how to get the input file name when perl script runs from the command line?

greatshots has asked for the wisdom of the Perl Monks concerning the following question:

monks,

perl -nle '/TABLE;(.*)/?($t=$1):/COLUMN;([^;]*)/&&print(qq{$t.$1})' *
[download]

The above commad run against 450 files. I would also like to print the input_file_name which is being processed inside the perl code. how to get that?

Comment on how to get the input file name when perl script runs from the command line? Download Code

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Re: how to get the input file name when perl script runs from the command line? by davido (Cardinal) on Oct 20, 2006 at 02:58 UTC
As described in perlvar, `$ARGV` "contains the name of the current file when reading from <>." When iterating over a file using the -n switch (as you're doing ... described in perlrun), the `<>` operator is silently being invoked behind the scenes, which is why `$ARGV` is relevant. Dave	[reply] [d/l] [select]
Re^2: how to get the input file name when perl script runs from the command line? by greatshots (Pilgrim) on Oct 20, 2006 at 03:10 UTC
`perl -nle '/<meta-property type=\"meta-attribute\">(.)\<\/meta-proper +ty\>/ and print "$ARGV:$1"' ` [download] worked as expected. thanks a bunch	[reply] [d/l]