Question

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?

Answer 1

Windows cmd.exe does not use ' as string delimiters, only '. What you’re doing is equivalent to:

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

so -w is complaining ‘you gave me a string but it does nothing’.

The solution is to use double quotes instead:

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

or to simply leave them off, since there’s no metacharacters in this command that would be interpreted by cmd.exe.

Addendum

cmd.exe is just a really troublesome beast, for anybody accustomed to sh-like shells. Here’s a few other common failures and workarounds regarding perl invocation.

 @REM doesn't work: perl -e'print' @REM works: perl -e 'print' @REM doesn't work: perl -e 'print \'Hello, world!\n\'' @REM works: perl -e 'print qq(Hello, world!\n)'