android-layout - 如何更改 StateListDrawable 中的颜色？

Question

我有以下按钮选择器(有 2 种状态，常规状态和按下状态):

<selector xmlns:android="http://schemas.android.com/apk/res/android">
<item android:state_pressed="true" >
 <shape>
  <solid
      android:color="#3498DB" />
  <stroke
      android:width="1dp"
      android:color="#2980B9" />
  <corners
      android:radius="0dp" />
  <padding
      android:left="12dp"
      android:top="12dp"
      android:right="12dp"
      android:bottom="12dp" />
 </shape>
</item>
<item>
 <shape>
  <solid
      android:color="#2980B9" />
  <stroke
      android:width="1dp"
      android:color="#2980B9" />
  <corners
      android:radius="0dp" />
  <padding
      android:left="12dp"
      android:top="12dp"
      android:right="12dp"
      android:bottom="12dp" />
 </shape>
</item>
</selector>

我的按钮具有以下内容，将背景指定为上述选择器:

<Button
  android:id="@+id/button_LaunchShiftsGame"
  android:layout_width="match_parent"
  android:layout_height="wrap_content"
  android:text="@string/ShiftsMode"
  android:background="@drawable/selector_Button"
  style="@style/Button_Text" />

当 Activity 加载时，我需要从代码中访问和更改两种状态的颜色。

我该怎么做？

Answer 1

     StateListDrawable gradientDrawable = (StateListDrawable) inflatedView.getBackground();
    DrawableContainerState drawableContainerState = (DrawableContainerState) gradientDrawable.getConstantState();
    Drawable[] children = drawableContainerState.getChildren();
    LayerDrawable selectedItem = (LayerDrawable) children[0];
    LayerDrawable unselectedItem = (LayerDrawable) children[1];
    GradientDrawable selectedDrawable = (GradientDrawable) selectedItem.getDrawable(0);
    GradientDrawable unselectedDrawable = (GradientDrawable) unselectedItem.getDrawable(0);
    selectedDrawable.setStroke(STORKE_SIZE, NOTIFICATION_COLOR);
    unselectedDrawable.setStroke(STORKE_SIZE, NOTIFICATION_COLOR);

我用它来改变笔画，它会很有帮助。