1

在我的登录活动中......我在询问用户名和密码......我想检查用户是否是授权用户......如果用户被授权，那么我的网络活动 menu.java 开始......否则他应该是重定向到同一个登录页面...带有警报...用户不存在...怎么做？

我已经写了我的活动.. 我怎样才能开始下一个活动？

import java.io.BufferedReader;
import java.io.InputStreamReader;
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.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;

import android.app.Activity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button; 
import android.widget.EditText;
import android.widget.TextView;

public class Login_Menu extends Activity {


@Override
protected void onCreate(Bundle savedInstanceState) {
    // TODO Auto-generated method stub
    super.onCreate(savedInstanceState);
    setContentView(R.layout.login_lay);
    final TextView tv=(TextView) findViewById(R.id.login_stat_tv);
    final EditText uname=(EditText)findViewById(R.id.uname);
    final EditText pass=(EditText)findViewById(R.id.pass);
    Button login=(Button)findViewById(R.id.login_but);
    Button cancel=(Button)findViewById(R.id.cancel_but);

    final HttpClient client = new DefaultHttpClient();
    String url="http://10.0.2.2:7001/f/json.jsp";
    final HttpPost post = new HttpPost(url);


    login.setOnClickListener(new View.OnClickListener() {


        public void onClick(View arg0) {

        try{

            List<NameValuePair> pairs = new ArrayList<NameValuePair>();   
            pairs.add(new BasicNameValuePair("username",uname.getText().toString()));   
            pairs.add(new BasicNameValuePair("password",pass.getText().toString()));   
            post.setEntity(new UrlEncodedFormEntity(pairs));   
            HttpResponse response = client.execute(post);


            BufferedReader rd = new BufferedReader(new InputStreamReader(
                    response.getEntity().getContent()));
            String line = null;

            while ((line = rd.readLine()) != null) {

                tv.append(line);

            }



            //startActivity(new Intent("com.campuspro.start.DEMO_RETRV"));

          }

        catch(Exception e)
        {
            e.printStackTrace();   

        }


        }
    });

    cancel.setOnClickListener(new View.OnClickListener() {

        public void onClick(View arg0) {

            uname.getText().clear();
            pass.getText().clear();


        }
    });



}



}
4

4 回答 4

2

在这里使用Aysntask类

哪个在doInBackground()结果中进行操作并给出结果onPostExecute()

public class Login_Menu extends Activity {


@Override
protected void onCreate(Bundle savedInstanceState) {
    // TODO Auto-generated method stub
    super.onCreate(savedInstanceState);
    setContentView(R.layout.login_lay);
    final TextView tv=(TextView) findViewById(R.id.login_stat_tv);
    final EditText uname=(EditText)findViewById(R.id.uname);
    final EditText pass=(EditText)findViewById(R.id.pass);
    Button login=(Button)findViewById(R.id.login_but);
    Button cancel=(Button)findViewById(R.id.cancel_but);

    final HttpClient client = new DefaultHttpClient();
    String url="http://10.0.2.2:7001/f/json.jsp";
    final HttpPost post = new HttpPost(url);
    new login().execute("");
 }


private class login extends AsyncTask<String, Void, Void>{
    ProgressDialog dialog = ProgressDialog.show(activity.this, "", "Loading, Please wait...");

    @Override
    protected int doInBackground(String... params) {
        // TODO Auto-generated method stub
        Log.i("thread", "Doing Something...");
       //authentication operation
try{

        List<NameValuePair> pairs = new ArrayList<NameValuePair>();   
        pairs.add(new BasicNameValuePair("username",uname.getText().toString()));   
        pairs.add(new BasicNameValuePair("password",pass.getText().toString()));   
        post.setEntity(new UrlEncodedFormEntity(pairs));   
        HttpResponse response = client.execute(post);

        BufferedReader rd = new BufferedReader(new InputStreamReader(
                response.getEntity().getContent()));
        String line = null;

        while ((line = rd.readLine()) != null) {

            tv.append(line);

        }
        //startActivity(new Intent("com.campuspro.start.DEMO_RETRV"));
      }

    catch(Exception e)
    {
        e.printStackTrace();   

    }

        return val;
    }

    protected void onPreExecute(){
        //dialog.dismiss();
        Log.i("thread", "Started...");
        dialog.show();
    }
    protected void onPostExecute(int result){
        Log.i("thread", "Done...");
        if(dialog!=null)
        dialog.dismiss();
        if(result){
            toast.setText("No User Found, please try again!");
            toast.show();
        }else{
        Intent myIntent = new Intent(ctx, main.class);
        myIntent.putExtra("user", user);
        startActivity(myIntent);
        }
    }

}
于 2012-04-06T05:55:32.970 回答
1

用于Aysntask 制作HttpPost按钮 click.for 启动新 Activity 这取决于用户成功授权时您的服务返回的内容。like

public void onClick(View arg0) {
            try{

                List<NameValuePair> pairs = new ArrayList<NameValuePair>();   
                pairs.add(new BasicNameValuePair("username",uname.getText().toString()));   
                pairs.add(new BasicNameValuePair("password",pass.getText().toString()));   
                post.setEntity(new UrlEncodedFormEntity(pairs));   
                HttpResponse response = client.execute(post);


                BufferedReader rd = new BufferedReader(new InputStreamReader(
                        response.getEntity().getContent()));
                String line = null;
                String strinfo="";  
                while ((line = rd.readLine()) != null) {

                    tv.append(line);
                    strinfo+=line;

