分析dataorg.json.JSONException时出错：java.lang.String类型的值

我正试图为我的论文制作android食品订单，因为这个错误，我的时间不多了：

logcat上的错误：

分析dataorg.json.JSONException时出错：无法将值转换为JSONObject org.json.JSONException:JSONObject的值

下面是我的JSONParser：

package com.makanan.restotradisional;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.util.List;

import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.utils.URLEncodedUtils;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONException;
import org.json.JSONObject;

import android.util.Log;

public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";

public JSONParser() {
}

// fungsi abil json url lewat method HTTP POST atau GET
public JSONObject makeHttpRequest(String url, String method,
        List<NameValuePair> params) {
    try {
        if (method == "POST") {
            // jika request method adalah POST
            // defaultHttpClient

            DefaultHttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(url);
            httpPost.setEntity(new UrlEncodedFormEntity(params));

            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();
        } else if (method == "GET") {
            // jika request method adalah GET

            DefaultHttpClient httpClient = new DefaultHttpClient();
            String paramString = URLEncodedUtils.format(params, "utf-8");
            url += "?" + paramString;
            HttpGet httpGet = new HttpGet(url);

            HttpResponse httpResponse = httpClient.execute(httpGet);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();
        }
    } catch (UnsupportedEncodingException e) {

        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
        is.close();
        json = sb.toString();

    } catch (Exception e) {

        Log.e("Buffer Error", "Error Converting result" + e.toString());
    }



    try {
        jObj = new JSONObject(json);
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data" + e.toString());
        e.printStackTrace();
    }
    return jObj;

}
}

这是我的PHP和Java：

这是我在phpmyadmin上的数据库：

请帮助我

从php文件中删除任何语句或echo语句，但用于传递json的语句除外

---

在浏览器中检查文件的输出，删除除json以外的所有不需要的内容。

请打印并检查字符串json的格式是否符合JSONObject构造函数的预期。根据文档，构造JSONObject的有效json字符串应该是-一个以{左大括号开始，以}右大括号结束的字符串

---

请参阅。

进入JSONParser内部，这样您就可以在logcat中看到php带来了什么。可能是php错误

try {
        jObj = new JSONObject(json);
    } catch (JSONException e) {
        //This line is what u need to add
        Log.d("Whats wrong?", json.toString());
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    }

请提供在JSONObject构造函数中传递的字符串