Question

import java.io.*;

public class Main {

    public static void main(String[] args) {


        // change file names in 'Directory':
        String absolutePath = "/storage/emulated/0/Gadm";
        File dir = new File(absolutePath);
        File[] filesInDir = dir.listFiles();
        int i = 0;



        for(File file:filesInDir) {

            i++;



            String[] iso = {
                "AFG",
                "XAD",
                "ALA",
                "ZWE"};


            String[] country = {
                "Afghanistan",
                "Akrotiri and Dhekelia",
                "Åland",
                "Zimbabwe"};


        String name = file.getName();
        String newName = name.replace(iso[i],country[i]);



        String newPath = absolutePath + "/" + newName;
        file.renameTo(new File(newPath));

          System.out.println(name + " has been changed to " + newName);
        }


    }
}

我有一个名为Gadm的目录，它包含一个文件列表，其名称如下，并带有国家/地区的iso代码，例如iso.kmz。 }}

存储在数组中的iso名称以及国家/地区名称和正确的顺序。

我在上面尝试了此代码，但没有用

Answer 1

我将使用单个HashMap而不是使用两个数组，其中的键是国家/地区ISO代码，值是关联的国家/地区名称。喜欢：

String absolutePath = "/storage/emulated/0/Gadm/";
HashMap<String, String> countryCodes = new HashMap<>();
countryCodes.put("AFG","Afghanistan");
countryCodes.put("XAD","Akrotiri and Dhekelia");
countryCodes.put("ALA","Åland");
countryCodes.put("ZWE","Zimbabwe");

for(Map.Entry<String, String> entry : countryCodes.entrySet()) {
    File file = new File(absolutePath + entry.getKey());
    if (file.renameTo(new File(absolutePath + entry.getValue()))) {
        System.out.println("Successfully renamed " + entry.getKey() + " to " + entry.getValue());
    } else {
        System.out.println("Failed to rename " + entry.getKey() + " to " + entry.getValue() +
                ". Please make sure filepath exists: " + absolutePath + entry.getKey());
    }
}

Answer 2

作为替代方案，您可以使用Path的{{1}} istead：

File

客户代码为：public static void rename(Path source) throws IOException { Map<String, String> countries = countries.get(); Files.list(source) .filter(path -> Files.isRegularFile(path)) .filter(path -> countries.containsKey(getFileName.apply(path))) .forEach(path -> { try { Files.move(path, source.resolve(countries.get(getFileName.apply(path)) + getFileExt.apply(path))); } catch(IOException e) { e.printStackTrace(); } }); } private static final Function<Path, String> getFileName = path -> { String fileName = path.getFileName().toString(); return fileName.substring(0, fileName.lastIndexOf('.')).toUpperCase(); }; private static final Function<Path, String> getFileExt = path -> { String fileName = path.getFileName().toString(); return fileName.substring(fileName.lastIndexOf('.')); }; private static Supplier<Map<String, String>> countries = () -> { Map<String, String> map = new HashMap<>(); map.put("AFG", "Afghanistan"); map.put("XAD", "Akrotiri and Dhekelia"); map.put("ALA", "Åland"); map.put("ZWE", "Zimbabwe"); return Collections.unmodifiableMap(map); };

使用Java将文件名列表重命名为另一个列表的文件名

2 个答案: