如何将文件从一个位置复制到另一个位置？

Question

我想将文件从一个位置复制到 Java 中的另一个位置。 做这个的最好方式是什么？

---

这是我到目前为止所拥有的：

import java.io.File;
import java.io.FilenameFilter;
import java.util.ArrayList;
import java.util.List;
public class TestArrayList {
    public static void main(String[] args) {
        File f = new File(
            "D:\\CBSE_Demo\\Demo_original\\fscommand\\contentplayer\\config");
        List<String>temp=new ArrayList<String>();
        temp.add(0, "N33");
        temp.add(1, "N1417");
        temp.add(2, "N331");
        File[] matchingFiles = null;
        for(final String temp1: temp){
            matchingFiles = f.listFiles(new FilenameFilter() {
                public boolean accept(File dir, String name) {
                    return name.startsWith(temp1);
                }
            });
            System.out.println("size>>--"+matchingFiles.length);

        }
    }
}

这不会复制文件，最好的方法是什么？

Answer 1

您可以使用此（或任何变体）：

Files.copy(src, dst, StandardCopyOption.REPLACE_EXISTING);

此外，我建议使用File.separator或/而不是\\\\以使其在多个操作系统中兼容， 此处提供有关此的问题/答案。

由于您不确定如何临时存储文件，请查看ArrayList ：

List<File> files = new ArrayList();
files.add(foundFile);

要将文件List移动到单个目录中：

List<File> files = ...;
String path = "C:/destination/";
for(File file : files) {
    Files.copy(file.toPath(),
        (new File(path + file.getName())).toPath(),
        StandardCopyOption.REPLACE_EXISTING);
}

Answer 2

使用流

private static void copyFileUsingStream(File source, File dest) throws IOException {
    InputStream is = null;
    OutputStream os = null;
    try {
        is = new FileInputStream(source);
        os = new FileOutputStream(dest);
        byte[] buffer = new byte[1024];
        int length;
        while ((length = is.read(buffer)) > 0) {
            os.write(buffer, 0, length);
        }
    } finally {
        is.close();
        os.close();
    }
}

使用频道

private static void copyFileUsingChannel(File source, File dest) throws IOException {
    FileChannel sourceChannel = null;
    FileChannel destChannel = null;
    try {
        sourceChannel = new FileInputStream(source).getChannel();
        destChannel = new FileOutputStream(dest).getChannel();
        destChannel.transferFrom(sourceChannel, 0, sourceChannel.size());
       }finally{
           sourceChannel.close();
           destChannel.close();
       }
}

使用 Apache Commons IO 库：

private static void copyFileUsingApacheCommonsIO(File source, File dest) throws IOException {
    FileUtils.copyFile(source, dest);
}

使用 Java SE 7 Files 类：

private static void copyFileUsingJava7Files(File source, File dest) throws IOException {
    Files.copy(source.toPath(), dest.toPath());
}

或者试试谷歌番石榴：

https://github.com/google/guava

文档： https : //guava.dev/releases/snapshot-jre/api/docs/com/google/common/io/Files.html

比较时间：

    File source = new File("/Users/sidikov/tmp/source.avi");
    File dest = new File("/Users/sidikov/tmp/dest.avi");


    //copy file conventional way using Stream
    long start = System.nanoTime();
    copyFileUsingStream(source, dest);
    System.out.println("Time taken by Stream Copy = "+(System.nanoTime()-start));
     
    //copy files using java.nio FileChannel
    source = new File("/Users/sidikov/tmp/sourceChannel.avi");
    dest = new File("/Users/sidikov/tmp/destChannel.avi");
    start = System.nanoTime();
    copyFileUsingChannel(source, dest);
    System.out.println("Time taken by Channel Copy = "+(System.nanoTime()-start));
     
    //copy files using apache commons io
    source = new File("/Users/sidikov/tmp/sourceApache.avi");
    dest = new File("/Users/sidikov/tmp/destApache.avi");
    start = System.nanoTime();
    copyFileUsingApacheCommonsIO(source, dest);
    System.out.println("Time taken by Apache Commons IO Copy = "+(System.nanoTime()-start));
     
    //using Java 7 Files class
    source = new File("/Users/sidikov/tmp/sourceJava7.avi");
    dest = new File("/Users/sidikov/tmp/destJava7.avi");
    start = System.nanoTime();
    copyFileUsingJava7Files(source, dest);
    System.out.println("Time taken by Java7 Files Copy = "+(System.nanoTime()-start));

Answer 3

在 Java >=7 中使用 New Java File 类。

创建以下方法并导入必要的库。

public static void copyFile( File from, File to ) throws IOException {
    Files.copy( from.toPath(), to.toPath() );
} 

