我有一个简单的Android应用程序,它从phpMyAdmin DB获取数据并在ListView中显示它。在我的数据库中,我在第2列中列出了一个名为place_name的名称,第二列名为numbers_of_visits。我可以毫无问题地从数据库中获取数据。
但是我在尝试根据最高访问次数(即数据库)对列表视图进行排序时遇到问题,因此访问次数最多的地方应该位于顶部,依此类推。
请大家帮我做这个
这是我的Jvav代码
package com.example.ems_project;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
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.Activity;
import android.os.AsyncTask;
import android.os.Bundle;
import android.widget.ListView;
import android.widget.SimpleAdapter;
import android.widget.Toast;
public class PlacesListView extends Activity {
private String jsonResult;
private String url = "http://ibrahimaldosari.t15.org/listview.php";
private ListView listView;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.listview_main2);
listView = (ListView) findViewById(R.id.listview);
accessWebService();
}
// 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("placestable");
for (int i = 0; i < jsonMainNode.length(); i++) {
JSONObject jsonChildNode = jsonMainNode.getJSONObject(i);
String name = jsonChildNode.optString("name");
int number = jsonChildNode.optInt("visit");
String outPut = name + "---" + number;
employeeList.add(createEmployee("places", 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[] { "places" }, 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;
}
}
这是我正在使用的php文件
<?php
$host="mysql.freehostingnoads.net"; //replace with database hostname
$username="u746805016_ibrr1"; //replace with database username
$password="113281188"; //replace with database password
$db_name="u746805016_ibrr1"; //replace with database name
$con=mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql = "select * from placestable";
$result = mysql_query($sql);
$json = array();
if(mysql_num_rows($result)){
while($row=mysql_fetch_assoc($result)){
$json['placestable'][]=$row;
}
}
mysql_close($con);
echo json_encode($json);
?>
答案 0 :(得分:0)
你可以用两种方式做到这一点。 1.在服务器端,通过稍微改变选择查询到这样的事情:
$sql = "select * from placestable ORDER BY visit DESC";
或 2.在应用程序端对此进行排序。但后来我会创建带有字段的Employee类:name和visit,并在那里实现Comparable接口。