AWK to process compressed files and printing original (compressed) file names - awk

Question

I would like to process multiple .gz files with gawk.
I was thinking of decompressing and passing it to gawk on the fly
but I have an additional requirement to also store/print the original file name in the output.
The thing is there's 100s of .gz files with rather large size to process.
Looking for anomalies (~0.001% rows) and want to print out the list of found inconsistencies ALONG with the file name and row number that contained it.
If I could have all the files decompressed I would simply use FILENAME variable to get this.
Because of large quantity and size of those files I can't decompress them upfront.
Any ideas how to pass filename (in addition to the gzip stdout) to gawk to produce required output?

Answer 1

Assuming you are looping over all the files and piping their decompression directly into awk something like the following will work.
for file in *.gz; do
gunzip -c "$file" | awk -v origname="$file" '.... {print origname " whatever"}'
done
Edit: To use a list of filenames from some source other than a direct glob something like the following can be used.
$ ls *.awk
a.awk e.awk
$ while IFS= read -d '' filename; do
echo "$filename";
done < <(find . -name \*.awk -printf '%P\0')
e.awk
a.awk
To use xargs instead of the above loop will require the body of the command to be in a pre-written script file I believe which can be called with xargs and the filename.

Answer 2

this is using combination of xargs and sh (to be able to use pipe on two commands: gzip and awk):
find *.gz -print0 | xargs -0 -I fname sh -c 'gzip -dc fname | gawk -v origfile="fname" -f printbadrowsonly.awk >> baddata.txt'
I'm wondering if there's any bad practice with the above approach…