使用json从php:mysql到android接收数据

我正在开发一个Android应用程序，它从我的数据库发送和接收简单的fname，Lname。我正在使用php和mysql进行Web服务。我在用这个 在JSONParser类中，如下所示

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 + "talha");
        }
        Log.i("Buffer Error", "Baby Baby! Yes Mama! ");

        is.close();
        json = sb.toString();
        Log.i("Buffer Error", "Eating suger! NO Mama! "+ json);




    } 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("MyJSON", "Error parsing data "+ e.toString());
    }



    // return JSON String
    return jObj;

}
}

---

我遇到异常JSONArray无法转换为JSONObject。请提供帮助，因为在其他每一个教程中，都会描述类似的过程。

是的，如果您直接创建该过程，您将遇到该问题。请在

json=b.toString（）之后添加一行

那么，这里的错误是什么呢？当它试图在jObj=newjsonobject（json）处解析到json对象时出现异常；放入错误日志。您在menifest文件中有internet权限吗？该错误仅仅意味着您正在获取JSONArray作为回复，但您正在尝试将其作为JSONObject使用。使用新的JSONArray（json）您可以在以下位置向我们显示输出：Log.i（“缓冲区错误”，“吃糖！没有妈妈！”+json）；现在我得到了这个异常->解析数据org.json.JSONException时出错：在{json_parse[[[{“name”：“fahad”}，{“name”：“talha”}，{“name”：“umer”}，{“name”：“ahmed”}，{“name”：“}，{“name”：“ahmed”}，{“abc”}，{“name”：“ahmed perv”}]talha]}
try {
     jObj = new JSONObject(json);
    } catch (JSONException e) {
        Log.e("MyJSON", "Error parsing data "+ e.toString());
    }

json = sb.toString();
json = "{json_parse" + "[" + json + "]" + "}";