英文:
How do I print the url path without parameters?
问题
我正在使用Gorilla/mux创建一个RESTful API，它在路径中使用参数。我创建的代码如下：
r := mux.NewRouter()
subr := r.PathPrefix("/v1").Subrouter()
subr.Handle("/organizations/{orgId}/projects/{projectId}", CreateProject()).Methods(http.MethodPost)
然而，当捕获请求并记录结果时，我不想记录如下内容：
/organizations/fff-555-aaa9999/projects/amazing-project
这是从解析r *http.Request.URL值得到的结果。
相反，我想要获取初始路由器值/organizations/{orgId}/projects/{projectId}，这样我就可以在日志中从那里进行过滤。
使用gorilla/mux，有没有一种方法可以获取路由路径而不包含实际参数，而是参数变量呢？
英文:
I have a RESTful API using Gorilla/mux. It is parameterized in the path. I am creating it as such:
r := mux.NewRouter()
subr := r.PathPrefix("/v1").Subrouter()
subr.Handle("/organizations/{orgId}/projects/{projectId}", CreateProject()).Methods(http.MethodPost)
However, when capturing the request and logging the outcome, I don't want to log i.e.
/organizations/fff-555-aaa9999/projects/amazing-project
which is what I'd get from parsing the r *http.Request.URL value.
Instead, I'd like to get the initial router value of /organizations/{orgId}/projects/{projectId} so I can filter from that in my logging.
Using gorilla/mux, is there a way to get the router path without the actual parameters, but the parameter variables instead?
答案1
得分: 3
在CreateProject函数内部，你可以使用以下代码：
func CreateProject(w http.ResponseWriter, r *http.Request) {
path, _ := mux.CurrentRoute(r).GetPathTemplate()
fmt.Println(path)
}
英文:
Inside CreateProject function you coud use this:
func CreateProject(w http.ResponseWriter, r *http.Request) {
path, _ := mux.CurrentRoute(r).GetPathTemplate()
fmt.Println(path)
}
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