Extract lines after a pattern - awk

Question

I have 50 files in a folder and all have a common pattern "^^". I want to print everything after "^^" and append the filename and print out all the extracted lines to one output file. While my code works fine with a single file it doesn't work on all the files.
awk '/\^^/{getline; getline; print FILENAME; print}' *.txt > output
Example
1.txt
ghghh hghg
ghfg hghg hjg
jhhkjh
kjhkjh kjh
^^
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
2.txt
hghjhg hgj
jhgj
jhgjh kjgh
jhg
^^
bbbbbbbbbbbbbbbbbbbbbbb
Desired output.txt
1.txt
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
2.txt
bbbbbbbbbbbbbbbbbbbbbbbbbbbbbb
My actual output
1.txt
ghghh hghg
1.txt
zzzzzzzzzzzzzzzzzzzzzzzzzzzzz

Answer 1

To print the line after ^^, try:
$ awk 'f{print FILENAME ORS $0; f=0} /\^\^/{f=1}' *.txt
1.txt
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
2.txt
bbbbbbbbbbbbbbbbbbbbbbb
How it works:
f{print FILENAME ORS $0; f=0}
If variable f is true (nonzero), print the filename, the output record separator, and the current line. Then set f back to zero.
/\^\^/{f=1}
If the current line contains ^^, set f to one.

Answer 2

$ awk 'FNR==1{print FILENAME; f=0} f; $1=="^^"{f=1}' *.txt
1.txt
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
2.txt
bbbbbbbbbbbbbbbbbbbbbbb

Answer 3

I like a more "bash(ish)" approach.
grep -Hn '^^' *.txt |
cut -d: -f1,2 --output-delimiter=' ' |
while read f n; do echo $f; tail $f -n+$((n+1)); done
grep -Hn will tell the line number of your pattern.
With cut we get only the needed fields, as we need.
In a loop we read the two informations into variables, to use they freely as we need.
The tail can read not only the last N lines, but also all lines from +N if you use the plus signal.
We can do arithmetic operation inside $((...)) to jump the pattern line.
And it solves your issue. And it can print all lines after the pattern, not only the next one.

Answer 4

use awk:
awk 'FNR==1{print FILENAME} FNR==1,/\^\^/{next}1' *.txt
Where:
print FILENAME when FNR == 1
FNR==1,/\^\^/{next}: all lines between FNR==1 and the first line matching ^^ will be skipped
1 at the end to print the rest of lines after the matched ^^ line

Answer 5

The following outputs only if we have a file that matches our pattern:
awk 'FNR==1 { f=0 }; f; /\^\^/ { f=1; print FILENAME }' *.txt > output
Reset flag f on every new file.
Print if f is set.
Set f and print FILENAME if we match our pattern.
This one prints out the FILENAME regardless of matching pattern:
awk 'FNR==1 { f=0; print FILENAME }; f; /\^\^/ { f=1 }' *.txt > output
We can adjust the pattern matching in step 3 in accord with whatever is required... exact matching for instance can be done with $0=="^^".

Answer 6

let your files' name are 1 to 50 with txt type
for f in {1..50}.txt
{
sed -nE "/^\^\^\s*$/{N;s/.+\n(.+)/$f\n\1/p}" $f>$f.result.txt
}

Answer 7

Stealing from some answers and comments to your previous question on this topic, you can also use grep -A and format the output with sed.
$ grep -A100 '^^' *.txt | sed '/\^^/d;/--/d;s/-/\n/'
1.txt
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
2.txt
bbbbbbbbbbbbbbbbbbbbbbb
Assuming 100 lines is sufficient, and that you don't have hyphens of your own.
If you only need one line, use -A1

Answer 8

This might work for you (GNU sed):
sed -s '1,/^^^/{/^^^/F;d}' file1 file2 file3 ... >fileOut