如何根据限制列出文件系统中的文件:java

Question

如何列出文件系统中可用的文件的起始编号和结束编号？

就好像有500文件在那里C:\Test\如何列出从1到20的文件一样

start number and end number根据此列表给出可用于特定文件路径的文件.

我在java中尝试这个

我试过这样的事情,它给了我给定路径的所有可用文件

public static List<String> loadAllFiles(String filesLocation) {

        //find OS
        //String osName = System.getProperty("os.name");

        //replace file path based on OS
        filesLocation = filesLocation.replaceAll("\\\\|/", "\\"+System.getProperty("file.separator"));

        List<String> pdfFiles = new ArrayList<String>();
        if (log.isDebugEnabled()) {
            log.debug("In loadAllFiles execute start");
        }

        File directoryList = new File(filesLocation);
        File[] filesList = directoryList.listFiles();

        try {
            for (int count = 0; count < filesList.length; count++) {
                if (!filesList[count].isDirectory() && filesList[count].getName().endsWith(SPLIT_AND_SAVE_WORKING_FILE_EXTENSION.trim())) { 
                    // load only PDF  files
                    pdfFiles.add(filesList[count].getName().replace(SPLIT_AND_SAVE_WORKING_FILE_EXTENSION.trim(), ""));
                }
            }


        } catch (Exception filesException) {
            filesException.printStackTrace();
            //TODO : Log the exception
        } finally {

            if (filesList != null)
                filesList = null;

            if (directoryList != null)
                directoryList = null;
        }

        log.debug("In loadAllFiles execute end");

        return pdfFiles;
    }

I think the question is misunderstood, Say if i have 1000 files[file names can be anything] and i want to restrict getting the files name like i will give starting Number and ending number. like 1 to 20 and i want to load those 20 files alone. 

Answer 1

没有使用普通Java 7的外部库的示例

import java.io.IOException;
import java.nio.file.DirectoryStream;
import static java.nio.file.DirectoryStream.Filter;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;

// list files starting with 1 till 20 "-.*"
public class FileNameFilter {

    private static final Filter<Path> fileNameFilter = new Filter<Path>() {
        @Override
        public boolean accept(Path entry) throws IOException {
            if (!Files.isRegularFile(entry)) {
                return false;
            }
            return entry.getFileName().toString().matches("^([1][0-9]{0,1}|2[0]{0,1})-.*");
        }
    };

    public static void main(String[] args) {
        final String filesLocation = "resources/";

        Path path = Paths.get(filesLocation);
        try (DirectoryStream<Path> dirStream = Files.newDirectoryStream(path, fileNameFilter)) {
            for (Path entry : dirStream) {
                System.out.printf("%-5s: %s%n", "entry", entry.getFileName());
            }
        } catch (IOException e) {
            // add your exception handling here
            e.printStackTrace(System.err);
        }
    }
}

编辑 Java 8版本

// list files starting with 1 till 20 "-.*"
public class FileNameFilter {
    public static void main(String[] args) {
        final String filesLocation = "resources/";

        try {
            Files.walk(Paths.get(filesLocation))
                    .filter(p -> p.getFileName().toString().matches("^([1][0-9]{0,1}|2[0]{0,1})-.*"))
                    .forEach(entry -> {System.out.printf("%-5s: %s%n", "entry", entry.getFileName());});
        } catch (IOException e) {
            // add your exception handling here
            e.printStackTrace(System.err);
        }
    }
}

编辑2 列出目录中前20个文件的示例.
note注意文件的顺序与运行ls或dir在目录中的顺序相同.

Java 7的例子

import java.io.IOException;
import java.nio.file.DirectoryStream;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;

public class FileListLimiter {

private static final int MAX_FILES_TO_LIST = 20;

    public static void main(String[] args) {
        final String filesLocation = "resources/";

        Path path = Paths.get(filesLocation);
        try (DirectoryStream<Path> dirStream = Files.newDirectoryStream(path)) {
            int fileCounter = 1;
            for (Path entry : dirStream) {
                System.out.printf("%-5s %2d: %s%n", "entry", fileCounter++, entry.getFileName());
                if (fileCounter > MAX_FILES_TO_LIST) {
                    break;
                }
            }
        } catch (IOException e) {
            // add your exception handling here
            e.printStackTrace(System.err);
        }
    }
}

Java 8的例子

import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Paths;

public class FileListLimiter {

private static final int MAX_FILES_TO_LIST = 20;

    public static void main(String[] args) {
        final String filesLocation = "resources/";

        try {
            Files.walk(Paths.get(filesLocation))
                    .filter(p -> p.toFile().isFile())
                    .limit(MAX_FILES_TO_LIST)
                    .forEach(entry -> {System.out.printf("%-5s: %s%n", "entry", entry.getFileName());});
        } catch (IOException e) {
            // add your exception handling here
            e.printStackTrace(System.err);
        }
    }
}