将图像从文件读/写到 BufferedImage 的最快方法? [英] Fastest way to read/write Images from a File into a BufferedImage?

问题描述

1. 在 Java/Grails 中将图像从文件读取到 BufferedImage 的最快方法是什么?
2. 在 Java/Grails 中将图像从 BufferedImage 写入文件的最快方法是什么?

我的变体(阅读):

byte [] imageByteArray = new File(basePath+imageSource).readBytes()
InputStream inStream = new ByteArrayInputStream(imageByteArray)
BufferedImage bufferedImage = ImageIO.read(inStream)

我的变体(写):

BufferedImage bufferedImage = // some image
def fullPath = // image page + file name
byte [] currentImage

try{

    ByteArrayOutputStream baos = new ByteArrayOutputStream();
    ImageIO.write( bufferedImage, "jpg", baos );
    baos.flush();
    currentImage = baos.toByteArray();
    baos.close();

    }catch(IOException e){
        System.out.println(e.getMessage());
    }       
   }    


def newFile = new FileOutputStream(fullPath)
newFile.write(currentImage)
newFile.close()

推荐答案

您的读取解决方案基本上是读取字节两次，一次是从文件中读取，一次是从 ByteArrayInputStream 中读取.不要这样

Your solution to read is basically reading the bytes twice, once from the file and once from the ByteArrayInputStream. Don't do that

用Java 7阅读

BufferedImage bufferedImage = ImageIO.read(Files.newInputStream(Paths.get(basePath + imageSource)));

用Java 7编写

ImageIO.write(bufferedImage, "jpg", Files.newOutputStream(Paths.get(fullPath)));

对 Files.newInputStream 的调用将返回一个 ChannelInputStream，它 (AFAIK) 没有被缓冲.你会想要包装它

The call to Files.newInputStream will return a ChannelInputStream which (AFAIK) is not buffered. You'll want to wrap it

new BufferedInputStream(Files.newInputStream(...));

减少对磁盘的 IO 调用，具体取决于您的使用方式.

So that there are less IO calls to disk, depending on how you use it.

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