Java 从文件创建一个BuffereImage,并将其类型设置为\u INT\u ARGB
我有一个透明的PNG文件,它被加载并存储在Java 从文件创建一个BuffereImage,并将其类型设置为\u INT\u ARGB,java,image,image-manipulation,bufferedimage,argb,Java,Image,Image Manipulation,Bufferedimage,Argb,我有一个透明的PNG文件,它被加载并存储在buffereImage中。我需要此buffereImage为类型\u INT\u ARGB。但是,当我使用getType()时,返回的值是0(TYPE\u CUSTOM),而不是2(TYPE\u INT\u ARGB) 这就是我加载.png的方式: public File img = new File("imagen.png"); public BufferedImage buffImg = new BufferedImage(240, 2
buffereImage
中。我需要此buffereImage
为类型\u INT\u ARGB
。但是,当我使用getType()
时,返回的值是0(TYPE\u CUSTOM
),而不是2(TYPE\u INT\u ARGB
)
这就是我加载.png
的方式:
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BufferedImage
中并使其键入\u INT\u ARGB
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();
我的图像文件的输出示例:
BufferedImage@5d391d: type = 5 ColorModel: #pixelBits = 24
numComponents = 3 color
space = java.awt.color.ICC_ColorSpace@50a649
transparency = 1
has alpha = false
isAlphaPre = false
ByteInterleavedRaster:
width = 800
height = 600
#numDataElements 3
dataOff[0] = 2
您可以运行System.out.println(对象);在任何对象上创建一个BuffereImage文件,并使其键入\u INT\u RGB
import java.io.*;
import java.awt.image.*;
import javax.imageio.*;
public class Main{
public static void main(String args[]){
try{
BufferedImage img = new BufferedImage(
500, 500, BufferedImage.TYPE_INT_RGB );
File f = new File("MyFile.png");
int r = 5;
int g = 25;
int b = 255;
int col = (r << 16) | (g << 8) | b;
for(int x = 0; x < 500; x++){
for(int y = 20; y < 300; y++){
img.setRGB(x, y, col);
}
}
ImageIO.write(img, "PNG", f);
}
catch(Exception e){
e.printStackTrace();
}
}
}
import java.io.*;
导入java.awt.image.*;
导入javax.imageio.*;
公共班机{
公共静态void main(字符串参数[]){
试一试{
BuffereImage img=新的BuffereImage(
500,500,BufferedImage.TYPE_INT_RGB);
文件f=新文件(“MyFile.png”);
int r=5;
int g=25;
int b=255;
int col=(r将public BufferedImage buffImg=new BufferedImage(240240,BufferedImage.TYPE_int_ARGB)
更改为public BufferedImage buffImg;
捕获(IOException e){}
到catch(IOException e){e.printStackTrace()}
。报告新的输出。System.Out.Println
无法编译。为了更快地获得更好的帮助,请发布一个。这是非常低效的,并且随着图像大小的增加而变得越来越不可行。呃,@alexantd,为什么它如此低效?是的,它需要O(宽x高)内存,可以使用恒定内存完成,但它与任何其他方法具有相同的时间复杂性,任何其他方法都会非常混乱。@tucuxi由于读者首先创建正确类型的BuffereImage,因此它的效率低于根本不需要任何转换方法
import java.io.*;
import java.awt.image.*;
import javax.imageio.*;
public class Main{
public static void main(String args[]){
try{
BufferedImage img = new BufferedImage(
500, 500, BufferedImage.TYPE_INT_RGB );
File f = new File("MyFile.png");
int r = 5;
int g = 25;
int b = 255;
int col = (r << 16) | (g << 8) | b;
for(int x = 0; x < 500; x++){
for(int y = 20; y < 300; y++){
img.setRGB(x, y, col);
}
}
ImageIO.write(img, "PNG", f);
}
catch(Exception e){
e.printStackTrace();
}
}
}
try{
BufferedImage img = new BufferedImage(
500, 500, BufferedImage.TYPE_INT_ARGB );
File f = new File("MyFile.png");
int r = 255;
int g = 10;
int b = 57;
int alpha = 255;
int col = (alpha << 24) | (r << 16) | (g << 8) | b;
for(int x = 0; x < 500; x++){
for(int y = 20; y < 30; y++){
img.setRGB(x, y, col);
}
}
ImageIO.write(img, "PNG", f);
}
catch(Exception e){
e.printStackTrace();
}