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Question

我正在设计一个带有PHP PDO的CMS，供用户更新信息，这部分我让用户更改业务图像，这将通过PHPMyAdmin上传到服务器和数据库。

有了这个，我可以上传图片，它也将取代它在数据库上的价值。我遇到的问题是，当用户只想更新其他信息（不是图像）而没有其他信息时，系统会提交图像值并替换相关行中没有值（空白上传）的数据库中的行？如果没有上传图片，我希望它不会影响这一行。

我一直在看if ($_FILES['image']['error'] === UPLOAD_ERR_NO_FILE)但不是运气，但我认为我在正确的区域。

HTML

<form action='edit_account.php' method='post' enctype='multipart/form-data'>
<h4>Business Image</h4>
<input type='file' name='image' id='fileToUpload' >
<input type='submit' value='Update Account' />
</form>

PHP

if(isset($_FILES['image'])) {
            $username = $_POST['username'];
            $target_dir = "images/users/$username/";
            $target_file = $target_dir.basename($_FILES['image']['name']);
            $upload0K = 1;
            $imageFileType = pathinfo($target_file, PATHINFO_EXTENSION);

            if(isset($_POST['submit'])) {
                $check = getimagesize($_FILES['image']['tmp_name']);
                if($check !== false) {
                    echo "<p>File is an image - ".$check["mime"]."</p>";
                    $upload0K = 1;
                } else {
                    echo "<p>File is not an image</p>";
                    $upload0K = 0;
                }
            }

            //if (file_exists($target_file)) {
                //echo "<p>Sorry, file already exists.</p>";
                //$upload0K = 0;
            //}

            if ($_FILES['image']['size'] > 5000000) {
                echo "<p>Sorry, your file is too large.</p>";
                $upload0K = 0;
            }
            ///
            if ($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg" && $imageFileType != "gif" ) {
                echo "<p>Sorry, only JPG, JPEG, PNG and GIF files are allowed.</p>";
                $upload0K = 0;
            }

            //if(!isset($_FILES['image']) || $_FILES['image']['error'] == UPLOAD_ERR_NO_FILE) {
                //echo "Error no file selected"; 
            //}

            ///
            if ($upload0K == 0) {
                echo "Sorry, your file was not uploaded.";
            }

            else {
                if (move_uploaded_file($_FILES['image']['tmp_name'], $target_file)) {
                    echo "<p>The file ". basename( $_FILES["image"]["name"])."has been uploaded.</p>";
                } else {
                    echo "<p>Sorry, there was an error uploading your file.</p>";
                }
              }
            }

SQL插入

$query_params = array( 
            ':email' => $_POST['email'],
            ':username' => $_POST['username'], 
            ':name' => $_POST['name'],
            ':image' => $_FILES['image']['name'],
            ':description' => $_POST['description'], 
            ':moderatedcheck' => $moderatedcheck, 
            ':user_id' => $_SESSION['user']['id']
        ); 

$query = " UPDATE users SET email = :email, username = :username, name = :name, image = :image, description = :description, moderatedcheck = :moderatedcheck";

Answer 1

执行此操作的最佳和正确方法是通过判断图像错误值，错误值4表示没有上载文件。对文件名称或其他属性的其他测试有些不精确。

$imageFile = $_FILES['image'];
if ($imageFile['error'] != 4) {
///this code only runs if a file has been given.
}

更新

我一直在看if（$ _FILES ['image'] ['error'] === UPLOAD_ERR_NO_FILE）但不是运气，但我认为我在正确的区域。

要检查$_FILES['image']['error']是正确的，但此值是数值而不是文本值。查看是否检查$_FILES['image']['error'] == 4是否更有效。

更新2

将文件中的if(isset($_FILES['image'])) {替换为：

if ($_FILES['image']['error'] != 4)}{

如果有文件，这将只运行以下代码，尽管可能仍然存在文件大小等错误。

另外SQL插入：

将原件替换为：

$query_params = array( 
            ':email' => $_POST['email'],
            ':username' => $_POST['username'], 
            ':name' => $_POST['name'],
            ':description' => $_POST['description'], 
            ':moderatedcheck' => $moderatedcheck, 
            ':user_id' => $_SESSION['user']['id']
        ); 
$query = " UPDATE users SET email = :email, username = :username, name = :name, description = :description, moderatedcheck = :moderatedcheck";

并添加限定符：

    if ($_FILES['image']['error'] == 0 ){
       $query_params[':image'] = $_FILES['image']['name'];
$query = " UPDATE users SET email = :email, username = :username, name = :name, image = :image, description = :description, moderatedcheck = :moderatedcheck";
    }

如果文件是i）存在，则只会将文件名数据添加到插入数组中; ii）没有错误。或者，如果有错误，将文件保存到数据库可能会毫无意义，但可以用if if ($_FILES['image']['error'] != 4 )替换if语句。

Answer 2

您的Sql可以动态构建，在代码开头定义$imageSql = '';之类的变量，当$_FILES['image']设置为覆盖您的$imageSql = ', image = :image';和$query_params[':image'] = $_FILES['image']['name']时，最后是您的$query = " UPDATE users SET email = :email, username = :username, name = :name".$imageSql.", description = :description, moderatedcheck = :moderatedcheck";查询将如下： {{1}}

Answer 3

我用这个：

   if(is_uploaded_file($_FILES["image"]["tmp_name"])){

        //query to update the data with the image

    }else{
       //query to update the data without the image
}

PHP表单类型文件 - 如果未选择任何映像，请不要更新数据库中的行

3 个答案:

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