Java 无需解压直接读取ZIP压缩包里的文件及内容

示例如下：

@Test
    public void test() throws Exception {
        //获取文件输入流
        FileInputStream input = new FileInputStream("D:\\2022-03-28.zip");

        //获取ZIP输入流(一定要指定字符集Charset.forName("GBK")否则会报java.lang.IllegalArgumentException: MALFORMED)
        ZipInputStream zipInputStream = new ZipInputStream(new BufferedInputStream(input), Charset.forName("GBK"));

        //定义ZipEntry置为null,避免由于重复调用zipInputStream.getNextEntry造成的不必要的问题
        ZipEntry ze;
        List<List<Object>> list;
        //循环遍历
        while ((ze = zipInputStream.getNextEntry()) != null) {
            ByteArrayOutputStream baos = new ByteArrayOutputStream();
            if (!ze.isDirectory() && ze.toString().endsWith("xls")) {
                //读取
                byte[] buffer = new byte[1024];
                int len;
                while ((len = zipInputStream.read(buffer)) > -1) {
                    baos.write(buffer, 0, len);
                }
                baos.flush();
                InputStream stream = new ByteArrayInputStream(baos.toByteArray()); //excel 流
                //根据excel输入流读取EXCEL中的数据
                ExcelReader excelReader = ExcelUtil.getReader(stream);
                list =  excelReader.read(2, excelReader.getRowCount());
                for(List<Object> objList : list){
                 objList.get(0);
                 objList.get(1); 
                 //获取到数据进行相关处理
                 ......
                }
            }
        }
        //一定记得关闭流
        zipInputStream.closeEntry();
        input.close();
    }