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我正在制作一个 android 应用程序，这是它抛出的第一个错误：

Error parsing data org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject

这是我的类功能：

package com.ernest.httppost;

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 = "";

    // constructor
    public JSONParser() {

    }

    // function get json from url
    // by making HTTP POST or GET method
    public JSONObject makeHttpRequest(String url, String method,
            List<NameValuePair> params) {

        // Making HTTP request
        try {

            // check for request method
            if(method == "POST"){
                // request method is 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"){
                // request method is 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 parse the string to a JSON object
        try {
            jObj = new JSONObject(json);
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data " + e.toString());
        }

        // return JSON String
        return jObj;

    }
}

我认为它jObj = new JSONObject(json);在最后一个try-catch块中返回了该错误。任何想法如何解决这个问题？

4

2 回答 2

3

看起来您的响应<br不是 JSON，因此预计会出现此解析错误。

您是否在请求中使用正确的有效负载访问了正确的 URL？尝试调试以查看您正在请求什么以及响应是什么 - 错误可能很明显。

于 2013-01-13T13:15:22.223 回答
2

评论提升回答：

我认为您用来构造 JSONObject 的字符串应该是有效的 json 字符串。由于错误状态值<br无法转换，我猜您正在尝试从 HTML 字符串构造一个 json 对象。确保您传递的是格式正确的 json 字符串。例如：

"{\"name\":\"John Ernest Guadalupe\", \"reputation\":181, \"messages\":[\"msg 1\",\"msg 2\",\"msg 3\"]}"

就像 davnicwil 所说：问题可能是由 URL 引起的，而不仅仅是返回格式正确的 JSON 字符串。

于 2013-01-13T13:21:43.710 回答