Extracting using awk in a specific column

Question

I have 4 fields and I want to start from last column and ignore the first 3 columns. How can I use NF in awk to do that?

root     pts/0        192.168.108.1    Mon Mar 19 08:45   still logged in
root     tty1                          Mon Mar 19 08:45   still logged in
reboot   system boot  3.10.0-693.el7.x Mon Mar 19 08:44 - 08:49  (00:04)
root     pts/0        192.168.108.1    Sun Mar 18 13:06 - crash  (19:37)
root     tty1                          Sun Mar 18 13:06 - crash  (19:38)
reboot   system boot  3.10.0-693.el7.x Sun Mar 18 13:01 - 08:49  (19:47)
root     pts/2        192.168.108.1    Sat Mar 17 12:38 - 13:09  (00:31)
root     pts/1        192.168.108.1    Sat Mar 17 10:49 - down   (02:20)
root     pts/1        192.168.108.1    Fri Mar 16 23:17 - 00:30  (01:13)
root     pts/0        192.168.108.1    Fri Mar 16 21:25 - 12:56  (15:31)
root     tty1                          Fri Mar 16 21:25 - 13:09  (15:44)
reboot   system boot  3.10.0-693.el7.x Fri Mar 16 21:04 - 13:09  (16:05)
root     pts/2        192.168.108.1    Fri Mar 16 08:45 - crash  (12:18)
root     tty1                          Fri Mar 16 08:45 - 17:25  (08:40)
syslog                                          **Never logged in**

I want to show only:

Sun Jan 19 13:52:08 -0800 2018
**Never logged in**
Fri Mar 16 08:45 - 17:25  (08:40)
Mon Mar 19 08:45   still logged in

Answer 1

You could do it with a simple awk command to print the last column contents, and using a multiple spaces as the field separator. Since the default separator in awk is a single white-space, using it directly on the last column would split the contents of the last column. The NR> condition takes care of skipping the first line in applying the actual actions defined for awk.

awk -F '[[:space:]][[:space:]]+' 'NR>1{print $NF}' file

or using sed too, assuming multiple spaces between columns and column separated words only in the last column.

sed '1d;s/^.*  //' file

---

OP apparently re-phrased the question to dump the output of the last column command from which they wanted the last column output. Since the intermediate columns themselves could have spaces, we do a match of the column containing the day name and print from there to the end of the line i.e.

last | awk ' {
     for( i=1;i<=NF;i++ ) {
         if ( $i ~ /Mon|Tue|Wed|Thu|Fri|Sat|Sun/ ) {
             j = 0
             str = ""
             for ( j=i; j<=NF;j++ ) {
                 str = ( str ? (str FS $j):$j )
             }
             print str
             break
         }
     }
 }'

Answer 2

For the original version of the question, I had liked and up-voted Inian's answer, and deleted this one, but now that the question has changed....

awk 'NR>1{if ($2~/^\*\*/) $2=foo;if ($3!~/^[[:digit:].]+$/) $3=foo; $1=$2=$3="";print}' file

The if statements should place dummy values in the second and third columns if the second is **Never or if the third isn't an IP address, but will cause the answer to fail if other funky possibilities exist...

For readability, let's re-post the one-liner since it's a bit long...

awk 'NR > 1 {
  if ( $2 ~ /^\*\*/ ) $2=foo
  if ( $3 !~ /^[[:digit:].]+$/ ) $3=foo
  $1=$2=$3=""
  print }' file