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我有以下代碼,我努力讓它工作。在java文件中有2個錯誤,我找不到修復程序。Android和PHP連接到MySQL數據庫
這裏是我的MySQLData.java文件
package com.example.qosmetre2;
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.Toast;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import org.apache.http.NameValuePair;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;
import android.app.ListActivity;
import android.app.ProgressDialog;
import android.content.Intent;
import android.os.AsyncTask;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.AdapterView;
import android.widget.AdapterView.OnItemClickListener;
import android.widget.ListAdapter;
import android.widget.ListView;
import android.widget.SimpleAdapter;
import android.widget.TextView;
public class MySQLData extends ListActivity {
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
String result = null;
InputStream is = null;
StringBuilder sb=null;
String result=null;
//http post
try{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://localhost/android_test/fetch_data2.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,"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 nam;
int rec_pow;
int tow;
int stat;
try{
JSONArray jArray = new JSONArray(result);
JSONObject json_data=null;
for(int i=0;i<jArray.length();i++){
json_data = jArray.getJSONObject(i);
nam=json_data.getString("name");
rec_pow=json_data.getInt("recieved_power");
tow=json_data.getInt("tower");
stat=json_data.getInt("status");
}
}catch(JSONException e1){
Toast.makeText(getBaseContext(), "Could not Parse Data", Toast.LENGTH_LONG).show();
}catch (ParseException e1){
e1.printStackTrace();
}
}
}
有以下2行錯誤:
String result=null;
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
的第一個錯誤說重複的局部變量的結果,第二個錯誤說namevaluepairs中無法解析爲變量。
我也有一個php文件,我把我的wamp服務器,我想在我的本地服務器上顯示的infro這是在應用程序中。我在某處閱讀我可能不得不向前移動,以便通過互聯網訪問應用程序,但我不知道。該應用程序不會運行,直到錯誤得到解決。
這是我的PHP文件fetch_data2.php
<?php
mysql_connect("888888","8888888","*******");
mysql_select_db("*******");
$q=mysql_query("SELECT * FROM users");
while($row=mysql_fetch_assoc($q))
$output[]=$row;
print(json_encode($output));
mysql_close();
?>
您用於Java的IDE是什麼IDE?一個體面的人會強調重複和未定義的變量。 – devnate 2013-04-04 23:57:24