Question

我正在尝试创建一个Android应用程序，目前处于登录阶段。我尝试在模拟器上运行我的应用程序，它没有任何问题或错误，除了控制台说活动管理器正在启动Intent，但它实际上没有启动它。有人可以帮我这个吗？我是android编程中的菜鸟。

这是我的主要活动

    package com.ysleps.simbayanan;

import java.util.ArrayList;
import java.util.List;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.HttpClient;
import org.apache.http.client.ResponseHandler;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.BasicResponseHandler;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;
import android.app.Activity;
import android.app.AlertDialog;
import android.app.ProgressDialog;
import android.content.DialogInterface;
import android.content.Intent;
import android.os.Bundle;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;

public class MainActivity extends Activity {
    Button b;
    EditText et,pass;
    TextView tv;
    HttpPost httppost;
    StringBuffer buffer;
    HttpResponse response;
    HttpClient httpclient;
    List<NameValuePair> nameValuePairs;
    ProgressDialog dialog = null;

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

        b = (Button)findViewById(R.id.Button01);  
        et = (EditText)findViewById(R.id.username);
        pass= (EditText)findViewById(R.id.password);
        tv = (TextView)findViewById(R.id.tv);

        b.setOnClickListener(new OnClickListener() {
            @Override
            public void onClick(View v) {
                dialog = ProgressDialog.show(MainActivity.this, "", 
                        "Validating user...", true);
                 new Thread(new Runnable() {
                        public void run() {
                            login();                          
                        }
                      }).start();               
            }
        });
    }

    void login(){
        try{            

            httpclient=new DefaultHttpClient();
            httppost= new HttpPost("http://10.0.2.2:81/login/check.php"); // make sure the url is correct.
            //add your data
            nameValuePairs = new ArrayList<NameValuePair>(2);
            // Always use the same variable name for posting i.e the android side variable name and php side variable name should be similar, 
            nameValuePairs.add(new BasicNameValuePair("username",et.getText().toString().trim()));  // $Edittext_value = $_POST['Edittext_value'];
            nameValuePairs.add(new BasicNameValuePair("password",pass.getText().toString().trim())); 
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            //Execute HTTP Post Request
            response=httpclient.execute(httppost);
            // edited by James from coderzheaven.. <span id="IL_AD10" class="IL_AD">from here</span>....
            ResponseHandler<String> responseHandler = new BasicResponseHandler();
            final String response = httpclient.execute(httppost, responseHandler);
            System.out.println("Response : " + response); 
            runOnUiThread(new Runnable() {
                public void run() {
                    tv.setText("Response from PHP : " + response);
                    dialog.dismiss();
                }
            });

            if(response.equalsIgnoreCase("User Found")){
                runOnUiThread(new Runnable() {
                    public void run() {
                        Toast.makeText(MainActivity.this,"Login Success", Toast.LENGTH_SHORT).show();
                    }
                });

            startActivity(new Intent(this, UserPage.class));
            }else{
                showAlert();                
            }

        }catch(Exception e){
            dialog.dismiss();
            System.out.println("Exception : " + e.getMessage());
        }
    }
    public void showAlert(){
        MainActivity.this.runOnUiThread(new Runnable() {
            public void run() {
                AlertDialog.Builder builder = new AlertDialog.Builder(MainActivity.this);
                builder.setTitle("Login Error.");
                builder.setMessage("User not Found.")  
                       .setCancelable(false)
                       .setPositiveButton("OK", new DialogInterface.OnClickListener() {
                           public void onClick(DialogInterface dialog, int id) {
                           }
                       });                     
                AlertDialog alert = builder.create();
                alert.show();               
            }
        });
    }
}

Answer 1

我建议你使用AsyncTask来做到这一点。这是运行后台任务的推荐方法，您也可以轻松显示进度条和警报。请参阅此http://developer.android.com/reference/android/os/AsyncTask.html此链接还包含有关AsyncTask http://www.vogella.com/tutorials/AndroidBackgroundProcessing/article.html

的一些有价值的信息

Answer 2

您应该确保在AndroidManifest.xml中添加了此代码。就像

一样

<application
    android:allowBackup="true"
    android:icon="@drawable/icon"
    android:label="@string/app_name"
    android:theme="@style/AppTheme" >
    <activity 
        android:name="com.abcd.yourmainactivity"
        android:screenOrientation="landscape"
        android:theme="@android:style/Theme.NoTitleBar.Fullscreen">
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />

            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>
</application>

意图不会开始：它表示在控制台上开始意图，但它不会启动

2 个答案: