From the Windows command prompt I generate a text file of all the files in a directory:

dir c:\logfiles /B > config.txt

Output:

0001_832ec657.log
0002_a7c8eafc.log

I need to feed the "config.txt" file to another executable, but before I do so, I need to modify the file to add some additional information that the executable needs. So I use the following command:

perl -p -i.bak -e 's/log/log,XYZ/g' config.txt

I'm expecting the result to be:

0001_832ec657.log,XYZ
0002_a7c8eafc.log,XYZ

However, the "config.txt" file is not modified. Using the "-w" option, I get the warning message:

Useless use of a constant in void context at -e line 1.

What am I doing wrong?

windows perl
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