AWK print based on FILENAME pattern - regex

Question

I have a directory of files with filenames of the form file000.txt to filennn.txt. I would like to be able to specify a range of file names and print the content of those files based on a match. I have achieved it with a single file pattern:
$ gawk 'FILENAME ~/file038.txt/ {print FILENAME, $0}' file*.txt
file038.txt Some 038 text here
But I cannot get a pattern that would allow me to specify a range of file names, for instance
gawk 'FILENAME ~/file[038-040].txt/ {print FILENAME, $0}' file*.txt
I'm sure I'm missing something simple here, I'm an AWK newbie. Any suggestions?

Answer 1

you can do some substitution on the filename, for example:
awk '{x=FILENAME;gsub(/[^0-9]/,"",x);x+=0}x>10&&x<50{your logic}' file*.txt
in this way, file file011.txt ~ file049.txt would be handled with "your logic"
You can adjust the part: x>10&&x<50 for example, handle only file with the number in the name as odd/even/.... just write boolean expressions there.

Answer 2

Odd way but something on these lines:
awk '{ if (match(FILENAME,/file0[3-4][0-8].txt/)) { print FILENAME, $0}}' file*.txt

Answer 3

Solution using gawk and a recent version of bash
There is a bash primitive to handle file[038-040].txt. It makes the code quite simple:
gawk 'FNR==1 {print FILENAME, $0} {quit}' file{038..040}.txt
Key points:
FNR==1 {print FILENAME, $0}
This prints the filename and the first line of each file
{quit}
This saves time by skipping directly to the next file.
file{038..040}.txt
The construct {038..040} is a bash feature called brace expansion. bash will replace this with the file names that you want. If you want to test out brace expansion to see how it works, try it on the command line with this simple statement:
echo file{038..040}.txt
UPDATE 1: Mac OSX currently uses bash v3.2 which does not support leading zeros in brace expansion.
UPDATE 2: If there are missing files and you have a modern gawk (v4.0 or better), use this instead:
gawk 'BEGINFILE{ if (ERRNO) nextfile} FNR==1 {print FILENAME, $0} {quit}' file{038..040}.txt
Solution using gawk with a plain POSIX shell
gawk '{n=0+substr(FILENAME,5,3)} FNR==1 && n>=38 && n<=40 {print FILENAME, $0} {quit}' file*.txt
Explanation:
n=0+substr(FILENAME,5,3)
Extract the number from the filename. 0+ is a trick to force awk to treat n as numeric.
n>=38 && n<=40 {print FILENAME, $0}
This selects the file based on its number and prints the filename and first line.
{quit}
As before, this saves time by stopping awk from reading the rest of each file.
file*.txt
This can be expanded by any POSIX shell to the list of file names.

Answer 4

Should work
awk '(x=FILENAME)~/(3[8-9]|40).txt$/{print x,$0;quit}' file*.txt
As quit doesn't work(atleast with my version of awk) here is another way
awk 'FNR==((x=FILENAME)~/(3[8-9]|40).txt$/){print x,$0}' file*.txt