我正在使用下面的簡單腳本通過php上傳zip文件,然後將其解壓縮到我的服務器上。PHP文件上傳然後顯示下載鏈接
該文件將是一個壓縮文件夾。上傳完成後,我想回顯一個指向新文件夾的鏈接。
因此,例如,如果我上傳包含名爲「bar」的文件夾的zip文件,在成功消息之後,我想回顯「http://foo.com/bar」。
任何幫助非常感謝。
<?php
if($_FILES["zip_file"]["name"]) {
$filename = $_FILES["zip_file"]["name"];
$source = $_FILES["zip_file"]["tmp_name"];
$type = $_FILES["zip_file"]["type"];
$name = explode(".", $filename);
$accepted_types = array('application/zip', 'application/x-zip-compressed',
'multipart/x-zip', 'application/x-compressed');
foreach($accepted_types as $mime_type) {
if($mime_type == $type) {
$okay = true;
break;
}
}
$continue = strtolower($name[1]) == 'zip' ? true : false;
if(!$continue) {
$message = "The file you are trying to upload is not a .zip file. Please try again.";
}
$target_path = "/home/var/foo.com/".$filename; // change this to the correct
site path
if(move_uploaded_file($source, $target_path)) {
$zip = new ZipArchive();
$x = $zip->open($target_path);
if ($x === true) {
$zip->extractTo("/home/var/foo.com/"); // change this to the correct site path
$zip->close();
unlink($target_path);
}
$message = "Your .zip file was uploaded and unpacked.";
} else {
$message = "There was a problem with the upload. Please try again.";
}
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1
/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Untitled Document</title>
</head>
<body>
<?php if($message) echo "<p>$message</p>"; ?>
<form enctype="multipart/form-data" method="post" action="">
<label>Choose a zip file to upload: <input type="file" name="zip_file" /></label>
<br />
<input type="submit" name="submit" value="Upload" />
</form>
</body>
</html>
你的代碼似乎正確,寫得很好。你試圖解決的問題究竟是什麼?此代碼是否顯示任何錯誤? – akanevsky 2012-07-09 01:03:19