我正在编辑，使问题更简单，希望有助于得到一个准确的答案。
假设我有如下椭圆形状:
<?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>
如何从一个活动类中以编程方式设置颜色?
当前回答
希望这对有同样问题的人有所帮助
GradientDrawable gd = (GradientDrawable) YourImageView.getBackground();
//To shange the solid color
gd.setColor(yourColor)
//To change the stroke color
int width_px = (int)TypedValue.applyDimension(TypedValue.COMPLEX_UNIT_DIP, youStrokeWidth, getResources().getDisplayMetrics());
gd.setStroke(width_px, yourColor);
其他回答
希望这对有同样问题的人有所帮助
GradientDrawable gd = (GradientDrawable) YourImageView.getBackground();
//To shange the solid color
gd.setColor(yourColor)
//To change the stroke color
int width_px = (int)TypedValue.applyDimension(TypedValue.COMPLEX_UNIT_DIP, youStrokeWidth, getResources().getDisplayMetrics());
gd.setStroke(width_px, yourColor);
注意:答案已经更新，以涵盖背景是ColorDrawable实例的场景。谢谢泰勒·普法夫指出这一点。
可绘制对象是一个椭圆形，是ImageView的背景
使用getBackground()从imageView获取Drawable:
Drawable background = imageView.getBackground();
检查通常的嫌疑:
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));
}
简洁版:
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));
}
注意，空检查不是必需的。
但是，如果在其他地方使用可绘制对象，则应该在修改它们之前对它们使用mutate()。(默认情况下，从XML加载的可绘制对象共享相同的状态。)
我的Kotlin扩展函数版本基于上述答案与Compat:
fun Drawable.overrideColor_Ext(context: Context, colorInt: Int) {
val muted = this.mutate()
when (muted) {
is GradientDrawable -> muted.setColor(ContextCompat.getColor(context, colorInt))
is ShapeDrawable -> muted.paint.setColor(ContextCompat.getColor(context, colorInt))
is ColorDrawable -> muted.setColor(ContextCompat.getColor(context, colorInt))
else -> Log.d("Tag", "Not a valid background type")
}
}
这个问题之前已经回答过了，但是可以通过重写为kotlin扩展函数来实现现代化。
fun Drawable.overrideColor(@ColorInt colorInt: Int) {
when (this) {
is GradientDrawable -> setColor(colorInt)
is ShapeDrawable -> paint.color = colorInt
is ColorDrawable -> color = colorInt
}
}
对于任何使用c# Xamarin的人来说，这里是一个基于Vikram代码片段的方法:
private void SetDrawableColor(Drawable drawable, Android.Graphics.Color color)
{
switch (drawable)
{
case ShapeDrawable sd:
sd.Paint.Color = color;
break;
case GradientDrawable gd:
gd.SetColor(color);
break;
case ColorDrawable cd:
cd.Color = color;
break;
}
}