                }
              //check here authorized user or not
               strinfo = strinfo.toLowerCase();
               String strlogin="login=true";
                if((strinfo.contains(strlogin)==true)
                {
                 Intent intent=new Intent(CurrentActivity.this,DEMO_RETRV.this);
                 startActivity(intent);
                //startActivity(new Intent("com.campuspro.start.DEMO_RETRV"));
                }
                else
                {

                }

              }

            catch(Exception e)
            {
                e.printStackTrace();   

            }


            }
于 2012-04-06T06:15:11.940 回答
0 
 package com.campuspro.start;

 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.entity.UrlEncodedFormEntity;
  import org.apache.http.client.methods.HttpPost;
  import org.apache.http.impl.client.DefaultHttpClient;
  import org.apache.http.message.BasicNameValuePair;
  import org.apache.http.util.EntityUtils;
  import org.json.JSONException;
  import org.json.JSONObject;

  import android.app.Activity;
  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.Button;
   import android.widget.EditText;
  import android.widget.TextView;


public class Login_Menu extends Activity {

EditText usname;
EditText pass;
TextView tv;
HttpClient client;
HttpPost post;
@Override
protected void onCreate(Bundle savedInstanceState) {
// TODO Auto-generated method stub
super.onCreate(savedInstanceState);
setContentView(R.layout.login_lay);
 tv=(TextView) findViewById(R.id.login_stat_tv);
 usname=(EditText)findViewById(R.id.uname);
 pass=(EditText)findViewById(R.id.pass);
Button login=(Button)findViewById(R.id.login_but);
Button cancel=(Button)findViewById(R.id.cancel_but);

client = new DefaultHttpClient();
String url="http://10.0.2.2:7001/proj/login.jsp";
post = new HttpPost(url);
login.setOnClickListener(new View.OnClickListener() {

    public void onClick(View arg0) {
        // TODO Auto-generated method stub
        new login().execute("");
    }
});

cancel.setOnClickListener(new View.OnClickListener() {

    public void onClick(View v) {
        // TODO Auto-generated method stub
        usname.getText().clear();
        pass.getText().clear();
    }
});

}



@Override
protected void onDestroy() {
// TODO Auto-generated method stub
super.onDestroy();
}



 @Override
 protected void onPause() {
// TODO Auto-generated method stub
super.onPause();
finish();
}

private class login extends AsyncTask<String, Void, JSONObject>{
ProgressDialog dialog = ProgressDialog.show(Login_Menu.this, "", "Authenticating,   Please wait...");

@Override
protected JSONObject doInBackground(String... params) {
    // TODO Auto-generated method stub
    Log.i("thread", "Doing Something...");
   //authentication operation
try{

    List<NameValuePair> pairs = new ArrayList<NameValuePair>();   
    pairs.add(new BasicNameValuePair("username",usname.getText().toString()));   
    pairs.add(new BasicNameValuePair("password",pass.getText().toString()));   
    post.setEntity(new UrlEncodedFormEntity(pairs));   
    HttpResponse response = client.execute(post);
    int status=response.getStatusLine().getStatusCode();

    if(status == 200)
    {
        HttpEntity e=response.getEntity();
        String data=EntityUtils.toString(e);
        JSONObject last=new JSONObject(data);
        return last;

    }

}

catch(Exception e)
{
    e.printStackTrace();   

}

    return null;
}

protected void onPreExecute(){
    //dialog.dismiss();
    Log.i("thread", "Started...");
    dialog.show();
}
protected void onPostExecute(JSONObject result){
    Log.i("thread", "Done...");
    String status;
    String name;
    try {
        status= result.getString("status");
        name=result.getString("uname");

       if(dialog!=null)
       {
         dialog.dismiss();
       }
       if(status.equalsIgnoreCase("yes"))
          {
        tv.setText("Login Successful...");

        Bundle newbundle=new Bundle();
        newbundle.putString("uname",name);

        Intent myIntent=new Intent(Login_Menu.this,Instruction.class);
        myIntent.putExtras(newbundle);

        startActivity(myIntent);

    }else{

    tv.setText("No User Found, please try again!");
    }
    } catch (JSONException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
}

}}
于 2012-07-06T17:26:52.563 回答
0

make http connection in a thread.

and send the response in a UI-Handler

like

//UI-Handler

final Handler handler = new Handler() 
     { 
          public void handleMessage(Message msg) 

          { 
                  switch(msg.what){

                   case 100:

                      String response = (String)msg.obj;
                       tv.setText(response);

                  }

               }

          }

//make Http Connection inside  button's onclick and send http response to handleMessage(Message msg) method to update UI

login.setOnClickListener(new View.OnClickListener() {

        public void onClick(View arg0) {

Thread t = new Thread()
 {

public void run()
   {
   try{
        final HttpClient client = new DefaultHttpClient();
        String url="http://10.0.2.2:7001/f/json.jsp";
        final HttpPost post = new HttpPost(url);

            List<NameValuePair> pairs = new ArrayList<NameValuePair>();   
            pairs.add(new BasicNameValuePair("username",uname.getText().toString()));   
            pairs.add(new BasicNameValuePair("password",pass.getText().toString()));   
            post.setEntity(new UrlEncodedFormEntity(pairs));   
            HttpResponse response = client.execute(post);
            StringBuffer sb = new StringBuffer()

            while ((line = rd.readLine()) != null) {

            sb.append(line);

            }
             //send the response in a UI-Handler
              Message message = new Message();
                  message.obj=sb.toString();
                  message.what=100;

            handler.sendMessage( message );

    }catch(Exception e){}

  }

 }


  t.start();

}
于 2012-04-06T11:39:52.183 回答