

sed -n '2,3s/:.*//p' file.csv

But I'm wondering if you really meant that you want the second and third lines, as opposed to some other criteria.

Answer 4

Perl solution:

perl -F: -lane 'print $F[0] if $. > 1' file

Prints first field of input if $. (line number variable) > 1

These command-line options are used:

• n loop around every line of the input file, do not automatically print every line

• l removes newlines before processing, and adds them back in afterwards

• a autosplit mode – perl will automatically split input lines into the @F array. Defaults to splitting on whitespace

• F autosplit modifier, in this example splits on ':'

• e execute the perl code

Answer 5

Given:

$ echo "$tgt"
A:qwe
B:ert
C:rty

You can use Perl and the 'flip-flop' operator to print the file or text between two line numbers:

$ echo "$tgt" | perl -lne 'print $1 if /(^[^:]+)/ and 2..3'
B
C

Which works for a file also (obviously)