[linux] How to print third column to last column?

I'm trying to remove the first two columns (of which I'm not interested in) from a DbgView log file. I can't seem to find an example that prints from column 3 onwards until the end of the line. Note that each line has variable number of columns.

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

The following awk command prints the last N fields of each line and at the end of the line prints a new line character:

awk '{for( i=6; i<=NF; i++ ){printf( "%s ", $i )}; printf( "\n"); }'

Find below an example that lists the content of the /usr/bin directory and then holds the last 3 lines and then prints the last 4 columns of each line using awk:

$ ls -ltr /usr/bin/ | tail -3
-rwxr-xr-x 1 root root       14736 Jan 14  2014 bcomps
-rwxr-xr-x 1 root root       10480 Jan 14  2014 acyclic
-rwxr-xr-x 1 root root    35868448 May 22  2014 skype

$ ls -ltr /usr/bin/ | tail -3 | awk '{for( i=6; i<=NF; i++ ){printf( "%s ", $i )}; printf( "\n"); }'
Jan 14 2014 bcomps 
Jan 14 2014 acyclic 
May 22 2014 skype

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

Jonathan Feinberg's answer prints each field on a separate line. You could use printf to rebuild the record for output on the same line, but you can also just move the fields a jump to the left.

awk '{for (i=1; i<=NF-2; i++) $i = $(i+2); NF-=2; print}' logfile

awk '{$1=$2=$3=""}1' file

NB: this method will leave "blanks" in 1,2,3 fields but not a problem if you just want to look at output.

awk '{ print substr($0, index($0,$3)) }'

solution found here:
http://www.linuxquestions.org/questions/linux-newbie-8/awk-print-field-to-end-and-character-count-179078/

If you want to print the columns after the 3rd for example in the same line, you can use:

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

For example:

Mar 09:39 20180301_123131.jpg
Mar 13:28 20180301_124304.jpg
Mar 13:35 20180301_124358.jpg
Feb 09:45 Cisco_WebEx_Add-On.dmg
Feb 12:49 Docker.dmg
Feb 09:04 Grammarly.dmg
Feb 09:20 Payslip 10459 %2828-02-2018%29.pdf

It will print:

20180301_123131.jpg
20180301_124304.jpg
20180301_124358.jpg
Cisco_WebEx_Add-On.dmg
Docker.dmg
Grammarly.dmg
Payslip 10459 %2828-02-2018%29.pdf

As we can see, the payslip even with space, shows in the correct line.

awk '{a=match($0, $3); print substr($0,a)}'

First you find the position of the start of the third column. With substr you will print the whole line ($0) starting at the position(in this case a) to the end of the line.

In AWK columns are called fields, hence NF is the key

all rows:

awk -F '<column separator>' '{print $(NF-2)}' <filename>

first row only:

awk -F '<column separator>' 'NR<=1{print $(NF-2)}' <filename>

In Bash you can use the following syntax with positional parameters:

while read -a cols; do echo ${cols[@]:2}; done < file.txt

Learn more: Handling positional parameters at Bash Hackers Wiki

awk '{$1=$2=""}1' FILENAME | sed 's/\s\+//g'

First two columns are cleared, sed removes leading spaces.

Perl solution:

perl -lane 'splice @F,0,2; print join " ",@F' file

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 – split input lines into the @F array. Defaults to splitting on whitespace

• -e execute the perl code

splice @F,0,2 cleanly removes columns 0 and 1 from the @F array

join " ",@F joins the elements of the @F array, using a space in-between each element

If your input file is comma-delimited, rather than space-delimited, use -F, -lane

Python solution:

python -c "import sys;[sys.stdout.write(' '.join(line.split()[2:]) + '\n') for line in sys.stdin]" < file

awk '{for (i=4; i<=NF; i++)printf("%c", $i); printf("\n");}'

prints records starting from the 4th field to the last field in the same order they were in the original file

awk -v m="\x0a" -v N="3" '{$N=m$N ;print substr($0, index($0,m)+1)}'

This chops what is before the given field nr., N, and prints all the rest of the line, including field nr.N and maintaining the original spacing (it does not reformat). It doesn't mater if the string of the field appears also somewhere else in the line, which is the problem with daisaa's answer.

Define a function:

fromField () { 
awk -v m="\x0a" -v N="$1" '{$N=m$N; print substr($0,index($0,m)+1)}'
}

And use it like this:

$ echo "  bat   bi       iru   lau bost   " | fromField 3
iru   lau bost   
$ echo "  bat   bi       iru   lau bost   " | fromField 2
bi       iru   lau bost 

Output maintains everything, including trailing spaces

Works well for files where '/n' is the record separator so you don't have that new-line char inside the lines. If you want to use it with other record separators then use:

awk -v m="\x01" -v N="3" '{$N=m$N ;print substr($0, index($0,m)+1)}'

for example. Works well with almost all files as long as they don't use hexadecimal char nr. 1 inside the lines.

What about following line:

awk '{$1=$2=$3=""; print}' file

Based on @ghostdog74 suggestion. Mine should behave better when you filter lines, i.e.:

awk '/^exim4-config/ {$1=""; print }' file

If its only about ignoring the first two fields and if you don't want a space when masking those fields (like some of the answers above do) :

awk '{gsub($1" "$2" ",""); print;}' file

A bit late here, but none of the above seemed to work. Try this, using printf, inserts spaces between each. I chose to not have newline at the end.

awk '{for(i=3;i<=NF;++i) printf("%s ",  $i) }'

Well, you can easily accomplish the same effect using a regular expression. Assuming the separator is a space, it would look like:

awk '{ sub(/[^ ]+ +[^ ]+ +/, ""); print }'