英文:
golang gin: router like django?
问题
gin能像django一样描述路由吗？
在所有的例子中，路由器都在一个地方，从未涉及附件。
我想在包中描述路由，然后在主文件中只需简单地写一些东西，例如：
r := gin.New()
r.Include("/main", here_imported_route.Route)
here_imported_route.go
package here_imported_route
Router := gin.New()
Router.Use(midl())
Router.Get("/test", hello)
然后在"/main/test"上我们得到"hello"。
英文:
Can gin describe route like django?
In all examples, the routers are in one place, never found about attachment.
I would like to describe the routes in the package, and in the main file is simply to write something like.
example:
r := gin.New()
r.Include("/main", here_imported_route.Route)
#here_imported_route.go
package here_imported_route
Router := gin.New()
Router.Use(midl())
Router.Get("/test", hello)
and then on "/main/test" we get "hello".
1: https://github.com/gin-gonic/gin
答案1
得分: 2
在主路由中，像这样：
package main
import (
"path_to_pkg/pkg"
"github.com/gin-gonic/gin"
)
var r *gin.Engine
func init() {
r = gin.New()
pkg.Concon(r.Group("/pkg"))
}
func main() {
r.Run(":8080")
}
在导入的包中创建连接函数：
pkg.go
package pkg
import "github.com/gin-gonic/gin"
func Concon(g *gin.RouterGroup) {
g.GET("/ping", func(c *gin.Context) {
c.String(200, "pong")
})
}
打开 127.0.0.1:8080/pkg/ping 并获取 "pong"。
英文:
in main route like here
package main
import (
"path_to_pkg/pkg"
"github.com/gin-gonic/gin"
)
var r *gin.Engine
func init() {
r = gin.New()
pkg.Concon(r.Group("/pkg"))
}
func main() {
r.Run(":8080")
}
in imported package create concatenation func
pkg.go
package pkg
import "github.com/gin-gonic/gin"
func Concon(g *gin.RouterGroup) {
g.GET("/ping", func(c *gin.Context) {
c.String(200, "pong")
})
}
open 127.0.0.1:8080/pkg/ping and get "pong"
答案2
得分: 1
如果我正确理解你的问题，我认为你可以通过路由分组来实现这个目标。你可以像这样设置：
r := gin.New()
main := r.Group("/main")
{
main.GET("/test", hello)
}
在这里可以查看更多详细信息：链接。
英文:
If I understand your question correctly, I think you can accomplish this with route grouping. So you would have something like this:
r := gin.New()
main := r.Group("/main")
{
main.GET("/test", hello)
}
See more details here.
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