在java中组合图像

Question

这是我的代码：

    Image partNumberImage = Toolkit.getDefaultToolkit().getImage("D:/partNumber.png");
    Image lotNumberImage = Toolkit.getDefaultToolkit().getImage("D:/lotNumber.png");
    Image dteImage = Toolkit.getDefaultToolkit().getImage("D:/dte.png");
    Image quantityImage = Toolkit.getDefaultToolkit().getImage("D:/quantity.png");

    BufferedImage combinedImage = new BufferedImage(486, 
                                          151, 
                                          BufferedImage.TYPE_INT_RGB);


    Graphics g = combinedImage.getGraphics();

    combinedImage.createGraphics().setBackground(Color.white);

    g.clearRect(0,0, 486, 151);
    g.drawImage(partNumberImage, x, 18, null);
    g.drawImage(lotNumberImage, x, 48, null);
    g.drawImage(dteImage, x, 58, null);
    g.drawImage(quantityImage, x, 68, null);


    g.dispose();

    Iterator writers = ImageIO.getImageWritersByFormatName("png");
    ImageWriter writer = (ImageWriter) writers.next();
    if (writer == null) {
        throw new RuntimeException("PNG not supported?!");
    }

     ImageOutputStream out = ImageIO.createImageOutputStream(
                                new File("D:/Combined.png" ));
    writer.setOutput(out);
    writer.write(combinedImage);
    out.close();
}

我的问题是代码将输出此图像：

我需要的是白色背景对于图像。谢谢！

Answer 1

这对我来说看起来很危险：

Graphics g = combinedImage.getGraphics(); // Graphics object #1

combinedImage.createGraphics().setBackground(Color.white);  // Graphics object #2
// so now you've set the background color for the second Graphics object only

g.clearRect(0,0, 486, 151);  // but clear the rect in the first Graphics object
g.drawImage(partNumberImage, x, 18, null);
g.drawImage(lotNumberImage, x, 48, null);
g.drawImage(dteImage, x, 58, null);
g.drawImage(quantityImage, x, 68, null);

在我看来，您可能正在创建两个非常不同的 Graphics 对象，一个是 Graphics2D 对象，另一个是 Graphics 对象。当您在 Graphics2D 对象中设置背景颜色时，您会清除 Graphics 对象中的矩形，因此它可以解释为什么您的背景不是白色的。为什么不直接创建一个 Graphics2D 对象并将其用于所有用途：

Graphics2D g = combinedImage.createGraphics(); 
g.setBackground(Color.white);

//  Now there is only one Graphics object, and its background has been set
g.clearRect(0,0, 486, 151);  // This now uses the correct background color
g.drawImage(partNumberImage, x, 18, null);
g.drawImage(lotNumberImage, x, 48, null);
g.drawImage(dteImage, x, 58, null);
g.drawImage(quantityImage, x, 68, null);

Answer 2

添加图像之前，绘制一个图像大小的白色矩形：

g.clearRect(0,0, 486, 151);
g.setColor(Color.white);
g.fillRect(0,0,486,151);

g.drawImage(partNumberImage, x, 18, null);
g.drawImage(lotNumberImage, x, 48, null);
g.drawImage(dteImage, x, 58, null);
g.drawImage(quantityImage, x, 68, null);


g.dispose();