Question

我在android中完全是绿色的。我想创建从服务器获取数据并在应用程序中显示它的App。谁能告诉我如何开始呢？我在下面尝试了这段代码。但唯一的例外就是没有找到食物。

private EditText outputStream;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    String result = null;
    InputStream input = null;
    StringBuilder sbuilder = null;
    outputStream = (EditText)findViewById(R.id.output);
    ArrayList <NameValuePair> nameValuePairs = new ArrayList <NameValuePair>();

    try{
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("ik.su.lt/~jbarzelis/index.php");
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = httpclient.execute(httppost);
        if (response.getStatusLine().getStatusCode() != 200) {
            Log.d("MyApp", "Server encountered an error");
        }
        HttpEntity entity = response.getEntity();
        input = entity.getContent();
    }
    catch(Exception e){
        e.printStackTrace();
    }
    try{
        BufferedReader reader = new BufferedReader(new InputStreamReader(input,"iso-8859-1"),8);
        sbuilder = new StringBuilder();

        String line = null;

        while((line = reader.readLine()) != null){
            sbuilder.append(line + "\n");
        }
        input.close();
        result = sbuilder.toString();
    }
    catch(Exception e){
        e.printStackTrace();
    }
    int fd_id;
    String fd_name;
    try{
        JSONArray jArray = new JSONArray(result);
        JSONObject json_data = null;
        for(int i=0;i<jArray.length();i++){
            json_data = jArray.getJSONObject(i);
            fd_id = json_data.getInt("FOOD_ID");
            fd_name = json_data.getString("FOOD_NAME");
            outputStream.append(fd_id +" " + fd_name + "\n");
        }


        }
    catch(JSONException e1){
        Toast.makeText(getBaseContext(), "No food found", Toast.LENGTH_LONG).show();
    }
    catch(ParseException e1){
        e1.printStackTrace();
    }
}

PHP代码是正确的，它显示数据。我认为上面的代码有些错误。

  <?php
    mysql_connect("localhost","********","**********");
    mysql_select_db("test");
    $sql = mysql_query("select FOOD_NAME as 'Maistas' from FOOD where FOOD_NAME like 'A%'");
    while($row = mysql_fetch_assoc($sql)) $output[]=$row;
    print(json_encode($output));
    mysql_close;
?>

Answer 1

package com.Valluru;

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
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.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

import android.app.ListActivity;
import android.content.Intent;
import android.net.ParseException;
import android.net.Uri;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.AdapterView;
import android.widget.AdapterView.OnItemClickListener;
import android.widget.ArrayAdapter;
import android.widget.ListView;
import android.widget.TextView;
import android.widget.Toast;

public class Food extends ListActivity {
String result = null;
InputStream is = null;
StringBuilder sb=null;
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
ListView list1;

@Override
public void onCreate(Bundle savedInstanceState) 
{
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);
    list1 = (ListView) findViewById(android.R.id.list);

    //http post
    try{
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://10.0.2.2/food.php");
        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);
        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
    String fd_id;
    String fd_name;
    try{
    JSONArray jArray = new JSONArray(result);
    JSONObject json_data=null;

    for(int i=0;i<jArray.length();i++){
            json_data = jArray.getJSONObject(i);
            fd_id =json_data.getString("FOOD_ID");

            fd_name = json_data.getString("FOOD_NAME");
            nameValuePairs.add(new list<String, String>(fd_id, fd_name));
    }

    list1.setAdapter(new ArrayAdapter<NameValuePair>(getApplicationContext(),android.R.layout.simple_expandable_list_item_1,nameValuePairs));

    }catch(JSONException e1){
        Toast.makeText(getBaseContext(), "No FOOD Found", Toast.LENGTH_LONG).show();
    }catch (ParseException e1){
        e1.printStackTrace();
    }

}
}

list.java 包com.Valluru;

import org.apache.http.NameValuePair;

import android.R.integer;

public class list<T,V> implements NameValuePair {
T data;
V text;

public list(T data, V text)
{
    this.data = data;
    this.text = text;
}

@Override
public String toString(){
    return text.toString();
}

@Override
public String getName() {
    return (String) data;
}

@Override
public String getValue() {
    return (String) text;
}
}

food.php

<?php
  mysql_connect("localhost","root");
  mysql_select_db("FOOD");
  $sql=mysql_query("select * from FOOD where FOOD_NAME like '%'");
  while($row=mysql_fetch_assoc($sql)) $output[]=$row;
  print(json_encode($output));
  mysql_close();
?>

Answer 2

要尝试的事情：

在服务器上，有没有以A开头的食物？
您是否尝试过从脚本返回一个简单的字符串，而不是运行查询？
您的Log来电是否会写入LogCat？如果是这样，它会说什么？
初始httpclient.execute是否有效或是否触发了异常？
在result = sbuilder.toString（）上设置断点，然后调试该值。那里有一个有效的JSON字典，还是别的东西？
如果步骤4和/或5失败，并且您不确定原因，请尝试使用Fiddler分析您的http流量。

如果您完成上述任何步骤，则必须从项目中退后一步，先了解这些内容。在需要的地方进行实验，学习并提出具体问题。

Answer 3

你解析了这些字段失败了：

fd_id = json_data.getInt("FOOD_ID");
fd_name = json_data.getString("FOOD_NAME")

这是服务器错误

Android，从服务器获取数据

3 个答案: