Question

我在phpmyadmin数据库中有 USERS 表。在表中我有一个列，其中包含用户个人资料图片的链接。但我认为我上传的图像并不好。当我打开它时，我看到一个像这样的小方块：

编辑：图片部分已修复

当我试图用piccaso获取它时，我什么都没得到。我上传图片的方式是将其编码为字符串，然后在register.php中对其进行解码

这是将图片转换为字符串的方法：

 public String getStringImage(Bitmap bmp) {
        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        bmp.compress(Bitmap.CompressFormat.JPEG, 100, baos);
        byte[] imageBytes = baos.toByteArray();
        String encodedImage = Base64.encodeToString(imageBytes, Base64.DEFAULT);
        return encodedImage;
    }

这是我上传图片的php：

$firstName = $_POST['firstname'];
$lastName = $_POST['lastname'];
$passWord = $_POST['password'];
$phoneNumber = $_POST['phonenumber'];
$apartmentNumber = $_POST['apartmentNumber'];
$image = $_POST['image']; //this is the encoded string

//upload image to image folder in the server    
$upload_folder = "upload";
$path = "$upload_folder/$phoneNumber.jpeg";
if(file_put_contents($path, base64_encode($image)) != flase){
    echo "uploaded";
}
$image_link = "my url/$path";
$result = mysqli_query($con,"INSERT INTO Users (firstName, lastname, password, phone , apartmentNum,image) 
        VALUES ('$firstName', '$lastName', '$EncryptPassword', '$phoneNumber', '$apartmentNumber' ,'$image_link')");

这是我在创建json对象的php：

$sql = ("select * from Users where phone = '$phone' and password = '$password' ");

$check = mysqli_fetch_array(mysqli_query($con,$sql));

if(isset($check)){
    $hasApartment = mysqli_query($con,$sql);
    $row = mysqli_fetch_assoc($hasApartment);
    $apartmentNumber = $row['apartmentNum'];
        $firstName = $row['firstName'];
        $lastName = $row['lastname'];
        $image = $row['image'];
 echo '{"query_result":"SUCCESS", "apartmentNumber":' . $apartmentNumber.' , "firstName":' . trim($firstName) .' ,"lastName":' . trim($lastName).', "image":' . $image.'}';

        }

我得到了这个例外：

org.json.JSONException: Unterminated object at character 102 of {"query_result":"SUCCESS", "apartmentNumber":54 , "firstName":a ,"lastName":a, "image":XXXXXX.XX.XX/upload/112.jpeg}

这是json对象：

{＆＃34; query_result＆＃34;：＆＃34; SUCCESS＆＃34;，＆＃34; apartmentNumber＆＃34;：54，＆＃34; firstName＆＃34;：a，＆＃34; lastName＆＃ 34;：a，＆＃34; image＆＃34;：XXXXXX.XX.XX / upload / 112.jpeg}

char 102它的＆＃34; u＆＃34;在上传

这是我做GET请求的时间。在onResponse（）中，我尝试将json文本作为json对象处理：

 JsonObjectRequest jsonObjectRequest = new JsonObjectRequest(Request.Method.GET, url, null,
                new Response.Listener<JSONObject>() {
                    @Override
                    public void onResponse(JSONObject response) {
                        try {

                            String query_result = response.getString("query_result");
                            switch (query_result){
                                case "SUCCESS": // user name and password exists and mathces and user associated with apartment
                                    int apartmentNum = Integer.parseInt(response.get("apartmentNumber").toString());

                                    Intent intent = new Intent(LoginActivityScreen.this , HomeActivityScreen.class);

                                    SharedPreferences.Editor editor = preferences.edit();

                                    editor.putString("phoneNumber" , PhoneValidationActivity.PHONE_NUMBER);
                                    editor.putString("firstName", response.getString("firstName"));
                                    editor.putString("lastName", response.getString("firstName"));
                                    editor.putBoolean("loggedIn", true);
                                    editor.putInt("apartmentNumber", apartmentNum);

                                    Bitmap profilePicture = Picasso.with(context).load("http://roomates.96.lt/upload/0547517307.jpeg").get();

                                    editor.apply();

                                    startActivity(intent);
                                    break;

                                case "FAILURE":  //user name or password are incorrect
                                    Toast.makeText(getApplicationContext(), "wrong name or password", Toast.LENGTH_SHORT).show();
                                    break;
                                default:
                                    break;
                            }

                        } catch (JSONException e) { //this exception is caught 
                            e.printStackTrace();
                            Log.e("e" , e.toString());
                        } catch (IOException e) {
                            e.printStackTrace();
                            Log.e("e", e.toString());
                        }
                    }
                }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {

            }
        });
        MySingleton.getInstance(context).addToRquestQueue(jsonObjectRequest);
    }

Answer 1

String query_result = response.getString("query_result")

您是否看到"query_result":"SUCCESS"和"image":XXXXXX.XX.XX/upload/112.jpeg之间存在差异？

语法上？

Answer 2

如果您在图像路径上遇到JSON响应错误，那么您应该只需 json_encode 您的响应，这样您的所有特殊字符都将被正确转义。

在您的JSON响应中，您的图片路径 XXXXXX.XX.XX / upload / 112.jpeg 应该像 XXXXXX.XX.XX / upload / 112.jpeg

希望它对你有所帮助。

将图像上传到服务器然后在android

2 个答案: