问题

在过去的几天里，我阅读了一些资料，取得了一些进展，以下是我编写的代码：

MainActivity:

    package com.example.appv_6;
    
    import androidx.appcompat.app.AppCompatActivity;
    
    import android.os.Bundle;
    import android.os.Handler;
    import android.os.Looper;
    import android.os.Message;
    import android.view.View;
    import android.widget.Button;
    import android.widget.TextView;
    
    public class MainActivity extends AppCompatActivity {
    
        Handler h1;
        Thread t1;
        TextView Text;
        Button Butt;
    
        @Override
        protected void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.activity_main);
    
            Text = (TextView) findViewById(R.id.tvText_1);
            Butt = (Button) findViewById(R.id.bButt_1);
    
            h1 = new Handler(Looper.getMainLooper()){
                @Override
                public void handleMessage(Message msg){
                    Text.setText(msg.obj.toString());
                }
            };
    
            Butt.setOnClickListener(new View.OnClickListener() {
                @Override
                public void onClick(View v) {
                    t1 = new Thread(new HTTPRequest(h1));
                    t1.start();
                }
            });
        }
    }

HTTPRequest

    package com.example.appv_6;
    
    import android.os.Bundle;
    import android.os.Handler;
    import android.os.Message;
    
    import java.io.BufferedInputStream;
    import java.io.BufferedOutputStream;
    import java.io.IOException;
    import java.io.InputStream;
    import java.io.OutputStream;
    import java.net.HttpURLConnection;
    import java.net.MalformedURLException;
    import java.net.URL;
    
    import javax.net.ssl.HttpsURLConnection;
    
    public class HTTPRequest implements Runnable {
    
        Handler h2;
    
        public HTTPRequest(Handler h) {
            h2 = h;
        }
    
        @Override
        public void run() {
            try {
                URL url = new URL("my url");
                HttpsURLConnection conn = (HttpsURLConnection) url.openConnection();
    
                conn.setRequestMethod("POST");
    
                conn.setDoInput(true);
                conn.setDoOutput(true);
    
                OutputStream out;
                InputStream in;
                conn.setRequestProperty("accept","text/html");
                conn.setRequestProperty("Cookie","ulogin=111111");
                conn.setRequestProperty("Cookie","upassword=222555");
    
                out = new BufferedOutputStream(conn.getOutputStream());
                in = new BufferedInputStream(conn.getInputStream());
                //String response = conn.getResponseMessage();
    
                Message msg = Message.obtain();
                msg.obj = in;
    
                h2.sendMessage(msg);
    
            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
    
            }
        }
    }

没有错误，一切都正常运行，但问题是 - 我构建了这段代码来测试是否可以登录我尝试登录的网站，但我无法从中获取任何信息。
在我按下按钮后，似乎发生了某些事情，我发送到我的UI线程的输入流给我这个： 
"java.io.BufferedInputStream@afe19b8"，
每次按下按钮后，它都在改变。
我尝试使用conn.getResponseMessage() 并通过handle发送它，但它只显示 "OK"，所以在那方面也没有运气。

我正在寻找的是连接后的网页的源代码，在发送了我的两个cookie后，它将能够显示我是否已登录。

package com.example.appv_6;
import androidx.appcompat.app.AppCompatActivity;
import android.os.Bundle;
import android.os.Handler;
import android.os.Looper;
import android.os.Message;
import android.view.View;
import android.widget.Button;
import android.widget.TextView;
public class MainActivity extends AppCompatActivity {
Handler h1;
Thread t1;
TextView Text;
Button Butt;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
Text = (TextView) findViewById(R.id.tvText_1);
Butt = (Button) findViewById(R.id.bButt_1);
h1 = new Handler(Looper.getMainLooper()){
@Override
public void handleMessage(Message msg){
Text.setText(msg.obj.toString());
}
};
Butt.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
t1 = new Thread(new HTTPRequest(h1));
t1.start();
}
});
}
}

package com.example.appv_6;
import android.os.Bundle;
import android.os.Handler;
import android.os.Message;
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import javax.net.ssl.HttpsURLConnection;
public class HTTPRequest implements Runnable {
Handler h2;
public HTTPRequest(Handler h) {
h2 = h;
}
@Override
public void run() {
try {
URL url = new URL("my url");
HttpsURLConnection conn = (HttpsURLConnection) url.openConnection();
conn.setRequestMethod("POST");
conn.setDoInput(true);
conn.setDoOutput(true);
OutputStream out;
InputStream in;
conn.setRequestProperty("accept","text/html");
conn.setRequestProperty("Cookie","ulogin=111111");
conn.setRequestProperty("Cookie","upassword=222555");
out = new BufferedOutputStream(conn.getOutputStream());
in = new BufferedInputStream(conn.getInputStream());
//String response = conn.getResponseMessage();
Message msg = Message.obtain();
msg.obj = in;
h2.sendMessage(msg);
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
}

答案1

得分: 0

请参考以下主题，了解如何处理HTTP响应。

https://stackoverflow.com/questions/61306061/logging-into-a-website-using-android-app-java