3 回答

TA貢獻1801條經驗 獲得超16個贊
<?php
$fileName = $_GET['file'];? // get file name in the URL param "file"
if (isset($fileName)) {? ? // if $fileName php variable is set than? ??
? ? $zip = new ZipArchive;? ? ? ? ? // create object
? ? $res = $zip->open($fileName);? ?// open archive
? ? if ($res === TRUE) {
? ? ? $zip->extractTo('./');? ? ? ? // extract contents to destination directory
? ? ? $zip->close();? ? ? ? ? ? ? ?//close the archieve? ??
? ? ? echo 'Extracted file "'.$fileName.'"';
? ? } else {
? ? ? echo 'Cannot find the file name "'.$fileName.'" (the file name should include extension (.zip, ...))';
? ? }
}
else {
? ? echo 'Please set file name in the "file" param';
}
?>

TA貢獻1796條經驗 獲得超4個贊
試試這個代碼。也$zip->open($file+".zip");改為$zip->open($file);. +(加號)不是 php 中的連接運算符
<?php
// $_GET["file"] is set to `a.zip`
$file = $_GET["file"];
$zip = new ZipArchive;
$res = $zip->open($file);
$zip->extractTo('./');
$zip->close();
?>
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