使用 JSON 和 PHP 从 Mysql DB 获取数据到 listView

Posted 2023-02-24

【中文标题】使用 JSON 和 PHP 从 Mysql DB 获取数据到 listView

【英文标题】：Fetching data from Mysql DB to listView using JSON and PHP

【发布时间】：2014-05-07 17:36:09

【问题描述】：

我正在尝试从 android listview 中的 mysql DB 获取数据。但不知何故，我无法在 listView 中显示数据

我在教程的帮助下尝试了这个，即http://codeoncloud.blogspot.in/2013/07/android-mysql-php-json-tutorial.html

下面是我的 MainActivity.java

public class MainActivity extends Activity 
 private String jsonResult;
 private String url = "http://10.0.2.2/markit/login.php";
 private ListView listView;
 private TextView textv1;

 @Override
 protected void onCreate(Bundle savedInstanceState) 
  super.onCreate(savedInstanceState);
  setContentView(R.layout.activity_main);
  listView = (ListView) findViewById(R.id.listView1);
  textv1=(TextView)findViewById(R.id.textView1);
  accessWebService();
 

 @Override
 public boolean onCreateOptionsMenu(Menu menu) 
  // Inflate the menu; this adds items to the action bar if it is present.
  getMenuInflater().inflate(R.menu.main, menu);
  return true;
 

 // Async Task to access the web
 private class JsonReadTask extends AsyncTask<String, Void, String> 
  @Override
  protected String doInBackground(String... params) 
   HttpClient httpclient = new DefaultHttpClient();
   HttpPost httppost = new HttpPost(params[0]);
   try 
    HttpResponse response = httpclient.execute(httppost);
    jsonResult = inputStreamToString(
      response.getEntity().getContent()).toString();
   

   catch (ClientProtocolException e) 
    e.printStackTrace();
    catch (IOException e) 
    e.printStackTrace();
   
   return null;
  

  private StringBuilder inputStreamToString(InputStream is) 
   String rLine = "";
   StringBuilder answer = new StringBuilder();
   BufferedReader rd = new BufferedReader(new InputStreamReader(is));

   try 
    while ((rLine = rd.readLine()) != null) 
     answer.append(rLine);
    
   

   catch (IOException e) 
    // e.printStackTrace();
    Toast.makeText(getApplicationContext(),
      "Error..." + e.toString(), Toast.LENGTH_LONG).show();
   
   return answer;
  

  @Override
  protected void onPostExecute(String result) 
   ListDrwaer();
  
 // end async task

 public void accessWebService() 
  JsonReadTask task = new JsonReadTask();
  // passes values for the urls string array
  task.execute(new String[]  url );
 

 // build hash set for list view
 public void ListDrwaer() 
  List<Map<String, String>> employeeList = new ArrayList<Map<String, String>>();

  try 
   JSONObject jsonResponse = new JSONObject(jsonResult);
   JSONArray jsonMainNode = jsonResponse.optJSONArray("emp_info");
   for (int i = 0; i < jsonMainNode.length(); i++) 
    JSONObject jsonChildNode = jsonMainNode.getJSONObject(i);
    String name = jsonChildNode.optString("employee_name");
    String number = jsonChildNode.optString("employee_no");
    String outPut = name + "-" + number;
    //textv1.setText(name);
    //textv1.setText(jsonResult);
    employeeList.add(createEmployee("employees", outPut));
   
   catch (JSONException e) 
   Toast.makeText(getApplicationContext(), "Error" + e.toString(),
     Toast.LENGTH_SHORT).show();
  

  SimpleAdapter simpleAdapter = new SimpleAdapter(this, employeeList,
    android.R.layout.simple_list_item_1,
    new String[]  "employees" , new int[]  android.R.id.text1 );
  listView.setAdapter(simpleAdapter);
 

 private HashMap<String, String> createEmployee(String name, String number) 
  HashMap<String, String> employeeNameNo = new HashMap<String, String>();
  employeeNameNo.put(name, number);
  return employeeNameNo;
 

下面是mainactivity.xml

<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:tools="http://schemas.android.com/tools"
    android:layout_
    android:layout_
    android:paddingBottom="@dimen/activity_vertical_margin"
    android:paddingLeft="@dimen/activity_horizontal_margin"
    android:paddingRight="@dimen/activity_horizontal_margin"
    android:paddingTop="@dimen/activity_vertical_margin"
    tools:context=".MainActivity" >

    <TableRow
        android:id="@+id/tableRow1"
        android:layout_
        android:layout_ >

        <TextView
            android:id="@+id/textView1"
            android:layout_
            android:layout_
            android:text="TextView" />

    </TableRow>

    <ListView
        android:id="@+id/listView1"
        android:layout_
        android:layout_
         >
    </ListView>

</LinearLayout>

Login.php

<?php
$host="localhost"; //replace with database hostname 
$username="root"; //replace with database username 
$password=""; //replace with database password 
$db_name="markit"; //replace with database name

$con=mysql_connect("localhost", "root", "")or die("cannot connect"); 
mysql_select_db("markit")or die("cannot select DB");
$sql = "select * from emp_info"; 
$result = mysql_query($sql);
$json = array();

if(mysql_num_rows($result))
while($row=mysql_fetch_assoc($result))
$json['emp_info'][]=$row;


mysql_close($con);
echo json_encode($json); 
?> 

【问题讨论】：

您是否遇到任何错误或异常？

没有错误，但数据在列表中不可见......并且它显示的“-”数量与数据库中的条目数量一样多

我试图打印 //textv1.setText(jsonResult);这导致我们在浏览器中执行 php 后发现的总 JSON 结果

尝试记录您的 jsonResult 并查看来自服务器的内容。它是有效的json吗？并在此处发布结果。

这是在 textView 中打印 jsonResult 的结果 "emp_info":["emp_no":"1","emp_name":"Ravi","emp_no":"2" ,"emp_name":"Div","emp_no":"3","emp_name":"Harry","emp_no":"4","emp_name":"Suresh","emp_no" :"5","emp_name":"Suu"]

【参考方案1】：

改变这个

 String name = jsonChildNode.optString("employee_name");
 String number = jsonChildNode.optString("employee_no");

到

 String name = jsonChildNode.getString("emp_name");
 String number = jsonChildNode.getString("emp_no");