Shell – How to use a shell command to only show the first column and last column in a text file

awksedshell

I need some help to figure out how to use the sed command to only show the first column and last column in a text file. Here is what I have so far for column 1:

cat logfile | sed 's/\|/ /'|awk '{print $1}'

My feeble attempt at getting the last column to show as well was:

cat logfile | sed 's/\|/ /'|awk '{print $1}{print $8}'

However this takes the first column and last column and merges them together in one list.
Is there a way to print the first column and last columns clearly with sed and awk commands?

Sample input:

foo|dog|cat|mouse|lion|ox|tiger|bar

Best Answer

Almost there. Just put both column references next to each other.

cat logfile | sed 's/|/ /' | awk '{print $1, $8}'

Also note that you don't need cat here.

sed 's/|/ /' logfile | awk '{print $1, $8}'

Also note you can tell awk that the column separators is |, instead of blanks, so you don't need sed either.

awk -F '|' '{print $1, $8}' logfile

As per suggestions by Caleb, if you want a solution that still outputs the last field, even if there are not exactly eight, you can use $NF.

awk -F '|' '{print $1, $NF}' logfile

Also, if you want the output to retain the | separators, instead of using a space, you can specify the output field separators. Unfortunately, it's a bit more clumsy than just using the -F flag, but here are three approaches.

  • You can assign the input and output field separators in awk itself, in the BEGIN block.

    awk 'BEGIN {FS = OFS = "|"} {print $1, $8}' logfile

  • You can assign these variables when calling awk from the command line, via the -v flag.

    awk -v 'FS=|' -v 'OFS=|' '{print $1, $8}' logfile

  • or simply:

    awk -F '|' '{print $1 "|" $8}' logfile
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