mcf*_*oft 9 android paint bitmap android-canvas
有没有人对以下问题有提示或解释?我画了一个带有位图的路径.当画布没有缩放时,它看起来不错(第一张图片).当我缩放(放大)时,位图处理器不会被缩放并且看起来非常难看.我在放大后尝试了几个重新创建bitmapshader的东西,但没有成功:-(.有人有提示吗?
没有缩放它看起来不错:

缩放它看起来很难看:

代码:
canvas.scale(scalex, scaley);
canvas.translate(itranslatex, itranslatey);
fillBMP = makePatternCross(fscalex, 1, Color.GREEN/*,fscalex,fscaley*/);
fillBMPshader = new BitmapShader(fillBMP, BitmapShader.TileMode.REPEAT, BitmapShader.TileMode.REPEAT);
paintshader = new Paint();
paintshader.setShader(fillBMPshader);
canvas.drawPath(cpath.path, paintshader);
private static Bitmap makePatternCross(float fSize, float fStrokewith,int iColor) {
Log.v("Create Patter makePatternCross","makePatternCross");
float fBitmapSizeOrig = 10;
fBitmapSizeOrig=fBitmapSizeOrig*fSize;
Bitmap bm = Bitmap.createBitmap((int)fBitmapSizeOrig,(int) fBitmapSizeOrig,Bitmap.Config.ARGB_8888);
Canvas c = new Canvas(bm);
//c.scale(200, 200);
c.drawColor(Color.WHITE);
Paint p = new Paint();
p.setColor(iColor);
//p.setStrokeWidth(iStrokewith);
p.setStrokeWidth(fStrokewith/fSize);
p.setStrokeWidth((float) 0.000001);
c.drawLine(0, 0, fBitmapSizeOrig, fBitmapSizeOrig, p);
c.drawLine(0, fBitmapSizeOrig, fBitmapSizeOrig, 0, p);
if (fSize != 1) {
int iNewSize = (int) (( fBitmapSizeOrig) * fSize);
bm = Bitmap.createScaledBitmap(bm, iNewSize, iNewSize, false);
}
int width = bm.getWidth();
int height = bm.getHeight();
for (int x = 0; x < width; x++) {
for (int y = 0; y < height; y++) {
if (bm.getPixel(x, y) == Color.WHITE) {
bm.setPixel(x, y, Color.TRANSPARENT);
} else {
// bm.setPixel(x, y, bm.getPixel(x, y));
}
}
}
return bm;
}
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不确定这是否是您想要的。但如果您使用矩阵来缩放位图,它会比正常缩放保留更高的质量。
Matrix matrix = new Matrix();
matrix.postScale(desiredScale, desiredScale);
Bitmap scaledBitmap = Bitmap.createBitmap(sampledSrcBitmap, 0, 0, sampledSrcBitmap.getWidth(), sampledSrcBitmap.getHeight(), matrix, true);
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另外,当从较低的分辨率变为较高的分辨率时,您也可以尝试以下操作:
Options options = new BitmapFactory.Options();
options.inScaled = false;
Bitmap source = BitmapFactory.decodeResource(a.getResources(), path, options);
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