Android应用中实现跳转外部浏览器打开链接功能

xmlCopy code
<Button
    android:id="@+id/open_browser_button"
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:text="在浏览器中打开链接"
    android:onClick="openLinkInBrowser" />

javaCopy code
public void openLinkInBrowser(View view) {
    String url = "https://www.example.com"; // 要打开的链接地址
    Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
    startActivity(intent);
}

xmlCopy code
<uses-permission android:name="android.permission.INTERNET" />

xmlCopy code
<Button
    android:id="@+id/open_browser_button"
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:text="在浏览器中打开学习网站"
    android:onClick="openLinkInBrowser" />

javaCopy code
import android.content.Intent;
import android.net.Uri;
import android.os.Bundle;
import android.view.View;
import androidx.appcompat.app.AppCompatActivity;
public class MainActivity extends AppCompatActivity {
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
    }
    public void openLinkInBrowser(View view) {
        String url = "https://www.example.com"; // 这里替换为实际的链接地址
        Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
        startActivity(intent);
    }
}

xmlCopy code
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.example.browsertest">
    <uses-permission android:name="android.permission.INTERNET" />
    <application
        <!-- 应用程序配置 -->
    </application>
</manifest>

<uses-permission android:name="android.permission.INTERNET" />