Question

我的图片上传表单效果很好。保存数据库的名称和路径。我想将完整的URL保存到图像沿着白名而不是名称。我的意思是现在在DB中保存file_name.jpg我要保存http://example.com/images/file_name.jpg。这是upload.php

            define('MAX_FILE_SIZE', 20000000430);
            $uploadDir = "../../img/";
            $permitted = array('image/jpeg', 'image/jpeg', 'image/png', 'image/gif');               

            $fileName  = $_FILES['image']['name'];
            $tmpName   = $_FILES['image']['tmp_name'];
            $fileSize  = $_FILES['image']['size'];
            $fileType  = $_FILES['image']['type'];

            // make a new image name
            $ext = substr(strrchr($fileName, "."), 1);
            // generate the random file name
            $randName = md5(rand() * time());

            // image name with extension
            $myFile = $randName . '.' . $ext;
            // save image path
            $path = $uploadDir . $myFile;
            if (in_array($fileType, $permitted)) 
            {
                $result = move_uploaded_file($tmpName, $path);
                    if (!$result) 
                    {
                            echo "Error uploading image file";
                            exit;
                    } 
                    else 
                    {                           
                        // keep track post values
                        $name = $_POST['name'];
                        $description = $_POST['description'];
                        // update data
                        $pdo = Database::connect();
                        $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
                        $sql = "UPDATE food set name = ?, image = ?, path = ?, description = ? 
                                WHERE id = ?";
                            $q = $pdo->prepare($sql);
                            $q->execute(array($name,$myFile,$path,$description,$id));
                            Database::disconnect();
                            echo "<code>Information updated!!</code>";
                    } 
                }

我尝试将URL放在$ uploadDir中。

    $uploadDir = "http://example.com/img/";

但是我收到了这个错误。

Warning: move_uploaded_file(http://example.com/img/75a13564a8f3305fb0a30ab95487b8de.jpg): failed to open stream: HTTP wrapper does not support writeable connections

也试过这样的事情并得到同样的错误

define('domainURL',         'http://'.$_SERVER['HTTP_HOST']); 
$path = domainURL . $uploadDir . $myFile;

Answer 1

move_uploaded_file函数不接受文件网址。

它接受图像实际路径。

因为，您的文件正在物理移动。

e.g。 $path = $_SERVER['DOCUMENT_ROOT'] . 'img/'

Answer 2

您可以使用相对或绝对路径。相对路径是“../../img/”; 绝对路径应该像“/ www / htdocs / img /”; （你可以在FTP客户端看到绝对路径）而你不能使用URL。对于DB中的商店，使用另一个带URL的变量

Answer 3

在php变量中分配上传路径，并使用$ path将其存储在数据库中..

$path = "www.sitename.com/uploads/".$filename ;

图像上传时保存数据库中的完整URL链接

3 个答案: