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Home/ Questions/Q 5989359
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T23:05:25+00:00 2026-05-22T23:05:25+00:00

I have made a script that adds files to a zip (after lots of

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I have made a script that adds files to a zip (after lots of parsing, etc..) and I have this function:

function _add_file($command)
{
    if (!isset($command["source"]) || !isset($command["dest"])) {
        return;
    }

    if (!get_file_info("files/".$command["source"])) {
        return;
    }

    $zip->addFile("files/".$command["source"], $command["destination"]);
}

It gives an error because $zip is not defined in _add_file. How can I let _add_file access the $zip defined in the function that calls it (without _add_file($command, $zip))?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-22T23:05:26+00:00Added an answer on May 22, 2026 at 11:05 pm

    Make it a class variable var $zip and access it with $this->zip

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