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Java加速读取复制超大文件

2019-11-26 09:02:47
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用文件通道(FileChannel)来实现文件复制,供大家参考,具体内容如下

不考虑多线程优化,单线程文件复制最快的方法是(文件越大该方法越有优势,一般比常用方法快30+%):

直接上代码:

package test; import java.io.BufferedInputStream; import java.io.BufferedOutputStream; import java.io.File; import java.io.FileInputStream; import java.io.FileOutputStream; import java.io.IOException; import java.io.InputStream; import java.nio.channels.FileChannel; public class Test {public static void main(String[] args) {File source = new File("E://tools//fmw_12.1.3.0.0_wls.jar");File target = new File("E://tools//fmw_12.1.3.0.0_wls-copy.jar");long start, end;start = System.currentTimeMillis();fileChannelCopy(source, target);end = System.currentTimeMillis();System.out.println("文件通道用时:" + (end - start) + "毫秒");start = System.currentTimeMillis();copy(source, target);end = System.currentTimeMillis();System.out.println("普通缓冲用时:" + (end - start) + "毫秒");}/**  * 使用文件通道的方式复制文件  * @param source 源文件  * @param target 目标文件  */ public static void fileChannelCopy(File source, File target) {   FileInputStream in = null;   FileOutputStream out = null;   FileChannel inChannel = null;   FileChannel outChannel = null;   try {    in = new FileInputStream(source);    out = new FileOutputStream(target);    inChannel = in.getChannel();//得到对应的文件通道   outChannel = out.getChannel();//得到对应的文件通道   inChannel.transferTo(0, inChannel.size(), outChannel);//连接两个通道,并且从inChannel通道读取,然后写入outChannel通道  } catch (IOException e) {    e.printStackTrace();   } finally {    try {     in.close();     inChannel.close();     out.close();     outChannel.close();    } catch (IOException e) {     e.printStackTrace();    }   }  } /*** 普通缓冲复制* @param source 源文件* @param target 目标文件*/public static void copy (File source, File target) {InputStream in = null;BufferedOutputStream out = null;try {in = new BufferedInputStream(new FileInputStream(source));out = new BufferedOutputStream(new FileOutputStream(target));byte[] buf = new byte[4096];int i;while ((i = in.read(buf)) != -1) {out.write(buf, 0, i);}} catch (Exception e) {e.printStackTrace();} finally {try {if (null != in) {in.close();}if (null != out) {out.close();}} catch (IOException e) {e.printStackTrace();}}} } }

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