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
我没有工作，但当我设置色调颜色，它工作在形状绘制
Drawable background = imageView.getBackground();
background.setTint(getRandomColor())
需要android 5.0 API 21
这样做:
ImageView imgIcon = findViewById(R.id.imgIcon);
GradientDrawable backgroundGradient = (GradientDrawable)imgIcon.getBackground();
backgroundGradient.setColor(getResources().getColor(R.color.yellow));
GradientDrawable gd = new GradientDrawable(
GradientDrawable.Orientation.TOP_BOTTOM,
new int[] {0xFF616261,0xFF131313});
gd.setCornerRadius(0f);
layout.setBackgroundDrawable(gd);
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;
}
}
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")
}
}
