标题翻译
How to get HttpUrlConnection full response
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。
英文翻译
After doing some reading during the last couple of days I have been able to make some progress and here is the code that I have come up with:
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();
}
}
}
No errors, everything runs just fine, but the problem is - I have built this code as a test if I can log in the website I am trying to log into, yet I am not able to get any information out of this.
After I press the button, it seems like something is happening and the InputStream that I am sending to my UI thread is giving me this:
"java.io.BufferedInputStream@afe19b8"
and after each button press, it keeps changing.
I tried using conn.getResponseMessage() and send it via the handle, but it just shows "OK", so no luck there as well.
What I am looking for is the source code of the webpage that I am connected to after sending two of my cookies which will be able to show if I have logged in or not.
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
英文翻译
Refer to this topic for details on how to handle HTTP response.
https://stackoverflow.com/questions/61306061/logging-into-a-website-using-android-app-java
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