Re: dependency check

On Fri, Jul 22, 2011 at 8:17 AM, Nilesh Govindarajan <contact@nileshgr.com>wrote: > On 07/22/2011 11:21 AM, Andreas Moroder wrote: > > Hallo, > > > > I have a PHP application made of many files ( php, images etc. ) > > I have a strong suspicion that many of the files in the application > > directory are no more in use, because of changes made on the application. > > Is there a tool that, starting from the entry point of the application, > > scans the files recursively for included/used files and lists them ? > > With this list I could delete the remaining files. > > > > Thanks > > Andreas > > > > > > You could write a python or even php script to do that, storing all the > files an array/list and then finding files (regex) which are not > included in any of the php files. > > Of course, this applies if and only if you haven't used __autoload() magic. > > -- > Regards, > Nilesh Govindarajan > @nileshgr on twitter/identica > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > Or you could just grep the directory, not saying you have to do this, but this was kind of fun to write anyways, if i spent more time on it, i could perfect it, but i dont have that kind of time, so this will still give you a few doubles, but it shouldn't give you false-positives as long as you have all the extensions in that grep regex (and you cant make it more generic without introducing false-positives)... grep -oiPR "[a-zA-Z0-9]+\.(php|js|png| jpg|css|htm|html)" directory | awk 'function getfiles(input, files, i, n, file) {result = ""; n=split(input, files, ":"); for(i=0; i<=n; i++) { if(files[i] !~ /^\s*$/) print files[i];}} {getfiles($0)}' | sort -biu This should give you all the files that reference files and the files they reference. -- The trouble with programmers is that you can never tell what a programmer is doing until it’s too late. ~Seymour Cray