Question

我有一个程序可以通过PHP表单登录到服务器（他用用户名和密码填写表单）然后PHP脚本生成一个非常简单的JSON文件：

{
	"success":1,
	"message":"PHP Login successful!",
	"userid":"1",
	"data":
	[
		{
			"id_data":"1",
			"data_name":"Mona Lisa",
			"data_date":"29.9. 2015",
			"pic_user_id_user":"1",
			"data_picture":‰PNG

   
IHDR         Ä¾‹   sRGB ®Îé   gAMA  ±üa   	pHYs  Ä  Ä•+   TIDAT(S]NÙ  b”FqtGqòªgýB °„¶„¨Ž5—ŽxB¼æÇ³A»ar
SHÞMÞÐæ5š7Ž†ÿÈ
Œ*Qæ92V¸pRÏªü(Ü	
…üZI÷Í    IEND®B`‚"
		},
	]
}

这一切都有效，将TextView设置为图像的名称和日期。但是当我想绘制图像时，logCat写入错误SkImageDecoder :: Factory返回null。我不想从任何网址下载图片，我只想从JSON中获取图片。有公共方法，当登录已经成功，然后设置公共变量，其中是JSON数据。然后在下一个Fragment（list_of_files.java）中使用公共变量（main.java）并在屏幕上写入数据。

// main.java
public static void login_success(JSONObject login_json) throws JSONException, UnsupportedEncodingException {
    status.setText("Login successful.");
    JSONArray data = login_json.getJSONArray("data");   // get array data witch contains all information about image
    int count_of_images = data.length();                // get count, how many images user have
    img_name = new String[count_of_images];             // then, create an array, with contains name of each image
    img_date = new String[count_of_images];             // and the date of image

    // {...} - JSONObject (work with normal methods getting value of coll, f.e. getString( "name of coll" )
    // [...] - JSONArray (work with index - numbers)
    for (int i = 0; i < count_of_images; i++){
        JSONObject under_data = data.getJSONObject(i);
        img_name[i] = under_data.getString("data_name");
        img_date[i] = under_data.getString("data_date");
        if ( i == 0 )
            img_img = under_data.getString("data_picture").getBytes();
    }
    Log.d("== IMAGE SIZE ==", String.valueOf(img_img.length) + " B");
}





// list_of_files.java
    // Update listView list_of_files
    if (list_of_files_str != null) {
        adapter = new ArrayAdapter<String>(this.getActivity(), android.R.layout.simple_list_item_1, list_of_files_str);
        list.setAdapter(adapter);
        list.setOnItemClickListener(new AdapterView.OnItemClickListener() {
            private FragmentManager supportFragmentManager;

            public FragmentManager getSupportFragmentManager() {
                return supportFragmentManager;
            }

            public void onItemClick(AdapterView<?> parent, View view,
                                    int position, long id) {
                // set image name and date in textView
                TextView image_name = (TextView)getView().findViewById(R.id.image_name);
                image_name.setText(list_of_files_str[position] + " // " + date_str[position]);

                // draw image with char array from main class Red Orchestra
                ImageView img = (ImageView)getView().findViewById(R.id.imageView);

                Bitmap image_to_draw;
                byte[] data;
                String str = "";
                try {
                    data = str.getBytes("UTF-8");
                    String base64 = Base64.encodeToString(data, Base64.DEFAULT);
                    byte[] imageAsBytes = Base64.decode(base64.getBytes(), Base64.URL_SAFE);
                    image_to_draw = BitmapFactory.decodeByteArray(data, 0, data.length);
                    img.setImageBitmap( image_to_draw );
                } catch (UnsupportedEncodingException e) {
                    e.printStackTrace();
                }

            }
        });

    } else { // if the user is not login, write him message
        String[] must_login = new String[] {"You must be sign in."};
        adapter = new ArrayAdapter<String>(this.getActivity(), android.R.layout.simple_list_item_1, must_login);
        list.setAdapter(adapter);
    }

    // Inflate the layout for this fragment
    return rootView;
}

我真的不知道，如何保存这个问题。我搜索了不同的论坛和网站，但没有任何效果。

PHP文件：

//load and connect to MySQL database stuff

需要（ “config.inc.php中”）; $ login_ok = false;

if（！empty（$ _ POST））{ $ query_1 =“SELECT * FROM pic_user WHERE user_name =：username”;

$query_params = array(
    ':username' => $_POST['username']
);

try {
    $stmt   = $db->prepare($query_1);
    $result = $stmt->execute($query_params);
}
catch (PDOException $ex) {
    $response["success"] = 0;
    $response["message"] = "PHP Database Error1. Please Try Again!";
    die(json_encode($response));
}

$validated_info = false;

$row = $stmt->fetch();
if ($row) {
    if ($_POST['password'] === $row['user_password']) {
        $login_ok = true;
    }
}

if ($login_ok) {
    $query_2   = "SELECT * FROM pic_data WHERE pic_user_id_user = :userid";

    $user_id   = $row['id_user'];
    $user_name = $row['user_name'];

    $query_params_2 = array( ':userid' => $user_id );
    $stmt         = $db->prepare($query_2);
    $result       = $stmt->execute($query_params_2);
    $row          = $stmt->fetchAll();

    $response["success"] = 1;
    $response["message"] = "PHP Login successful!";
    $response["userid"]  = $user_id;
    $response['data']    = $row;

    die(json_encode($response));

} else {
    $response["success"] = 0;
    $response["message"] = "PHP Invalid Credentials!";
    die(json_encode($response));
}

Answer 1

我明白了！有没有错误的base64转换。当我删除代码中的转换并将图像转换为base64（http://www.motobit.com/util/base64-decoder-encoder.asp）并将base64字符串添加到数据库中而不是通过JSON下载它时，它可以工作。

BitmapFactory decodeByteArray SkImageDecoder Factory返回null - 来自JSON

1 个答案: