PHP: $zip->addFile issues

Question

I have written a simple script to add files to an archive. After much head scratching I can't get the script to work.

I have a php file here which reads from a check-box, the file/s selected in the check-box are added to the array $file.

$path = ('a path');
$dir = scandir('another path');

    if(isset($_POST["submit"])){
    if(!isset($_POST["File"])){
        echo 'A file must be selected to archive it';
        } else { 
            require_once('zip_function.php');
            $file = $_POST['File'];
            $goZipper = goZipper($file, $path);
            if($goZipper == false) {
                echo ("Error");
            } else { 
                echo ("Success");
            }
        }   
    }

The function goZipper is called and $file and the destination are passed to it, the function is below.

function goZipper($files, $destination = '',$overwrite = true) {
    if(file_exists($destination) && !$overwrite) {
        echo"File exists";
        return false; 
        }
    if(count($files)){
        $zip = new ZipArchive();
            if($zip->open($destination,$overwrite ? ZIPARCHIVE::OVERWRITE : ZIPARCHIVE::CREATE) !== true){
                echo"Error opening destination";
                return false;
            } 
            foreach($files as $file){
                $zip->addFile($file,basename($file));

                    echo("<pre>");
                    var_dump($zip);
                    exit();

            }
            $zip->close();
            return file_exists($destination);
            }
            else{
                return false;
            }
    }

The function is returning true every time it is called. But no files are being added. Can anyone spot an obvious error here?

lukassteiner · Accepted Answer · 2015-11-13T09:57:55.740

3

ZipArchive::addFile() expects the first param to be a string containing the filename. But you are passing a value of $_POST['File'] to it which is an array as $_POST['File'] is a two dimensional array.
See here for the contents of $_POST['File'].

What you need to do is change this line:

$zip->addFile($file,basename($file));

To:

$zip->addFile($file['tmp_name'],$file['name']);

edited Nov 13 '15 at 09:57

answered Nov 13 '15 at 09:28

lukassteiner

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Apologies for what is probably a dumb response, I'm only three weeks in to coding. But when I loop through the $_POST['File'] array; "foreach($files as $file)" am I not then passing individual string values to addFile()? – Liam Fell Nov 13 '15 at 09:50
Don't worry about asking. We all started at one point. When you loop through $_POST['File'], you still get an array in $file. As $_POST['File'] is a 2 dimensional array. So you need to do this: $file['tmp_name'] – lukassteiner Nov 13 '15 at 09:54
I would also recommend you to have a look at how xdebug works. With that, you can debug your code and check the contents of the variables at runtime. – lukassteiner Nov 13 '15 at 10:00

score 1 · Answer 2 · answered Nov 13 '15 at 09:53

Zip::addFile needs absolute path in the first parameter, and the second parameter would be file name as mention in PhpDOc

$zip->addFile('/path/to/index.txt', 'newname.txt');

And make sure you are getting $file variable with absolute path. if you are uploading file from browser then you should use $_FILE['file']['tmp_name'] in the $zip->addFile method on first parameter

PHP: $zip->addFile issues

2 Answers2