从文件创建BufferedImage并使其成为TYPE_INT_ARGB [英] Create a BufferedImage from file and make it TYPE_INT_ARGB

问题描述

我有一个带透明度的PNG文件，加载并存储在 BufferedImage 中。我需要 BufferedImage 为 TYPE_INT_ARGB 。但是，当我使用 getType（）时，返回的值为0（ TYPE_CUSTOM ）而不是2（ TYPE_INT_ARGB ）。

I have a PNG file with transparency that is loaded and stored in a BufferedImage. I need this BufferedImage to be of TYPE_INT_ARGB. However, when I use getType() the returned value is 0 (TYPE_CUSTOM) instead of 2 (TYPE_INT_ARGB).

这是我加载 .png 的方法：

public File img = new File("imagen.png");

public BufferedImage buffImg = 
    new BufferedImage(240, 240, BufferedImage.TYPE_INT_ARGB);

try { 
    buffImg = ImageIO.read(img ); 
} 
catch (IOException e) { }

System.out.Println(buffImg.getType()); //Prints 0 instead of 2

如何加载.png，保存在<$ c中$ c> BufferedImage 并将其设为 TYPE_INT_ARGB ？

How can I load the .png, save in the BufferedImage and make it TYPE_INT_ARGB?

推荐答案

BufferedImage in = ImageIO.read(img);

BufferedImage newImage = new BufferedImage(
    in.getWidth(), in.getHeight(), BufferedImage.TYPE_INT_ARGB);

Graphics2D g = newImage.createGraphics();
g.drawImage(in, 0, 0, null);
g.dispose();

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