Question

我有一组文件，如＆＃34; f-1.txt＆＃34; ，＆＃34; f-2.txt＆＃34;，.....，＆＃34; f-30.txt＆＃34;，＆＃34; g-1.txt＆＃34; ，＆＃34; g-2.txt＆＃34;，.....，＆＃34; g-23.txt＆＃34;，＆＃34; h-1.txt＆＃34; ，＆＃34; h-2.txt＆＃34;，.....，＆＃34; h-35.txt＆＃34; ..等等在一个文件夹中。我想为每个文件夹添加一些内容并将它们重命名为＆＃34; f-1new.txt＆＃34; ，＆＃34; g-2new.txt＆＃34;。我怎样才能在java中引用它们，最好是使用通配符并适当地重命名它们？

对于一个特殊文件，我使用BufferedReader来读取它的内容，然后使用Printwriter将修改后的内容写入一个新的文件名。但是，如果名称变化太大（如果有的话），我怎样才能（迭代地）读取所有文件中的内容如上所述？

我已经改回this但它没有帮助我如何获取数组中每个文件的文件名（帖子中的第一个答案）..

Answer 1

尝试以下方法：

//This method will return files with matching pattern in the specified directory
public File[] getMatchingFiles(String yourDirectoryWithFiles){
    File directoryWithFiles= new File(yourDirectoryWithFiles);

    return directoryWithFiles.listFiles(new FilenameFilter() { 
        public boolean accept(File dir, String filename)
        {     //Make this dynamic with passing the pattern as an argument 
              return filename.endsWith("f.*txt"); 
        }
    } );
}

//Iterate over the files and rename them
public void iterateFiles(String yourDirectoryWithFiles){
    File[] fileList=getMatchingFiles(yourDirectoryWithFiles);

    for(File oldFile:fileList){
        boolean success=createNewFile(oldFile);
        //Case 1 :Deleting the old file if file creation was successful
        if(success)
           oldFile.delete();

        //If using Case 2: return the newFileObject and call: oldFile.renameTo(newFile);
    }
}
public boolean createNewFile(File oldFile){
    //Case 1: create a new file object here and perform your name changing operations 

    //Case 2: If you don't want to create another file , write to the existing file
    //but you would still need to create an file object to perform rename operation
}

Answer 2

这是一个解决方案。

public class Main {

public static void main(String[] args) throws FileNotFoundException, IOException {


    String PATH_2_FOLDER = "path_2_folder";

    //listing all files in the desired folder
    File myDirectory = new File(PATH_2_FOLDER);
    File[] allFiles = myDirectory.listFiles();

    System.out.println(allFiles.length);

    for (int l = 0; l < allFiles.length; l++) {

        if (allFiles[l].getName().endsWith(".txt")) {


            //read the input file
            String thisPathIn = PATH_2_FOLDER+allFiles[l].getName();
            BufferedReader thisBR = new BufferedReader(new FileReader(thisPathIn));


            //create the output file
            String newName = allFiles[l].getName().replace(".txt", "").concat(".new.txt");
            String thisPathOut = PATH_2_FOLDER+newName;
            BufferedWriter thisBW = new BufferedWriter(new FileWriter(thisPathOut));


            //read the contents of the inputfile
            String s = "";
            while((s = thisBR.readLine()) != null){

            //process the content
            //...
            //create new content


            thisBW.write("new_content\n");

            }

            thisBW.flush();
            thisBW.close();

        }

    }

}
}

引用具有相似名称的文件夹，编辑，然后单独重命名

2 个答案: