Question

I would like a command line function that I can run on any file to change the include("myinc.inc"); PHP statement to include 'myfile.inc'; I have made a start by adding the following to my ~/.bashrc file:

function makestandard() {
    perl -p -i -e 's/include\("([\w\.]+)"\)/include '$1'/g' $*
}

I source ~/.bashrc; and run the command at the command line as follows:

$ makestandard myfile.php

I get myfile.php modified but instead of capturing the included file name, the included file name is corrupted to be the name of the current file. As a uninformed guess, I think the bash $1 variable is interfering with the $1 perl regexp variable.

How can I fix this?

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Background info (no need to read): We have started using PHP_CodeSniffer (phpcs) to sniff PHP code and report any bad “smells”. Unfortunatly, phpcs doesn’t fix the non-standard code, it only reports it. Hence I would like to make a script that fixes some of the easy and common non-standard elements of our PHP code. I plan to fill my makestandard bash function with a bunch of perl pie.

Shell environment: Whatever is out-of-the-box in Ubuntu 10.04.

Answer 1

It isn’t the bash variable, it is the single quotes in your single quoted bash string.