Find and replace and move a line that contains a specific string - awk

Question

Assuming I have the following text file:
a b c d 1 2 3
e f g h 1 2 3
i j k l 1 2 3
m n o p 1 2 3
How do I replace '1 2 3' with '4 5 6' in the line that contains the letter (e) and move it after the line that contains the letter (k)?
N.B. the line that contains the letter (k) may come in any location in the file, the lines are not assumed to be in any order
My approach is
Remove the line I want to replace
Find the lines before the line I want to move it after
Find the lines after the line I want to move it after
append the output to a file
grep -v 'e' $original > $file
grep -B999 'k' $file > $output
grep 'e' $original | sed 's/1 2 3/4 5 6/' >> $output
grep -A999 'k' $file | tail -n+2 >> $output
rm $file
mv $output $original
but there is a lot of issues in this solution:
a lot of grep commands that seems unnecessary
the argument -A999 and -B999 are assuming the file would not contain lines more than 999, it would be better to have another way to get lines before and after the matched line
I am looking for a more efficient way to achieve that

Answer 1

Using sed
$ sed '/e/{s/1 2 3/4 5 6/;h;d};/k/{G}' input_file
a b c d 1 2 3
i j k l 1 2 3
e f g h 4 5 6
m n o p 1 2 3

Answer 2

Here is a GNU awk solution:
awk '
/\<e\>/{
s=$0
sub("1 2 3", "4 5 6", s)
next
}
/\<k\>/ && s {
printf("%s\n%s\n",$0,s)
next
} 1
' file
Or POSIX awk:
awk '
function has(x) {
for(i=1; i<=NF; i++) if ($i==x) return 1
return 0
}
has("e") {
s=$0
sub("1 2 3", "4 5 6", s)
next
}
has("k") && s {
printf("%s\n%s\n",$0,s)
next
} 1
' file
Either prints:
a b c d 1 2 3
i j k l 1 2 3
e f g h 4 5 6
m n o p 1 2 3
This works regardless of the order of e and k in the file:
awk '
function has(x) {
for(i=1; i<=NF; i++) if ($i==x) return 1
return 0
}
has("e") {
s=$0
sub("1 2 3", "4 5 6", s)
next
}
FNR<NR && has("k") && s {
printf("%s\n%s\n",$0,s)
s=""
next
}
FNR<NR
' file file

Answer 3

This awk should work for you:
awk '
/(^| )e( |$)/ {
sub(/1 2 3/, "4 5 6")
p = $0
next
}
1
/(^| )k( |$)/ {
print p
p = ""
}' file
a b c d 1 2 3
i j k l 1 2 3
e f g h 4 5 6
m n o p 1 2 3

Answer 4

This might work for you (GNU sed):
sed -n '/e/{s/1 2 3/4 5 6/;s#.*#/e/d;/k/s/.*/\&\\n&/#p};' file | sed -f - file
Design a sed script by passing the file twice and applying the sed instructions from the first pass to the second.
Another solution is to use ed:
cat <<\! | ed file
/e/s/1 2 3/4 5 6/
/e/m/k/
wq
!
Or if you prefer:
<<<$'/e/s/1 2 3/4 5 6/\n.m/k/\nwq' ed -s file