使通知大图标变为圆形

14

我想将用户联系人中的圆形头像作为通知的大图标显示 - 就像接收短信或邮件时那样。当我将大图标设置为该联系人的图像时,结果是一个方形的图标。

我想让类似于顶部图标(方形头像)的东西看起来像电子邮件通知中的大图标(圆形头像):

enter image description here

如何使它变成圆形?


我会参考这个链接:https://dev59.com/OGct5IYBdhLWcg3wpO_H - Vijay
1
我认为这里真正的问题是,你喜欢三明治吗? - behelit
4个回答

20

由于setLargeIcon()接受一个Bitmap,所以您需要从源创建一个圆形Bitmap

以下是来自在Android中创建圆形位图的代码(我自己没有尝试过)。

private Bitmap getCircleBitmap(Bitmap bitmap) {
    final Bitmap output = Bitmap.createBitmap(bitmap.getWidth(),
            bitmap.getHeight(), Bitmap.Config.ARGB_8888);
    final Canvas canvas = new Canvas(output);

    final int color = Color.RED;
    final Paint paint = new Paint();
    final Rect rect = new Rect(0, 0, bitmap.getWidth(), bitmap.getHeight());
    final RectF rectF = new RectF(rect);

    paint.setAntiAlias(true);
    canvas.drawARGB(0, 0, 0, 0);
    paint.setColor(color);
    canvas.drawOval(rectF, paint);

    paint.setXfermode(new PorterDuffXfermode(PorterDuff.Mode.SRC_IN));
    canvas.drawBitmap(bitmap, rect, rect, paint);

    bitmap.recycle();

    return output;
}

我想这可能是做这件事的方法。谢谢你提供的链接,它确实有效。我会稍微调整一下代码。 - Itai Hanski
2
我遇到了一些问题,与宽度大于高度或高度大于宽度的图像有关,它们显示为椭圆形而不是圆形。我该如何解决这个问题?有人遇到过这个问题吗? - Sandra

15

被接受的答案要求输入的位图必须是正方形(高度和宽度相同)。如果你的位图是矩形形状,它将返回一个椭圆形。我已经修改了代码以接受任何形状的位图,并返回在输入位图中心位置居中的圆形。

public static Bitmap getCircleBitmap(Bitmap bitmap) {
    Bitmap output;
    Rect srcRect, dstRect;
    float r;
    final int width = bitmap.getWidth();
    final int height = bitmap.getHeight();

    if (width > height){
        output = Bitmap.createBitmap(height, height, Bitmap.Config.ARGB_8888);
        int left = (width - height) / 2;
        int right = left + height;
        srcRect = new Rect(left, 0, right, height);
        dstRect = new Rect(0, 0, height, height);
        r = height / 2;
    }else{
        output = Bitmap.createBitmap(width, width, Bitmap.Config.ARGB_8888);
        int top = (height - width)/2;
        int bottom = top + width;
        srcRect = new Rect(0, top, width, bottom);
        dstRect = new Rect(0, 0, width, width);
        r = width / 2;
    }

    Canvas canvas = new Canvas(output);

    final int color = 0xff424242;
    final Paint paint = new Paint();

    paint.setAntiAlias(true);
    canvas.drawARGB(0, 0, 0, 0);
    paint.setColor(color);
    canvas.drawCircle(r, r, r, paint);
    paint.setXfermode(new PorterDuffXfermode(PorterDuff.Mode.SRC_IN));
    canvas.drawBitmap(bitmap, srcRect, dstRect, paint);

    bitmap.recycle();

    return output;
}

java.lang.IllegalStateException: 无法对已回收的位图进行打包。 - behelit
这是最好的答案(请参见完全平方问题)。 - Mike Casan Ballester

