Question

我是android新手。那么我在下面的代码中尝试做的是将http请求发送到json获取响应的php文件。在php文件中，我试图显示从A开始的城市名称。每当我运行代码时，我收到此错误："org.json.jsonexception value doctype of type java.lang.string cannot be converted to jsonarray“。所有The related {{3}没有帮助，或者我无法理解那里给出的解决方案。这是我的Java代码

的 MainActivity.java

package com.list;

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.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

import android.app.Activity;
import android.net.ParseException;
import android.os.Bundle;
import android.util.Log;

import android.widget.TextView;
import android.widget.Toast;

public class MainActivity extends Activity {

JSONArray jArray;
String result = null;
InputStream is = null;
StringBuilder sb=null;


int ct_id;
String ct_name;



@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);


TextView textview1 = (TextView)findViewById(R.id.textView1);
textview1.setText("Erqem");

 ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
//http post
try{
     HttpClient httpclient = new DefaultHttpClient();
     HttpPost httppost = new HttpPost("http://10.0.2.2/city.php");
     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,"UTF-8"),8);
       sb = new StringBuilder();
       sb.append(reader.readLine() + "\n");

       String line="0";
       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());
        }

//paring data
try{
      jArray = new JSONArray(result);
      JSONObject json_data = new JSONObject();
      json_data=null;
      for(int i=0;i<jArray.length();i++){
             json_data = jArray.getJSONObject(i);
             ct_id=json_data.getInt("CITY_ID");
             ct_name=json_data.getString("CITY_NAME");

             textview1.setText(ct_name);
             }

      }

      catch(JSONException e1){
          Toast.makeText(getBaseContext(), "No City Found"+e1.toString() ,Toast.LENGTH_LONG).show();
          Log.e("log_tag", "Error converting result "+e1.getMessage() );
      } catch (ParseException e1) {
            e1.printStackTrace();
    }
}
}

php文件

<?php
header('Content-type=application/json; charset=utf-8');
mysql_connect("","root","");
 mysql_select_db("Deal");
 $sql=mysql_query("select * from CITY where CITY_NAME like 'A%'");
 while($row=mysql_fetch_assoc($sql)) 
 $output[]=$row; 
 print(json_encode($output));
 mysql_close();

?>

任何帮助都将不胜感激。谢谢。

Answer 1

也许您使用的地址不正确，请尝试10.0.2.2:8080/city.php

无法从JSON获得响应

1 个答案: