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我正在开发一个 Android 应用程序，它从我的数据库中发送和接收简单的 fname,Lname。我正在使用 php 和 mysql 进行 web 服务。我在 JSONParser 类中使用本教程 ，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;

}
}

当我试图将字符串解析为 json 对象时

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

我收到异常 JSONArray 无法转换为 JSONObject。请帮助，因为在所有其他教程中，都描述了类似的过程。

4

1 回答 1

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是的，如果你直接创建它，你会遇到这个问题。在后面添加一行json = b.toString();

json = sb.toString();
json = "{json_parse" + "[" + json + "]" + "}";
于 2013-03-07T07:18:52.160 回答