有时在0的字符0处抛出输入的结尾

时间:2014-06-14 04:37:18

标签: php json

我正在尝试开发一个在PHP和我的移动应用程序之间使用JSON解析器的应用程序,我发现大多数时候应用程序正常运行,但很少次,少于10%会抛出结束时的错误以字符0输入。任何人都可以简要解释一下我遇到了什么吗?

以下是我从PHP获取数据的代码:

public class getDataClass {

 public void onCreate(Bundle savedInstanceState) {
        StrictMode.ThreadPolicy policy = new      StrictMode.ThreadPolicy.Builder().permitAll().build();
        StrictMode.setThreadPolicy(policy);      }
    public static String executeQuery() {
        String result = "";

        try {
            HttpParams httpParameters = new BasicHttpParams();
            HttpConnectionParams.setConnectionTimeout(httpParameters, 6000);
            HttpConnectionParams.setSoTimeout(httpParameters, 6000);
            //check IF IP timeout
            HttpClient httpClient = new DefaultHttpClient(httpParameters);

            HttpPost httpPost = new HttpPost("somephp.php");
            //ArrayList<NameValuePair> params = new ArrayList<NameValuePair>();
           // params.add(new BasicNameValuePair("query_string", query_string));
        //    httpPost.setEntity(new UrlEncodedFormEntity(params, HTTP.UTF_8));
            HttpResponse httpResponse = httpClient.execute(httpPost);
            //view_account.setText(httpResponse.getStatusLine().toString());
            HttpEntity httpEntity = httpResponse.getEntity();
            InputStream inputStream = httpEntity.getContent();

            BufferedReader bufReader = new BufferedReader(new InputStreamReader(inputStream, "utf-8"), 8);
            StringBuilder builder = new StringBuilder();
            String line = null;
            while((line = bufReader.readLine()) != null) {
                builder.append(line + "\n");
            }
            inputStream.close();
            result = builder.toString();
        } catch(Exception e) {
            // Log.e("log_tag", e.toString());
        }

        return result;
    }
}

以下是我在主要活动中使用的方法:

private String DLdata(){

    String result = getDataClass.executeQuery();          
    if(result.startsWith("[")){             
    }else{
    result = result.substring(1);
    }
    return result;

}

0 个答案:

没有答案