g.Meta Add middleware to a single address when using canonical routing

[yayoyanyue]

Is your feature request related to a problem?

Option Yes

Describe the solution you'd like

规范路由请求：

type CreateTaskReq struct {
	g.Meta    `path:"/create" method:"post" tags:"任务" summary:"新建任务"`
	xxx
}

主函数：

func main() {
	s := g.Server()
	s.Use(
		ghttp.MiddlewareCORS,
		ghttp.MiddlewareHandlerResponse,
		middleware.MiddlewareLog,
	)
	s.Group("/task", func(group *ghttp.RouterGroup) {
		group.Bind(
			task.NewV1(),
		)
	})
	s.Run()
}

Describe alternatives you've considered

需求：
我想给 /task 这个地址下 /task/create 这个路由添加一个鉴权的中间件
希望能在 g.Meta 中有一个 middleware 中间件的数组 就很方便了
获取有其他的解决办法

Additional

No response

[gqcn]

其实有好办法，在元数据中加个标识，通过统一的中间件读取这个标识做判断。我搞个example和文档。

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In fact, there is a good way to add an identifier to the metadata and use unified middleware to read this identifier to make judgments. I'll make an example and documentation.