问题

我正在制作一个简单的Spring Boot应用程序，使用Thymeleaf。我有一个控制器用于上传图像，它保存在项目文件夹*uploads*中。请问有人可以解释一下，我如何在上传后立即显示它？我应该使用@PostMapping还是一些Thymeleaf标签？

**我的控制器类**
```java
@Controller
public class UploadFileController {
    public static String uploadDirectory = System.getProperty("user.dir") + "/uploads";

    @RequestMapping("/")
    public String uploadPage(Model model) {
        return "uploadview";
    }

    @RequestMapping("/upload")
    public String upload(Model model, @RequestParam("files")MultipartFile[] files) {
        StringBuilder fileNames = new StringBuilder();

        for(MultipartFile file : files) {
            Path fileNameAndPath = Paths.get(uploadDirectory, file.getOriginalFilename());
            fileNames.append(file.getOriginalFilename());
            try {
                Files.write(fileNameAndPath, file.getBytes());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        model.addAttribute("msg", "Uploaded files: " + fileNames.toString());
        return "uploadstatusview";
    }

更新
我正在添加如何应该显示的屏幕截图。

1. 选择文件
2. 选择的文件，例如test1.png - 上传文件
3. 然后它会将我发送到另一个具有上传端点的HTML模板。到这一步它是可以工作的。图片保存在项目中，但现在我想看到这张图片，就像屏幕截图中展示的那样想要是这样的。

<details>
<summary>英文:</summary>

I&#39;m making a simple spring boot app with Thymeleaf. I have a controller to upload images, it&#39;s saving in project folder *uploads*. Can please someone explain to me how can I display it instantly after uploading it? Should I use @PostMapping or just some Thymeleaf tag? 

**My controller class**

@Controller
public class UploadFileController {
public static String uploadDirectory = System.getProperty("user.dir") + "/uploads";

@RequestMapping("/")
public String uploadPage(Model model) {
    return "uploadview";
}

@RequestMapping("/upload")
public String upload(Model model, @RequestParam("files")MultipartFile[] files) {
    StringBuilder fileNames = new StringBuilder();

    for(MultipartFile file : files) {
        Path fileNameAndPath = Paths.get(uploadDirectory, file.getOriginalFilename());
        fileNames.append(file.getOriginalFilename());
        try {
            Files.write(fileNameAndPath, file.getBytes());
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    model.addAttribute("msg", "Uploaded files: " + fileNames.toString());
    return "uploadstatusview";
}

**UPDATE**
I'm adding screenshots how it should look. 
1. Choose file
2. Chosen file e.g test1.png - Upload file
3. Than its sending me on another html template with upload endpoint. To this point it's working. Picture it's saving in project but now I wanna see this picture like it's on screenshot [ want it to be like that][1]


  [1]: https://i.stack.imgur.com/ylYhf.png

</details>


# 答案1
**得分**: 0

如果我理解正确，您想通过请求获取新上传的图像，位于单独的 URL。您可以在处理上传时将图像存储到列表或队列中，然后通过请求为存储在列表中的每个图像编译模板。

您的模板可能如下所示：

```html
<img src="${imgPath}" />

<img src="${imgPath}" />