解析数据时出错org.json.JSONException:值<! - ?类型为java.lang.String的xml无法转换为JSONArray - >

时间:2012-04-22 21:56:59

标签: android eclipse parsing sdk

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

我已经从教程中下载了代码,请耐心等待我已经掌握了一些基本的PHP知识,而且很少有java知识。我已经测试了php脚本,它运行完美,所以我不会费心去附上它。

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; } }

我非常感谢你能给我的任何帮助。

1 个答案:

答案 0 :(得分:3)

正如您对问题的评论中指出的那样,您的服务器似乎正在返回XML而不是JSON。您只需输出result

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

如果它是XML,它几乎肯定是必须的,那么你需要让服务器输出JSON,或者你需要解析它发送给你的XML。