Extract specific words from a line

Question

I hope someone here can help me. I have a line in a text file looking like this:

Jan  8 14:12:56 kernel: SRC=1.2.3.4 DST=255.255.255.255 LEN=104 TOS=0x00 PREC=0x00 TTL=64 ID=0 DF PROTO=UDP SPT=44224 DPT=14000 LEN=84

I want to extract the words starting with SRC=, PROTO= and DPT=. My goal is to end up with a line looking something like this:

1.2.3.4 UDP 14000

I would prefer the solution being bash using sed, awk or similar if possible.

Solution 1

Use Sed with groups:

sed -r 's/.*SRC=(\S+).*PROTO=(\S+).*DPT=(\S+).*/\1 \2 \3/'

Solution 2

One way using awk:

awk 'BEGIN { FS = "[ =]" } { print $7, $22, $26 }' infile

Output:

1.2.3.4 UDP 14000

Solution 3

If the output is generated in a fixed order, then you could simply use shell builtins.

grep SRC= /var/log/messages |
while read mon day time kernel src dst len tos prec ttl id if proto spt dpt etc; do
    echo ${src#*=} ${proto#*=} ${dpt#*=}
done

If you have the data in $string and the desired parameters are at fixed positions, you could also

set -- $string
echo ${5#SRC=} ${13#PROTO=} ${15#DPT=}

If your shell can't handle positional parameters beyond $9 you will need a few shifts.

Comments

[potong]

You might replace [^ ]* by \S*