Set android shape color programmatically

2019-01-02 15:07发布

问题:

I am editing to make the question simpler, hoping that helps towards an accurate answer.

Say I have the following oval shape:

<?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>

How do I set the color programmatically, from within an activity class?

回答1:

Note: Answer has been updated to cover the scenario where background is an instance of ColorDrawable. Thanks Tyler Pfaff, for pointing this out.

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

Get the Drawable from imageView using getBackground():

Drawable background = imageView.getBackground();

Check against usual suspects:

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));
}

Compact version:

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));
}

Note that null-checking is not required.

However, you should use mutate() on the drawables before modifying them if they are used elsewhere. (By default, drawables loaded from XML share the same state.)



回答2:

Do like this:

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


回答3:

Try this:

 public void setGradientColors(int bottomColor, int topColor) {
 GradientDrawable gradient = new GradientDrawable(Orientation.BOTTOM_TOP, new int[]  
 {bottomColor, topColor});
 gradient.setShape(GradientDrawable.RECTANGLE);
 gradient.setCornerRadius(10.f);
 this.setBackgroundDrawable(gradient);
 }

for more detail check this link this

hope help.



回答4:

hope this will help someone with the same issue

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);


回答5:

Expanding on Vikram's answer, if you are coloring dynamic views, like recycler view items, etc.... Then you probably want to call mutate() before you set the color. If you don't do this, any views that have a common drawable (i.e a background) will also have their drawable changed/colored.

public static void setBackgroundColorAndRetainShape(final int color, final Drawable background) {

    if (background instanceof ShapeDrawable) {
        ((ShapeDrawable) background.mutate()).getPaint().setColor(color);
    } else if (background instanceof GradientDrawable) {
        ((GradientDrawable) background.mutate()).setColor(color);
    } else if (background instanceof ColorDrawable) {
        ((ColorDrawable) background.mutate()).setColor(color);
    }else{
        Log.w(TAG,"Not a valid background type");
    }

}


回答6:

A simpler solution nowadays would be to use your shape as a background and then programmatically change its color via

view.getBackground().setColorFilter(Color.parseColor("#343434"), PorterDuff.Mode.SRC_OVER)



回答7:

this is the solution that works for me...wrote it in another question as well: How to change shape color dynamically?

//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)


回答8:

This question was answered a while back, but it can modernized by rewriting as a kotlin extension function.

fun Drawable.overrideColor(@ColorInt colorInt: Int) {
    when (this) {
        is GradientDrawable -> setColor(colorInt)
        is ShapeDrawable -> paint.color = colorInt
        is ColorDrawable -> color = colorInt
    }
}


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