Awk multiple pattern match and print in single line

awktext processing

I have the following file:

$ cat disk.out
disk0
fcs0
text
text
text
disk1
fcs1
text
text
text
text
...

What I am trying to achieve is match "disk" + "fcs" and then print the pair in one line, like this:

disk0,fcs0
disk1,fcs1
...

So I am matching "disk" and "fcs" with awk and changing the output record separator to ",".`

$ awk '/disk|fcs/' ORS="," disk.out
disk0,fcs0,disk1,fcs1,

The problem is, it will print all the matches on one line and with a trailing ,.
How can I print only per match in one line? Like this:

disk0,fcs0
disk1,fcs1
...

You have to save the "disk" line (without printing it) until you find the next "fcs" line:

awk '/disk/{ DISK=$0; next } /fcs/{ print DISK "," $0 }'

The problem with your approach is that it prints either line matching "disk" or "fcs", without combining those lines.

Edit: the script of sp asic is more robust, in that it ignores

disk3
text
fcs3

My script would happily print "disk3,fcs3" in this case.

Though the expected output is not clear, I will try. If you have the gnu utils on Solaris, you can use gawk and ggrep in a very simple manner:

gawk 'NF > 0' fileName  |ggrep -C5 pattern

You can add a | ggrep -v somePattern at the end to filter out unwanted words.

One way is to use flags and counter to keep track

$ awk '/b/{f=1; c=0} f; /d/ && ++c==2{f=0}' file.txt
b
c
d
e
f
d

/b/{f=1; c=0} set flag for starting match and initialize counter
f; print input record as long flag is set
/d/ && ++c==2{f=0} clear the flag if ending string is matched for second time
can simplify to awk '/b/{f=1} f; /d/ && ++c==2{exit}' if there is only one set to be extracted

Note that if ending match doesn't have 2 matches, it will print until end of input

$ seq 10 | awk '/4/{f=1; c=0} f; /6/ && ++c==2{f=0}'
4
5
6
7
8
9
10