我正在根据像素比较两个图像,需要围绕不匹配的像素值绘制一个矩形。它正确地绘制了第一个不匹配的像素周围的矩形,但在病房里并没有。
for (int row = 0; row < height; row++) {
for (int col = 0; col < width; col++) {
result[row][col] = img1.getRGB(col, row);
result1[row][col] = img2.getRGB(col, row);
if(result[row][col] != result1[row][col]){
try {
g.drawImage(temp, 0, 0, null);
g.setColor(Color.red);
g.setStroke(new BasicStroke(2.0F));
g.drawRect(col-5, row-12, 35, 35);
} finally {
g.dispose();
}
}
}
}
答案 0 :(得分:1)
g.drawImage(temp, 0, 0, null);
将绘制以前绘制的任何内容,而不是环路中的好主意g.dispose
可以防止将来更新/绘制任何内容。你完全应该在完成后才打电话作为快速测试,我将以下图像作为输入...
通过......
BufferedImage img1 = ImageIO.read(new File("..."));
BufferedImage img2 = ImageIO.read(new File("..."));
int width = img1.getWidth();
int height = img1.getHeight();
BufferedImage temp = new BufferedImage(width, height, BufferedImage.TYPE_INT_ARGB);
Graphics2D g = temp.createGraphics();
g.drawImage(img1, 0, 0, null);
for (int row = 0; row < height; row++) {
for (int col = 0; col < width; col++) {
int result = img1.getRGB(col, row);
int result1 = img2.getRGB(col, row);
if (result != result1) {
g.setColor(Color.red);
g.setStroke(new BasicStroke(2.0F));
g.drawRect(col - 5, row - 12, 35, 35);
}
}
}
g.dispose();
JOptionPane.showMessageDialog(null, new ImageIcon(temp));
哪个输出......