java 按指定文件夹下的文件的前14个字符串批量打包成zip的代码

以下是可以回答的代码：

import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.zip.ZipEntry;
import java.util.zip.ZipOutputStream;

public class ZipFiles {

    public static void main(String[] args) {
        String folderPath = "指定文件夹路径";
        String zipFilePath = "生成的zip文件路径";

        try {
            FileOutputStream fos = new FileOutputStream(zipFilePath);
            ZipOutputStream zos = new ZipOutputStream(fos);

            File folder = new File(folderPath);
            for (File file : folder.listFiles()) {
                if (file.isFile() && file.getName().length() >= 14) {
                    String fileName = file.getName();
                    String prefix = fileName.substring(0, 14);
                    ZipEntry ze = new ZipEntry(prefix + ".txt");
                    zos.putNextEntry(ze);
                    FileInputStream fis = new FileInputStream(file);

                    byte[] buffer = new byte[1024];
                    int len;
                    while ((len = fis.read(buffer)) > 0) {
                        zos.write(buffer, 0, len);
                    }

                    fis.close();
                }
            }

            zos.closeEntry();
            zos.close();
            fos.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

}