Using awk to print all columns from the nth to the last

Question

Using awk to print all columns from the nth to the last

This code worked until I had whitespace in the second area.

svn status | grep '\!' | gawk '{print $2;}' > removedProjs

Is there a way to have awk print everything in $2 or greater? ($3, $4.. until we don't have any more columns?)

I suppose I should add that I'm doing this in a Windows environment with Cygwin.

awk linux

3 years ago by Kajalsi45d

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**espadacoder11** · Answer 1 · 2021-06-16T13:04:22

Print all columns:



awk '{print $0}' somefile

Print all but the first column:



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

Print all but the first two columns:



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

**sandhya6gczb** · Answer 2 · 2021-06-22T10:42:44

You could use a for-loop to loop through printing fields $2 through $NF (a built-in variable that represents the number of fields on the line).

Since "print" appends a newline, you'll want to buffer the results:



awk '{out=""; for(i=2;i<=NF;i++){out=out" "$i}; print out}'

Alternatively, use printf:



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

**kshitijrana14** · Answer 3 · 2021-06-23T04:48:06

kshitijrana14

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

3 years ago

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Using awk to print all columns from the nth to the last

3 Answers

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