CODEWITHSUNDEEP
perlzip
Chani
Chani

Reputation: 5175

How is this zip command being used in perl?

I found this piece of code in perl 

system("zip $ZIP_DEBUG -r -9 itvlib.zip $include $exclude");

However I don't understand how it is working. I mean system() is used to fire 'system' commands right ? So is this 'zip' command used here a 'system' command ? But I tried firing just the following on the command prompt; 

zip $ZIP_DEBUG -r -9 itvlib.zip arg1 arg2

It didn't work ! it gave the following error:

'zip' is not recognized as an internal or external command,
operable program or batch file.

Well this shouldn't have happened, since the command seems to use 'zip' as a system command. So this makes the command 'zip' mysterious

Can you please help me to understand this command with all its parameters?

Upvotes: 2

Views: 2836

Answers (3)

cjm
cjm

Reputation: 62109

system("zip $ZIP_DEBUG -r -9 itvlib.zip $include $exclude");

is running a program named zip (probably zip.exe) somewhere on the path. $ZIP_DEBUG, $include, and $exclude are Perl variables that are interpolated into the command line before the command is run.

If the system call works in the Perl script, but zip -? gives the 'zip' is not recognized as an internal or external command, operable program or batch file error, then the PATH of the Perl script must be different than the PATH in your command prompt. Or, there might be a zip command in the current directory when Perl executes the system command. (In Windows, the current directory is an implicit member of your PATH.)

To see what the PATH is for the Perl script, you can add a print "$ENV{PATH}\n"; before the system command. To see what the PATH is in your command prompt, type PATH.

Upvotes: 4

paxdiablo
paxdiablo

Reputation: 882296

It's probably not working since you're not replacing things like $ZIP_DEBUG with their equivalent real values. Within Perl, they will be replaced with the values of the variables before being passed to the system call.

If you print out those Perl variables (or even the entire command) before you execute that system call, you'll find out those real values that you need to use. You can use the following transcript to guide you:

$ perl -e '
>     $ZIP_DEBUG = "xyzzy";
>     $include = "inc_files";
>     $exclude = "exc_files";
>     print "zip $ZIP_DEBUG -r -9 itvlib.zip $include $exclude";
> '
zip xyzzy -r -9 itvlib.zip inc_files exc_files

For details on how system works, see here. For details on what zip needs to function, you should just be able to run:

man zip

from a command line shell (assuming you're on Linux or its brethren). If, instead, you're on a different operating system (like Windows), you'll have to figure out how to get the zip options out. This may well be as simple as zip -? of zip -h but there's no guarantee that will work.

If it's the same as the Info-ZIP zip under Linux (and it may be if you have the -9 and -r options and your exclude variable starts with -x), then zip -h will get you basic help and zip -h2 will give you a lot more.

Upvotes: 5

tuomassalo
tuomassalo

Reputation: 9121

Yes, zip is a system command. The variables $ZIP_DEBUG and such are perl variables that are interpolated to the command before launching zip.

To debug what the actual call is, try adding:

print("zip $ZIP_DEBUG -r -9 itvlib.zip $include $exclude\n");

See perldoc for details on system.

Upvotes: 3

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