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希望增加个named函数 #879
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Salvo 中的 Router 本身是嵌套的,路径只是 Router 中的一个 Filter,所以,是给 Router 命名还是给 Router 中的路径命名?如果Router 不包含 Path 的 Filter 呢?如果多个 Router 存在相同的路径,是都起同样的名字? 会不会很怪异?会不会路径不同又有相同名字?如何检测冲突? 感觉不好实现,也没太大的意义。 |
很有意义的,对于微服务还好,路由少,如果是一个相对复杂的项目,比如一个电商吧,涉及的路由就很多了,后端经常需要生成url,返还给前端,这个时候,去拼凑uri就很麻烦了, 很多类似的框架都有这种功能,其实有个简单的办法,就是PathFilter里的字段给pub,或者将PathParser给设置成pub,这样就可以重新parse,类似: #[handler]
// https://0.0.0.0/edit/<user_id>/<article_id>?ref=test_user&ref_code=abc123
async fn handle_edit(req: &mut Request, res: &mut Response) {
// before url https://0.0.0.0/edit/10/5?ref=test_user&ref_code=abc123
let params = HashMap::from([("user_id", "1"), ("article_id","100")]);
let queries = HashMap::from([("ref", "admin"), ("ref_code","11111111")]);
let url = req.make_uri(req, params, queries);
// or let url = req.make_uri(“/edit/<user_id>/<article_id>”, params, queries);
// after url = https://0.0.0.0/edit/1/100?ref=admin&ref_code=11111111
res.render(Text::Json(
json!({
"code": 200,
"success": true,
"payload": {
"redirect_url": url
}
})
.to_string(),
))
} |
感觉不如用oapi |
可以给router加个named函数嘛,类似
Router::new().path("edit/id:num").post(edit).named("user_edit")
这样,在后续的handler中可以获取到,并且做uri赋值的时候,方便生成uri
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