Question

我正在学习如何将Android APP连接到mysql数据库的教程。该应用程序有2个功能：显示产品（不起作用）创造产品（工程）当我制作产品时，它成功注册到数据库，但是我的JSOn解析器出现了一些错误。

我发现很多帖子都处理同样的问题，但我仍然遇到麻烦。

E / JSON Parser（1216）：解析数据时出错[java.lang.String类型的值p无法转换为JSONObject] get_all_products.php

这是我的解析器

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 mehtod
    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 {
            jObj = new JSONObject(json.substring(19));
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data [" + e.getMessage()+"] "+json);
        }

        // return JSON String
        return jObj;

    }

}

Answer 1

我遇到了同样的问题，我发现因为我离开了p 在PHP文件开始之前。

解析数据时出错[java.lang.String类型的值p无法转换为JSONObject

1 个答案: