java中快速读写图片到BufferedImage对象
java7读取文件到BufferedImage对象
BufferedImage bufferedImage = ImageIO.read(Files.newInputStream(Paths.get(basePath + imageSource)));
java7写入文件到图片对象
ImageIO.write(bufferedImage, "jpg", Files.newOutputStream(Paths.get(fullPath)));
The call to Files.newInputStream
will return a ChannelInputStream
which (AFAIK) is not buffered. You'll want to wrap it
new BufferedInputStream(Files.newInputStream(...));
So that there are less IO calls to disk, depending on how you use it.
意译:据我所知,调用Files.newInputStream将会返回一些ChannelInputStream对象。如果你想对他进行封装,使用以下代码
new BufferedInputStream(File.newInputStream(Paths.get(fullPath)));
如此一来,磁盘IO的调用频次将会降低,具体看你怎么用了。
工具类:
import lombok.extern.slf4j.Slf4j;import sun.misc.BASE64Decoder;
import sun.misc.BASE64Encoder;
import javax.imageio.ImageIO;
import java.awt.image.BufferedImage;
import java.io.BufferedInputStream;
import java.io.ByteArrayInputStream;
import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Paths;
@Slf4j
public final class GraphUtil {
/**
* Encode Image to Base64 String
* @param image
* @param type
* @return
*/
public static String encodeToString(BufferedImage image, String type) {
String imageString = null;
ByteArrayOutputStream bos = new ByteArrayOutputStream();
try {
ImageIO.write(image, type, bos);
byte[] imageBytes = bos.toByteArray();
BASE64Encoder encoder = new BASE64Encoder();
imageString = encoder.encode(imageBytes);
bos.close();
} catch (IOException e) {
e.printStackTrace();
}
return imageString;
}
/***
* Decode Base64 String to Image
* @param imageString
* @return
*/
public static BufferedImage decodeToImage(String imageString) {
BufferedImage image = null;
byte[] imageByte;
try {
BASE64Decoder decoder = new BASE64Decoder();
imageByte = decoder.decodeBuffer(imageString);
ByteArrayInputStream bis = new ByteArrayInputStream(imageByte);
image = ImageIO.read(bis);
bis.close();
} catch (Exception e) {
e.printStackTrace();
}
return image;
}
public static BufferedImage getBufferedImage(String basePath, String imageSource){
try {
return ImageIO.read(new BufferedInputStream(Files.newInputStream(Paths.get(basePath, imageSource))));
} catch (IOException e) {
log.error("读取图片出错:{}",e);
return null;
}
}
}
参考来源:
https://stackoverflow.com/questions/18522398/fastest-way-to-read-write-images-from-a-file-into-a-bufferedimage
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