如何将InputStream转换为FilePart？

import org.apache.commons.fileupload.FileItem;
import org.apache.commons.fileupload.disk.DiskFileItem;
import org.apache.commons.fileupload.FileItemFactory;
import org.apache.commons.fileupload.servlet.ServletFileUpload;

@WebServlet("/upload")
public class UploadServlet extends HttpServlet {
    protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        // 检查请求是否为文件上传类型
        boolean isMultipart = ServletFileUpload.isMultipartContent(request);
        if (isMultipart) {
            // 创建FileItemFactory对象
            FileItemFactory factory = new DiskFileItemFactory();

            // 创建ServletFileUpload对象
            ServletFileUpload upload = new ServletFileUpload(factory);

            try {
                // 解析请求，获取所有的FileItem对象
                List<FileItem> items = upload.parseRequest(request);

                // 遍历FileItem对象
                for (FileItem item : items) {
                    // 判断当前FileItem是否为文件类型
                    if (!item.isFormField()) {
                        // 获取文件名
                        String fileName = item.getName();

                        // 创建临时文件
                        File tempFile = File.createTempFile("temp", null);

                        // 将InputStream写入临时文件
                        item.write(tempFile);

                        // 创建FilePart对象
                        FilePart filePart = new FilePart(fileName, tempFile);

                        // 使用FilePart对象进行后续处理
                        // ...
                    }
                }
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
    }
}