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

Question

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

Answer 1

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
  

Answer 2

You are using awk anyway:

awk '{ print $1, $NF }' file

Answer 3

Just replace from the first to last | with a | (or space if you prefer):

sed 's/|.*|/|/'

Note that though there's no sed implementation where | is special (as long as extended regular expressions are not enabled via -E or -r in some implementations), \| itself is special in some like GNU sed. So you should not escape | if you intend it to match the | character.

If replacing with space and if the input may already contain lines with only one |, then, you'll have to treat that specially as |.*| won't match on those. That could be:

sed 's/|\(.*|\)\{0,1\}/ /'

(that is make the .*| part optional) Or:

sed 's/|.*|/ /;s/|/ /'

or:

sed 's/\([^|]*\).*|/\1 /'

If you want the first and eighth fields regardless of the number of fields in the input, then it's just:

cut -d'|' -f1,8

^{(all those would work with any POSIX compliant utility assuming the input forms valid text (in particular, the sed ones will generally not work if the input has bytes or sequences of bytes that don't form valid characters in the current locale like for instance printf 'unix|St\351phane|Chazelas\n' | sed 's/|.*|/|/' in a UTF-8 locale)).}

Answer 4

If you find yourself awk- and sed-less, you can achieve the same thing with coreutils:

paste <(           cut -d'|' -f1  file) \ 
      <(rev file | cut -d'|' -f1 | rev)

Answer 5

It seems like you are try to get the first and last fields of text which are delimited by |.

I assumed your log file contains the text like below,

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

And you want the output like,

foo bar
bar foo

If yes, then here comes the command for your's

Through GNU sed,

sed -r 's~^([^|]*).*\|(.*)$~\1 \2~' file

Example:

$ echo 'foo|dog|cat|mouse|lion|ox|tiger|bar' | sed -r 's~^([^|]*).*\|(.*)$~\1 \2~'
foo bar

Answer 6

You should probably do it with sed - I would anyway - but, just cause no one has written this one yet:

while IFS=\| read col1 cols
do  printf %10s%-s\\n "$col1 |" " ${cols##*|}"
done <<\INPUT
foo|dog|cat|mouse|lion|ox|tiger|bar
INPUT

OUTPUT

     foo | bar