Question: I'm doing some testing to ensure that the all in one zip file that i created using a script file will produce the same output as the content of a few zip files that i must manually click and create via web interface. Therefore the zip will have different folder structure. 

Of course i can manually extracted them out and using my powerful eyeball technique to scan them or even lazier i can write a script to do that, but before i invest more time and get accused by my boss for company time robbery, i'm asking if there's a better way to do this?

I'm using perl LAMP stack by the way.
thanks.

Answer: 
You can use perl's [Archive::ZIP](http://search.cpan.org/~adamk/Archive-Zip-1.26/lib/Archive/Zip.pm) or Python's [zipfile](http://docs.python.org/library/zipfile.html) to extract the filenames, sizes and CRC checksums of the files in the archives. Create a file which contains the results sorted by file name (ignore the path).

For your smaller ZIPs, merge the results of the script (`cat list1 list2 list3 | sort`).

Now, you can use `diff` to compare the results.
I can wholeheartly recommend [Beyond Compare](http://www.scootersoftware.com/). Unless you're really getting underpaid, it's the biggest bang for your (bosses) buck. 

[Edit] I seem to have scanned over the *different* folder structure, sorry about that.Beyond Compare can compare all files in folders with the same folderstructure. It does not have (I believe) the *intelligence* to go searching for matches in files in different folders.

Regards,  

Lieven