android - 以编程方式设置 android 形状颜色

Question

我正在编辑以使问题更简单，希望这有助于获得准确的答案。

假设我有以下 oval 形状:

<?xml version="1.0" encoding="utf-8"?>
<shape xmlns:android="http://schemas.android.com/apk/res/android" android:shape="oval">
    <solid android:angle="270"
           android:color="#FFFF0000"/>
    <stroke android:width="3dp"
            android:color="#FFAA0055"/>
</shape>

如何在 Activity 类中以编程方式设置颜色？

Answer 1

注意:答案已更新以涵盖 background 是 ColorDrawable 实例的情况>。谢谢 Tyler Pfaff , 指出这一点。

The drawable is an oval and is the background of an ImageView

使用 getBackground() 从 imageView 获取 Drawable:

Drawable background = imageView.getBackground();

检查通常的嫌疑人:

if (background instanceof ShapeDrawable) {
    // cast to 'ShapeDrawable'
    ShapeDrawable shapeDrawable = (ShapeDrawable) background;
    shapeDrawable.getPaint().setColor(ContextCompat.getColor(mContext,R.color.colorToSet));
} else if (background instanceof GradientDrawable) {
    // cast to 'GradientDrawable'
    GradientDrawable gradientDrawable = (GradientDrawable) background;
    gradientDrawable.setColor(ContextCompat.getColor(mContext,R.color.colorToSet));
} else if (background instanceof ColorDrawable) {
    // alpha value may need to be set again after this call
    ColorDrawable colorDrawable = (ColorDrawable) background;
    colorDrawable.setColor(ContextCompat.getColor(mContext,R.color.colorToSet));
}

精简版:

Drawable background = imageView.getBackground();
if (background instanceof ShapeDrawable) {
    ((ShapeDrawable)background).getPaint().setColor(ContextCompat.getColor(mContext,R.color.colorToSet));
} else if (background instanceof GradientDrawable) {
    ((GradientDrawable)background).setColor(ContextCompat.getColor(mContext,R.color.colorToSet));
} else if (background instanceof ColorDrawable) {
    ((ColorDrawable)background).setColor(ContextCompat.getColor(mContext,R.color.colorToSet));
}

请注意，不需要检查空值。

但是，如果在其他地方使用它们，您应该在修改它们之前在可绘制对象上使用 mutate()。 (默认情况下，从 XML 加载的可绘制对象共享相同的状态。)