Question

我有一个文件上传脚本，可以[终于]工作，但希望让用户将图片提交到一个文件夹中，我希望将其创建为用户名。我以为我可能只是提交文件然后它会创建它，但它没有，所以我怎么能这样做？这是我的脚本： PHP代码：

<?php 
// Configuration part  
$dbhost = "localhost"; // Database host  
$dbname = "users"; // Database name  
$dbuser = "dbuser"; // Database username  
$dbpass = ""; // Database password  

// Connect to database  
$db = mysql_connect($dbhost, $dbuser, $dbpass) or die("Error: Couldn't connect to database");  
mysql_select_db($dbname, $db) or die("Error: Couldn't select database.");  

if($IsLoggedIn) 
{ $userid = $_GET["userid"]; 
    if($_POST['Submit']) 
        { 

            if($userid) 
            { $uploadName = mysql_query("SELECT * FROM users WHERE userId ='" 
                             . $userid . "'"); 
            while ($upName = mysql_fetch_row($uploadName)) 
                { $userName2 = $upName[1]; 
                   $userPic1 = $upName[11];  
                   echo "$userName2"; } } 

//If the Submitbutton was pressed do:  

            if ($_FILES['imagefile']['type'] == 'image/pjpeg') 
                {  

//this is where the pic is put into a folder by the username 
    copy ($_FILES['imagefile']['tmp_name'],  
    "pics/profiles/" . $userName2 . "/" . $_FILES['imagefile']['name'])  
    or die ("Could not copy");  

        echo "";  
        echo "Name: ".$_FILES['imagefile']['name']."";  
        echo "Size: ".$_FILES['imagefile']['size']."";  
        echo "Type: ".$_FILES['imagefile']['type']."";  
        echo "Copy Done....";  
        }  


        else {  
            echo "";  
            echo "Could Not Copy, Wrong Filetype (".$_FILES['imagefile']['name'].")";  
        }  
}  

?> 
<form name="form1" method="post" action="index.php?page=classupload&userid=<?php echo "$userid"; ?>" enctype="multipart/form-data">  
<input type="file" name="imagefile">  

<input type="submit" name="Submit" value="Submit">  
</form> 
<?php } else { ?>How did you get here? You aren't <a href="index.php?page=login">Logged In</a>. 
<?php } ?>

有没有办法做到这一点？也许即使用户制作？谢谢！

Answer 1

$directoryPath = "/home/domain/public_html/site_name/pics/profiles/".$userName2;
mkdir($directoryPath, 0644);

Answer 2

您可以进一步检查您所在的目录是否允许您创建新目录。以下是一些验证示例，可帮助您提高应用的稳健性。

$parentDir = '/home/domain/public_html/site_name/pics/profiles/';
if(!is_dir($parentDir)) { // Check if the parent directory is a directory
    die('Invalid path specified');
}

if(!is_writable($parentDir)) { // Check if the parent directory is writeable
    die('Unable to create directory, permissions denied.');
}

if(mkdir($parentDir . $userName2) === false) { // Create the directory
    die('Problems creating directory.');
}
die('Created directory successfully'); // Success point

Answer 3

“如何使用PHP创建文件夹”的答案是mkdir()。我无法理解所有代码与主题有什么关系。

Answer 4

也许你应该看一下mkdir命令。

Answer 5

这应该if(!file_exists($targetPath)) mkdir($targetPath);，但你需要拥有正确的权限

用php创建新文件夹

5 个答案: