Filedrop.js无成功信息

Question

我正在使用filedrop.js的图像上传脚本。我在这里资助了一个脚本：https://tutorialzine.com/2011/09/html5-file-upload-jquery-php

在Prject中有一个file_post.php，我想要修改它，以便将一些信息(比如文件名)保存到数据库中。

这是我的post_file.php：

    <?php

// If you want to ignore the uploaded files, 
// set $demo_mode to true;

$demo_mode = false;
$upload_dir = 'uploads/tmp/';
$allowed_ext = array('jpg','jpeg','png','gif');


if(strtolower($_SERVER['REQUEST_METHOD']) != 'post'){
    exit_status('Error! Wrong HTTP method!');
}


if(array_key_exists('pic',$_FILES) && $_FILES['pic']['error'] == 0 ){

    $pic = $_FILES['pic'];

    if(!in_array(get_extension($pic['name']),$allowed_ext)){
        exit_status('Only '.implode(',',$allowed_ext).' files are allowed!');
    }   

    if($demo_mode){

        // File uploads are ignored. We only log them.

        $line = implode('       ', array( date('r'), $_SERVER['REMOTE_ADDR'], $pic['size'], $pic['name']));
        file_put_contents('log.txt', $line.PHP_EOL, FILE_APPEND);

        exit_status('Uploads are ignored in demo mode.');
    }


    // Move the uploaded file from the temporary 
    // directory to the uploads folder:

    if(move_uploaded_file($pic['tmp_name'], $upload_dir.$pic['name'])){

        //My added code 

包括(‘/var/www/html/board/ssI.php’)；

$userName = $context['user']['name'];
$content_id = $_COOKIE["contentid"];
$pic_name = $pic['name'];
$pic_code = $content_id;
$pic_path = $pic_name;


$db_host = "******";
$db_name = "******";
$db_user = "******";
$db_pass = "******";

$db = mysqli_connect("$db_host","$db_user","$db_pass","$db_name") or die("Error " . mysqli_error($db));

$stmt = $db->prepare("INSERT INTO `User_pics` (content_id, path, user_id,  user_name) VALUES (?, ?, ?, ?)");
$stmt->bind_param('ssss', $pic_code,
$pic_path,
$context['user']['id'],
$context['user']['name']);
$stmt->execute();
$stmt->close();

        //end of my added code





        exit_status('File was uploaded successfuly!');
    }

}

exit_status('Something went wrong with your upload!');


// Helper functions

function exit_status($str){
    echo json_encode(array('status'=>$str));
    exit;
}

function get_extension($file_name){
    $ext = explode('.', $file_name);
    $ext = array_pop($ext);
    return strtolower($ext);
}
?>

在我添加mysqli部分之后，就不再显示成功消息了。在图像上传上，Progressbar停止在大约50%。文件被上传，信息被ssaved输入到DB中，但是我没有得到成功的响应，这需要处理接下来的步骤。救命啊！谢谢。

Answer 1

谢谢你的插曲，两天我为解决问题而工作，我在这里问过之后，找到了答案;)

问题是在UTF-8中编码的SSI.php的包含。这就是Json编码响应中出现错误的原因。

SyntaxError: JSON.parse: JSON数据第1列中的意外字符

我的解决方案是:我为用户ID和用户名创建了一个cookie，以便能够删除包含。在这之后，每件事都很好。如果有人有相同的错误，只需创建一个cookie，您将显示您的上传表单，上传后删除他们。

感谢社区，祝您节日愉快;)