spring-boot - 在没有模板引擎的情况下有条件地提供静态内容 [英] spring-boot - Conditionally serve static content without template engine

问题描述

我有一个静态页面，我想在特定 URL 上有条件地提供该页面.
使用 spring-boot，我可以将页面放在 static 或 public 资源目录中，并将它们提供给所有人，但是如果我想通过功能标志那么这不合适.
使用模板引擎，我可以将页面作为模板加载并返回对视图的引用.但是，我的应用程序相当简单，我不想在不需要模板引擎的情况下使用模板引擎.
我希望能够使用控制器来确定页面是否被提供的天气.让控制器返回静态页面的最简单方法是什么?

I have a static page that I want to conditionally serve on a particular URL.
With spring-boot I can place pages in the static or public resource directories and have them served to everyone, but if I want to restrict access or disable access to them via a feature flag then this is not suitable.
Using a template engine I can load the page as a template and return a reference to the view. However my application is fairly simple and I don't want to utilize a template engine when I otherwise have no need for one.
I want to be able to use a controller to determine weather the page is served or not. What is the simplest way to have a controller return a static page?

推荐答案

我发现从控制器返回静态内容的最简单方法是返回一个 Resource.Resource 类有多种实现，但是 ClassPathResource 最适合 spring-boot 应用.
在常规 spring-boot 应用程序中，当 mycondition() 为 true 时，以下示例将显示 src/main/resources/path/to/mypage.html代码>，否则返回 404.

The simplest way I have found to return static content from a controller is to return a Resource. There are multiple implementations of the Resource class but ClassPathResource makes the most sense for a spring-boot app.
In a regular spring-boot application the below example will display src/main/resources/path/to/mypage.html when mycondition() is true and return a 404 otherwise.

@RequestMapping("mypage.html")
public Resource myPage() {
    if(mycondition()) {
        return new ClassPathResource("path/to/mypage.html");
    } else {
        throw new ResourceNotFoundException();
    }
}

@ResponseStatus(HttpStatus.NOT_FOUND)
private static class ResourceNotFoundException extends RuntimeException {
    public ResourceNotFoundException() {}
}

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