在Android中以编程方式设置布局的大小

Question

I want to create a layout depend on size of device screen. This is my MainActivity

public class LoginActivity extends Activity {

int width_device;
int height_device;
LinearLayout layout_login_content;
LinearLayout.LayoutParams param;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login);

    // Get screen size
    width_device = Utility.getDeviceWidth(this);
    height_device = Utility.getDeviceHeight(this);

    // Get widget from layout
    layout_login_content = (LinearLayout) findViewById(R.id.layout_login_content);

    // Set up layout size
    param = (LinearLayout.LayoutParams) layout_login_content
            .getLayoutParams();
    param.width = width_device * 60 / 100;
    param.height = height_device * 60 / 100;
    layout_login_content.setLayoutParams(param);
}

Don't worry about how to get screen size. I did it and it worked.

This is my main_fragment.xml

<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:tools="http://schemas.android.com/tools"
android:id="@+id/layout_login_content"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:layout_gravity="center"
android:orientation="vertical"
tools:context="login.LoginActivity" >
</LinearLayout>

And this is error message

java.lang.ClassCastException: android.widget.FrameLayout$LayoutParams cannot be cast to android.widget.LinearLayout$LayoutParams

I dont know what happen here. I declare LinearLayout.LayoutParams , get params from LinearLayout in xml. But why it can't cast correctly?

Sorry about my english. I hope everyone know what I mean :( Thank you so much!

Answer 1

Looks like your LinearLayout is inheriting some FrameLayout params.

Try simply using FrameLayout.LayoutParams instead of LinearLayout.LayoutParams :

param = (FrameLayout.LayoutParams) layout_login_content.getLayoutParams();