9

Egor的回答非常好。在这里发布代码以防链接消失:

private Bitmap getCircleBitmap(Bitmap bitmap) {
 final Bitmap output = Bitmap.createBitmap(bitmap.getWidth(),
  bitmap.getHeight(), Bitmap.Config.ARGB_8888);
 final Canvas canvas = new Canvas(output);

 final int color = Color.RED;
 final Paint paint = new Paint();
 final Rect rect = new Rect(0, 0, bitmap.getWidth(), bitmap.getHeight());
 final RectF rectF = new RectF(rect);

 paint.setAntiAlias(true);
 canvas.drawARGB(0, 0, 0, 0);
 paint.setColor(color);
 canvas.drawOval(rectF, paint);

 paint.setXfermode(new PorterDuffXfermode(PorterDuff.Mode.SRC_IN));
 canvas.drawBitmap(bitmap, rect, rect, paint);

 bitmap.recycle();

 return output;
}

最好更新一个仅包含链接的答案。无论如何,您的努力是受到赞赏的。 :) - Sufian
@Sufian 謝謝你,Bruce。 - Micro
如果您正在使用它来在通知中显示图像,则需要删除 bitmap.recycle()。在我的代码中,看起来像这样:.setLargeIcon(getCircleBitmap(Picasso.with(this.context).load(pictureUrl).get()))。希望这可以帮助到您。 - Isaac Bosca

0

我认为当输入的位图是一个完美的正方形(高度和宽度相同)时,它会很好地工作,从而产生完美的圆形,但如果您的位图是矩形或宽度大于高度,则会出现问题。我已经将代码修改为Kotlin,并添加了一个边框,您只需要提供位图作为输入即可。

private fun getCircleBitmap(bitmap: Bitmap): Bitmap 

  {

    var srcRect: Rect

    var dstRect: Rect

    var r: Float

    var paint = Paint();

    var width: Int = bitmap.getWidth()

    var height: Int = bitmap.getHeight()

    var widthToGenerate = 100F

    var heightToGenerate = 100F

    var borderWidth: Float = 1.toFloat()

    var output: Bitmap

    var canvas: Canvas

    if (width > height) {

        output = Bitmap.createBitmap(widthToGenerate.toInt(), heightToGenerate.toInt(), Bitmap.Config.ARGB_8888);

        canvas = Canvas(output);
        val scale: Float = (widthToGenerate / width)


        var xTranslation = 0.0f
        var yTranslation: Float = (heightToGenerate - height * scale) / 2.0f;


        var transformation = Matrix();
        transformation.postTranslate(xTranslation, yTranslation)
        transformation.preScale(scale, scale)

        var color: Int = Color.WHITE
        paint.setAntiAlias(true)
        paint.setColor(color)

        canvas.drawBitmap(bitmap, transformation, paint)

    } else {
        output = Bitmap.createBitmap(width, width, Bitmap.Config.ARGB_8888)
        canvas = Canvas(output);
        var top: Int = (height - width) / 2
        var bottom: Int = top + width
        srcRect = Rect(0, top, width, bottom)
        dstRect = Rect(0, 0, width, width);
        r = (width / 2).toFloat()


        var color: Int = Color.GRAY
        paint.setAntiAlias(true)
        canvas.drawARGB(0, 0, 0, 0)
        paint.setColor(color)


        canvas.drawCircle(r + borderWidth, r + borderWidth, r + borderWidth, paint)
        canvas.drawCircle(r, r, r, paint)
        paint.setXfermode(PorterDuffXfermode(PorterDuff.Mode.SRC_IN))

        canvas.drawBitmap(bitmap, srcRect, dstRect, paint)

        bitmap.recycle()
    }

    return output

}

notifiation_drawable.xml 文件用于边框

<shape xmlns:android="http://schemas.android.com/apk/res/android"
android:innerRadius="0dp"
android:shape="ring"
android:thicknessRatio="2.0"
android:useLevel="false" >
<solid android:color="@android:color/transparent" />
<stroke
    android:width="2dp"
    android:color="@android:color/darker_gray" />

</shape>

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