How to print the next or previous line using awk? - awk

Question

I have a file with 8 columns
1743 abc 04 10 29 31 34 35
1742 def 11 19 21 23 27 52
1741 ghi 15 18 20 32 48 49
and I also have a awk line that print the complete line that contains some specific numbers. The code is
awk -v col=1 '{ delete c; for (i=col; i<=NF; ++i) ++c[$i];
if (c['"$1"']>0 && c['"$2"']>0 && c['"$3"']>0 && c['"$4"']>0) print }'
< input_file
(the variables $1,$2,$3 and $4 is because I'm using it on bash).
In the previous example, when I put the numbers 11 21 27 and 52 I'll get the line 1742.
How can I print the next or the previous line? Like in the previous example, if I use the numbers, 11 21 27 and 52 how I take the line 1743 or the line 1741?

Answer 1

$ cat a.sh
echo "BEFORE"
awk -v p1="$1" -v p2="$2" -v p3="$3" -v p4="$4" -v col=1 -f before.awk file
echo "AFTER"
awk -v p1="$1" -v p2="$2" -v p3="$3" -v p4="$4" -v col=1 -f after.awk file
Quoting #triplee: "To print the previous line, remember the previous line in a variable."
$ cat before.awk
prev { delete c;
for (i=col; i<=NF; ++i) ++c[$i]
if (c[p1]>0 && c[p2]>0 && c[p3]>0 && c[p4]>0) print prev
}
{ prev = $0 }
Again, #triplee: "To print the next line, remember that you want to, and print and reset this variable on the next iteration."
$ cat after.awk
f { print; f = 0 }
{
delete c;
for (i=col; i<=NF; ++i) ++c[$i]
if (c[p1]>0 && c[p2]>0 && c[p3]>0 && c[p4]>0) f = 1
}
$ ./a.sh 11 21 27 52
BEFORE
1743 abc 04 10 29 31 34 35
AFTER
1741 ghi 15 18 20 32 48 49

Answer 2

a different approach with double scanning
$ awk -v search="11 21 27 52" -v offset=-1 '
NR==FNR {n=split(search,s);
for(i=1;i<=n;i++) if(FS $0 FS !~ FS s[i] FS) next;
line=NR; next}
FNR==line+offset' file{,}
1743 abc 04 10 29 31 34 35
you can set offset to any value (not just -1,0,1).
N.B. It only find one match though, if there are multiple matches only the last one will be reported. This can be handled by keeping the matched line numbers in an array instead of a scalar value (here line variable).