如何将数组中的文件连接到新文件夹中？ [复制]

Question

我正在尝试编写一个 java 程序，它将接受两个参数，dirName 和 fileName。该程序将搜索dirName 中以.java 结尾的所有文件，然后将它们连接到一个名为fileName 的新文件夹中。到目前为止，我有一种方法可以在 dirName 中搜索 .java 文件，然后我将它们放在一个名为 list 的文件数组中，但现在我正在努力将这个数组中的文件迭代地添加到我的新文件夹 fileName .这是我目前所拥有的：

import java.io.File;
import java.io.FileInputStream;
import java.io.FilenameFilter;
import java.util.ArrayList;

public class TwoFiles {

    File dir;
    File name;

    public TwoFiles(File dirName, File fileName) {

        dir = dirName;
        name = fileName;

    }

    public void setDir(File m) {
        this.dir = m;
    }

    public File getDir() {
        return dir;
    }

    public void setNewFolder(File n) {
        this.name = n;
    }

    public File getNewFolder() {
        return name;
    }

    public File[] Finder(File dir) {

        dir = getDir();

        return dir.listFiles(new FilenameFilter() {

                public boolean accept(File dir, String filename) {
                    return name.endsWith(".java"); }

        } );
    }

    public static void main(String[] args) {

        File folder = null;
        File newFolder = null;
        Integer b = null;

        TwoFiles tf = new TwoFiles(folder, newFolder);

        folder = tf.getDir();
        newFolder = tf.getNewFolder();

        File[] list = tf.Finder(folder); //add to an array


//here is where I've been experimenting to add files in `list` to new folder, `fileName`.

        for (File file : list)
          {
              FileInputStream inFile = new FileInputStream(file);

              while ((b = inFile.read()) != -1)
                  newFolder.write(b);
              inFile.close();
          }

        //copy files from array (list) into newFolder 

    }

}

感谢您的宝贵时间。

Answer 1

您的 newFolder 变量是文件类型。你不能写进去。我假设，您的代码甚至无法编译。您必须在循环前创建一个输出流：

FileOutputStream fos = new FileOutputStream( newFolder);
try
{
    for (File file : list)
    {
        FileInputStream inFile = new FileInputStream(file);

          while ((b = inFile.read()) != -1)
              fos.write(b);
          inFile.close();
      }
}
finally
{
    fos.close();
}

Answer 2

您可以使用Apache Commons IO copyDirectory() 和IOFileFilter（用于.java 扩展名）将文件从一个目录复制到另一个目录。在此之前，您可以确保使用forceMkdir() 为您的filename 创建一个新目录。

Answer 3

这是你的问题的我的版本： 我创建了其他构造函数，您可以在其中只放置您想要连接文件的目录/文件夹的路径，以及连接结果的文件。

public class TwoFiles {

private File dir;
private File name;

public TwoFiles(File dirName, File fileName) {
    dir = dirName;
    name = fileName;
}

public TwoFiles(String dirName, String destinationFileName) throws IOException{
    dir=new File(dirName);
    if(!dir.isDirectory()){
        throw new FileNotFoundException();//here your exception in case when dirName is file name instead folder name
    }
    name=new File(destinationFileName);
    if(!name.exists()){
        name.createNewFile();
    }
}

public void setDir(File m) {
    this.dir = m;
}

public File getDir() {
    return dir;
}

public void setNewFolder(File n) {
    this.name = n;
}

public File getNewFolder() {
    return name;
}

public void concatenateFiles() throws IOException{
    File[] files=dir.listFiles();
    for(File file: files){
        if(file.getName().endsWith(".java")){ //check is right file
            prescribe(name, file);
        }
    }
}

/** prescribe file to new destination */
private void prescribe(File destination, File file) throws IOException {
    FileInputStream inFile = new FileInputStream(file);
    FileOutputStream writer=new FileOutputStream(destination, true); //true means next file will be write beginning from end of the file
    int x;
    while((x=inFile.read())!=-1){
        writer.write(x);
    }
    String test="\n";   //next line in file
    writer.write(test.getBytes());
    writer.close();
    inFile.close();
}

public static void main(String...strings){
    String dirName="C/myApp/model/entity";
    String fileName="C:/Users/Dell/Desktop/temp/test.java";
    try {
        new TwoFiles(dirName, fileName).concatenateFiles();

    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
}

}