Question

我正在尝试使用volley将JsonArray从php提取到android。但我收到以下错误。

org.json.JSONException: Value <!-- of type java.lang.String cannot be converted to JSONObject

以下是我从php获得的JsonArray响应。

{
"result": [
    {
        "subject_name": "Entrepreneurship",
        "percentage": 1.1999999999999999555910790149937383830547332763671875
    },
    {
        "subject_name": "Advanced Computing Platform",
        "percentage": 1.1399999999999999023003738329862244427204132080078125
    },
    {
        "subject_name": "Enterprise Application Platforms",
        "percentage": 0
    },
    {
        "subject_name": "Major Project",
        "percentage": 0.340000000000000024424906541753443889319896697998046875
    }
]
}

以下是我用来从db获取JsonArray的PHP代码。

$get = mysqli_query($conn, "SELECT * FROM subject_dtls WHERE course_sem='$student_course'");
while($row = mysqli_fetch_array($get))
{
    $sub_id = $row['subject_id'];
    $subject_name = $row['subject_name'];
    $total_hours = $row['total_hours'];


    $get12 = mysqli_query($conn, "SELECT sum(duration) as du_sum FROM attendance_count WHERE subject_id='$sub_id' AND roll_numbers = '$official_id'");
    while($row12 = mysqli_fetch_array($get12))
    {
        $duration = $row12['du_sum'];

        if ($duration=='') 
        {
            $duration = 0;
        }

        $percentage = 0;
        $percentage = ($duration * $total_hours)/100;

        $temp = array();
        $temp["subject_name"] = $subject_name;
        $temp["percentage"] = $percentage;

        $value["result"][] = $temp;

    }
}

echo json_encode($value);

以下是我用来获取Json响应的Java代码。

public void onResponse(JSONObject response) {

            try {

                JSONArray result = response.getJSONArray("result");

                for (int i = 0; i < result.length(); i++) {
                    JSONObject jsonObject = result.getJSONObject(i);

                    String subject_name = jsonObject.getString("subject_name");
                    String percentage = jsonObject.getString("percentage");

                    // Each time create a new instance and then save the value.
                    CheckAttendanceAdapter caa = new CheckAttendanceAdapter();

                    caa.setSubject_name(subject_name);
                    caa.setPercentage(percentage);
                    arrayList.add(caa);
                }

                data = new String[arrayList.size()][10];
                for (int j = 0; j < arrayList.size(); j++) {
                    CheckAttendanceAdapter c = arrayList.get(j);
                    data[j][0] = c.getSubject_name();
                    data[j][1] = c.getPercentage();

                    tableView.setDataAdapter(new SimpleTableDataAdapter(getActivity(), data));
                }


            } catch (JSONException e) {
                e.printStackTrace();
                Toast.makeText(getActivity(), e.getMessage(), Toast.LENGTH_SHORT).show();
            }
        }

以下是错误的屏幕截图。

请告诉我，我做错了什么。

Answer 1

我不知道您是否需要手动解析JSON响应，或者只是您不知道还有其他方式来使用REST服务。

我建议您使用Retrofit库来构建REST客户端。它易于编写且易于维护。要使用Retrofit使用REST服务，您需要编写服务接口和相关模型：

Fruit

和REST客户端：

public class Result {
  public String subject_name;
  public double percentage;
}

public class YourResponse {
  List<Result> result;
}

调用您服务的代码：

public interface YourService {
  @GET("result")
  Call<YourResponse> getResult();
}

有关改造的更多信息是here。

无法获得JsonArray值

1 个答案: