grep -A1 'blah' logfile
Thanks to this command for every line that has 'blah' in it, I get the output of the line that contains 'blah' and the next line that follows in the logfile. It might be a simple one but I can't find a way to omit the line that has 'blah' and only show next line in the output.
I don't know of any way to do this with grep, but it is possible to use awk to achieve the same result:
awk '/blah/ {getline;print}' < logfile
It looks like you're using the wrong tool there. Grep isn't that sophisticated, I think you want to step up to awk as the tool for the job:
awk '/blah/ { getline; print $0 }' logfile
If you get any problems let me know, I think its well worth learning a bit of awk, its a great tool :)
p.s. This example doesn't win a 'useless use of cat award' ;) http://porkmail.org/era/unix/award.html
perl one-liner alert
just for fun... print only one line after match
perl -lne '$next = ($.+1) if /match/; $. == $next && print' data.txt
even more fun... print the next ten lines after match
perl -lne 'push @nexts, (($.+1)..($.+10)) if /match/; $. ~~ @nexts && print' data.txt
kinda cheating though since there's actually two commands
In general, I agree you're asking a lot of grep here, and that another tool may be the better solution. But in an embedded environment, I may not want to have sed or awk just to do this. I found the following solution works (as long as they're not contiguous matches):
grep -A1 AT\+CSQ wvdial.out | grep -v AT\+CSQ
Basically, match them, appending 1 line of context for each match, and then pipe that through an inverse match of your original pattern to strip those out. This of course means you can assume that your pattern doesn't show up in the "next" line.
Many good answers have been given to this question so far, but I still miss one with awk not using getline. Since, in general, it is not necessary to use getline, I would go for:
awk ' f && NR==f+1; /blah/ {f=NR}' file #all matches after "blah"
or
awk '/blah/ {f=NR} f && NR==f+1' file #matches after "blah" not being also "blah"
The logic always consists in storing the line where "blah" is found and then printing those lines that are one line after.
Sample file:
$ cat a
0
blah1
1
2
3
blah2
4
5
6
blah3
blah4
7
Get all the lines after "blah". This prints another "blah" if it appears after the first one.
$ awk 'f&&NR==f+1; /blah/ {f=NR}' a
1
4
blah4
7
Get all the lines after "blah" if they do not contain "blah" themselves.
$ awk '/blah/ {f=NR} f && NR==f+1' a
1
4
7
Great answer from raim, was very useful for me. It is trivial to extend this to print e.g. line 7 after the pattern
awk -v lines=7 '/blah/ {for(i=lines;i;--i)getline; print $0 }' logfile
you can use grep, then take lines in jumps:
grep -A1 'blah' logfile | awk 'NR%3==2'
you can also take n lines after match, for example:
seq 100 | grep -A3 .2 | awk 'NR%5==4'
15
25
35
45
55
65
75
85
95
explanation -
here we want to grep all lines that are *2 and take 3 lines after it, which is *5.
seq 100 | grep -A3 .2 will give you:
12
13
14
15
--
22
23
24
25
--
...
the number in the modulo (NR%5) is the added rows by grep (here it's 3 by the flag -A3), +2 extra lines because you have current matching line and also the -- line that the grep is adding.
If you want to stick to grep:
grep -A1 'blah' logfile | grep -v "blah"
or alternatively with sed:
sed -n '/blah/{n;p;}' logfile
grep /Pattern/ | tail -n 2 | head -n 1
Tail first 2 and then head last one to get exactly first line after match.
Piping is your friend...
Use grep -A1 to show the next line after a match, then pipe the result to tail and only grab 1 line,
cat logs/info.log | grep "term" -A1 | tail -n 1
If that next lines never contain 'blah', you can filter them with:
grep -A1 blah logfile | grep -v blah
The use of cat logfile | ... is not needed.
Source: Stackoverflow.com