2012-02-03 210 views

回答

3

設置新的形式:

<form enctype="multipart/form-data" action="uploader.php" method="POST"> 
Choose a file to upload: <input name="file" type="file" /><br /> 
<input type="submit" value="Upload File" /> 
</form> 

在uploader.php你應該有像成才:

$folder = "uploads/"; 

$path = $folder . basename($_FILES['file']['name']); 

if(move_uploaded_file($_FILES['file']['tmp_name'], $path)) { 
    echo "The file ". basename($_FILES['file']['name']). " has been uploaded"; 
} else{ 
    echo "There was an error uploading the file, please try again!"; 
} 

這是將文件上傳到服務器的最簡單方法之一。但是,您可能想要自定義文件上傳表單並添加更多信息。

0

您可以在PHP中使用ftp函數。我可以給你我幾個月前寫的課。它是爲Spawn Framework編寫的,但您可以獨立使用它。 link