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

这样做:

    ImageView imgIcon = findViewById(R.id.imgIcon);
    GradientDrawable backgroundGradient = (GradientDrawable)imgIcon.getBackground();
    backgroundGradient.setColor(getResources().getColor(R.color.yellow));

其他回答

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
    }
}

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。)

我们可以创建这个kotlin函数。

fun View.updateViewBGSolidColor(colorString: String) {
    when (val background: Drawable = this.background) {
        is ShapeDrawable -> {
            background.paint.color = Color.parseColor(colorString)
        }
        is GradientDrawable -> {
            background.setColor(Color.parseColor(colorString))
        }
        is ColorDrawable -> {
            background.color = Color.parseColor(colorString)
        }
    }
}

然后像下面这样使用它:

yourTextView.updateViewBGSolidColor("#FFFFFF")

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);

这是对我有效的解决方案……也写在另一个问题里: 如何动态改变形状颜色?

//get the image button by id
ImageButton myImg = (ImageButton) findViewById(R.id.some_id);

//get drawable from image button
GradientDrawable drawable = (GradientDrawable) myImg.getDrawable();

//set color as integer
//can use Color.parseColor(color) if color is a string
drawable.setColor(color)