4

我正在编写一个打算在android设备上运行的应用程序。该应用程序应该通过php读取Mysql数据库中的信息,但是当我运行该应用程序时,Log cat提示错误'解析数据时出错org.json.JSONException:值

我得到的代码是从教程中下载的,请耐心等待我对php有一些基础知识,在java中很少。我已经测试了 php 脚本并且它运行完美,所以我不会费心附加它。

main.java 代码:

 

package test.an2mysql; import java.io.BufferedReader; import java.io.InputStream; import java.io.InputStreamReader; import java.util.ArrayList; import org.apache.http.HttpEntity; import org.apache.http.HttpResponse; import org.apache.http.NameValuePair; import org.apache.http.client.HttpClient; import org.apache.http.client.entity.UrlEncodedFormEntity; import org.apache.http.client.methods.HttpPost; import org.apache.http.impl.client.DefaultHttpClient; import org.apache.http.message.BasicNameValuePair; import org.json.JSONArray; import org.json.JSONException; import org.json.JSONObject; import android.app.Activity; import android.os.Bundle; import android.util.Log; import android.widget.LinearLayout; import android.widget.TextView; public class main extends Activity { /** Called when the activity is first created. */ TextView txt; @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.main); // Create a crude view - this should really be set via the layout resources // but since its an example saves declaring them in the XML. LinearLayout rootLayout = new LinearLayout(getApplicationContext()); txt = new TextView(getApplicationContext()); rootLayout.addView(txt); setContentView(rootLayout); // Set the text and call the connect function. txt.setText("Connecting..."); //call the method to run the data retreival txt.setText(getServerData(KEY_121)); } public static final String KEY_121 = "http://10.1.1.19/cms/test/android2mysql/read.php"; //i use my real ip here private String getServerData(String returnString) { InputStream is = null; String result = ""; //the data to send ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(); nameValuePairs.add(new BasicNameValuePair("country","undefined")); //http post try{ HttpClient httpclient = new DefaultHttpClient(); HttpPost httppost = new HttpPost(KEY_121); httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs)); HttpResponse response = httpclient.execute(httppost); HttpEntity entity = response.getEntity(); is = entity.getContent(); }catch(Exception e){ Log.e("log_tag", "Error in http connection "+e.toString()); } //convert response to string 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(); result=sb.toString(); }catch(Exception e){ Log.e("log_tag", "Error converting result "+e.toString()); } //parse json data try{ JSONArray jArray = new JSONArray(result); for(int i=0;i<jArray.length();i++){ JSONObject json_data = jArray.getJSONObject(i); Log.i("log_tag","id: "+json_data.getInt("id")+ ", country: "+json_data.getString("country")+ ", documentn: "+json_data.getInt("documentn") ); //Get an output to the screen returnString += "\n\t" + jArray.getJSONObject(i); } }catch(JSONException e){ Log.e("log_tag", "Error parsing data "+e.toString()); } return returnString; } }

如果您能给我任何帮助,我将不胜感激。


正如对您问题的评论中所指出的,您的服务器似乎返回的是 XML 而不是 JSON。您只需输出即可轻松确认result

}catch(JSONException e){
    Log.e("log_tag", "Error parsing data "+e.toString());
    Log.e("log_tag", "Failed data was:\n" + result);
}

如果是 XML(几乎可以肯定),那么您要么需要让服务器输出 JSON,要么需要解析它发送给您的 XML。

4

1 回答 1

3

正如对您问题的评论中所指出的,您的服务器似乎返回的是 XML 而不是 JSON。您只需输出即可轻松确认result

}catch(JSONException e){
    Log.e("log_tag", "Error parsing data "+e.toString());
    Log.e("log_tag", "Failed data was:\n" + result);
}

如果是 XML(几乎可以肯定),那么您要么需要让服务器输出 JSON,要么需要解析它发送给您的 XML。

于 2012-06-05T10:24:00.967 回答