Question

我正在开发一个服务项目，其UI允许其用户上传文件。我需要编写一个服务，可以将此文件上传到服务器并读取和显示该文件的内容。谁能告诉我怎么做？

//Controller definition begins
@RequestMapping(value = "/uploadFile", method = RequestMethod.POST)
    public @ResponseBody
    String uploadFileHandler(@RequestParam("name") String name,
            @RequestParam("file") MultipartFile file) {

        // Create a new file upload handler
        ServletFileUpload upload = new ServletFileUpload();


        if (!file.isEmpty()) {
            try {
                byte[] bytes = file.getBytes();

                // Creating the directory to store file
                String rootPath = System.getProperty("catalina.home");
                File dir = new File(rootPath + File.separator + "tmpFiles");
                if (!dir.exists())
                    dir.mkdirs();

                // Create the file on server
                File serverFile = new File(dir.getAbsolutePath()
                        + File.separator + name);
                BufferedOutputStream stream = new BufferedOutputStream(
                        new FileOutputStream(serverFile));
                stream.write(bytes);
                stream.close();

                logger.info("Server File Location="
                        + serverFile.getAbsolutePath());
        return "You successfully uploaded file=" + name;    

            } catch (Exception e) {
                return "You failed to upload " + name + " => " + e.getMessage();
            }
        } else {
            return "You failed to upload " + name
                    + " because the file was empty.";
        }
    }

现在我想知道如何显示上传文件的内容。它被转换为FILE格式并存储在tomcat服务器的tempfiles目录中，并且数据是非人类可读的。我需要将其转换回xlsx（正在上传的文件是xlsx）或者能够从中读取数据它直接更新Db.Also我在Spring MVC中使用apache commons-io和文件上传，你可以从上面的代码中看到。

Answer 1

您需要一个能够读取xlsx文件类型的库，例如Apache POI

如果您选择此库，则会有very good example section in the example secion of it's website

Apache-commons-fileupload：如何在spring mvc

1 个答案: