in reply to Path Of an extracted Files

Use the extractMember() function and provide a second argument, which will be used as the output name.

PS: RTFM!

    -- Chip Salzenberg, Free-Floating Agent of Chaos

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Re: Re: Path Of an extracted Files
by Zitoune (Beadle) on Mar 11, 2004 at 18:02 UTC
    I know that. But you can't give a specific path when doing this.
    I tried it and it was giving me thsi error :
    mkdir AFCUArchive::Zip::.: Invalid argument at C:/Perl/site/lib/Archive/Zip.pm line 1760
[reply]
      You can give a specific path, that's what the FM says, which you should R. Maybe you're giving an invalid path, which would explain why the mkdir() is failing.

          -- Chip Salzenberg, Free-Floating Agent of Chaos

[reply]
[d/l]
        This is how i'm doing it : 
                     print "Y:\\autoupload\\Site_Dict\\Inbox\\$y\n";
            $zip->extractMemberWithoutPaths($y,"Y:\\autoupload\\Site_D
+ict\\Inbox\\"."$y");

        [download]
        Oliver
[reply]
[d/l]

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