在 main 中使用如下创建的方法：

File dirFrom = new File(fileFrom);
File dirTo = new File(fileTo);

try {
        copyFile(dirFrom, dirTo);
} catch (IOException ex) {
        Logger.getLogger(TestJava8.class.getName()).log(Level.SEVERE, null, ex);
}

注意：- fileFrom 是您要复制到不同文件夹中的新文件 fileTo 的文件。

积分 - @Scott： 在 Java 中复制文件的标准简洁方法？

Answer 4

  public static void copyFile(File oldLocation, File newLocation) throws IOException {
        if ( oldLocation.exists( )) {
            BufferedInputStream  reader = new BufferedInputStream( new FileInputStream(oldLocation) );
            BufferedOutputStream  writer = new BufferedOutputStream( new FileOutputStream(newLocation, false));
            try {
                byte[]  buff = new byte[8192];
                int numChars;
                while ( (numChars = reader.read(  buff, 0, buff.length ) ) != -1) {
                    writer.write( buff, 0, numChars );
                }
            } catch( IOException ex ) {
                throw new IOException("IOException when transferring " + oldLocation.getPath() + " to " + newLocation.getPath());
            } finally {
                try {
                    if ( reader != null ){                      
                        writer.close();
                        reader.close();
                    }
                } catch( IOException ex ){
                    Log.e(TAG, "Error closing files when transferring " + oldLocation.getPath() + " to " + newLocation.getPath() ); 
                }
            }
        } else {
            throw new IOException("Old location does not exist when transferring " + oldLocation.getPath() + " to " + newLocation.getPath() );
        }
    }  

Answer 5

Files.exists()

Files.createDirectory()

文件.copy()

覆盖现有文件： Files.move()

文件.delete()

Files.walkFileTree()在此处输入链接描述

Answer 6

您可以使用Java 8 Streaming API 、 PrintWriter和Files API 来实现

try (PrintWriter pw = new PrintWriter(new File("destination-path"), StandardCharsets.UTF_8)) {
    Files.readAllLines(Path.of("src/test/resources/source-file.something"), StandardCharsets.UTF_8)
         .forEach(pw::println);
}

如果您想在复制时即时修改内容，请查看此链接以获取扩展示例https://overflowed.dev/blog/copy-file-and-modify-with-java-streams/

Answer 7

我修改了其中一个答案以使其更有效率。

public void copy(){
    InputStream in = null;
    try {
        in = new FileInputStream(Files);
    } catch (FileNotFoundException e) {
        e.printStackTrace();
    }
    try {
        OutputStream out = new FileOutputStream();
        try {
            // Transfer bytes from in to out
            byte[] buf = new byte[1024];
            while (true) {
                int len = 0;
                try {
                    if (!((len = in.read(buf)) > 0)) break;
                } catch (IOException e) {
                    e.printStackTrace();
                }
                try {
                    out.write(buf, 0, len);
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
        } finally {
            try {
                out.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    } finally {
        try {
            in.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

private void moveFile() {
    copy();
    File dir = getFilesDir();
    File file = new File(dir, "my_filename");
    boolean deleted = file.delete();
}

Answer 8

将文件从一个位置复制到另一个位置意味着需要将整个内容复制到另一个位置。 Files.copy(Path source, Path target, CopyOption... options) throws IOException此方法期望源位置是原始文件位置，目标位置是目标位置相同类型文件（与原始文件相同）的新文件夹位置。 要么目标位置需要存在于我们的系统中，否则我们需要创建一个文件夹位置，然后在该文件夹位置我们需要创建一个与原始文件名同名的文件。然后使用复制功能，我们可以轻松地从一个位置复制文件到其他。

 public static void main(String[] args) throws IOException {
                String destFolderPath = "D:/TestFile/abc";
                String fileName = "pqr.xlsx";
                String sourceFilePath= "D:/TestFile/xyz.xlsx";
                File f = new File(destFolderPath);
                if(f.mkdir()){
                    System.out.println("Directory created!!!!");
                }
                else {
                    System.out.println("Directory Exists!!!!");
                }
                f= new File(destFolderPath,fileName);
                if(f.createNewFile())   {

                    System.out.println("File Created!!!!");
                }   else {
                    System.out.println("File exists!!!!");
                }

                Files.copy(Paths.get(sourceFilePath), Paths.get(destFolderPath, fileName),REPLACE_EXISTING);
                System.out.println("Copy done!!!!!!!!!!!!!!");


            }

Answer 9

您可以使用

FileUtils.copy(sourceFile, destinationFile);

https://commons.apache.org/proper/commons-io/apidocs/org/apache/commons/io/FileUtils.ZFC35FDC70D5FC69D269883A822C7A53