how do you pass the client's IP into the method of publishing?

[fromiuan]

如何将将客户端的ip传入发布的方法中，

example

package main

import (
	"github.com/hprose/hprose-golang/rpc"
	"net/http"
)

type Member struct{}

type Login struct {
	Username string
	Password string
}

type LoginRsp struct {
	Status bool
	Token  string
}

func (m *Member) Login(req *Login, rsp *LoginRsp, context *rpc.HTTPContext) error {
	// How to get the client IP?
	return nil
}

func (m *Member) Regiest(req *Login, rsp *LoginRsp) error {
	// How to get the client IP?
	return nil
}

func main() {
	service := rpc.NewHTTPService()
	service.AddNetRPCMethods(new(Member))
	http.ListenAndServe(":8080", service)
}

[andot]

func (m *Member) Login(req *Login,rsp *LoginRsp, context *rpc.HTTPContext) error{
        ip := context.Request.RemoteAddr
	return nil
}

最后的参数，客户端调用时，不需要传入，服务器会自动传入。

[fromiuan]

When I add the context *rpc.HTTPContext, that Login function is not released
当我增加context *rpc.HTTPContext的时候，login方法没有被发布，但是我使用AddFunction发布的可以，结果如下

Fa3{u#s7"Regiest"s12"member_login"}z

package main

import (
	"fmt"
	"github.com/hprose/hprose-golang/rpc"
	"net/http"
)

type Member struct{}

type Login struct {
	Username string
	Password string
}

type LoginRsp struct {
	Status bool
	Token  string
}

func (m *Member) Login(req *Login, rsp *LoginRsp, context *rpc.HTTPContext) error {
	// How to get the client IP?
	return nil
}

func (m *Member) Regiest(req *Login, rsp *LoginRsp) error {
	// How to get the client IP?
	return nil
}

func MemberLogin(req *Login, context *rpc.HTTPContext) (rsp *LoginRsp) {
	fmt.Println(context.Request.RemoteAddr)
	return &LoginRsp{
		Status: true,
		Token:  "token",
	}
}

func main() {
	service := rpc.NewHTTPService()
	service.AddNetRPCMethods(new(Member))
	service.AddFunction("member_login", MemberLogin)
	http.ListenAndServe(":8080", service)
}

[andot]

你应该使用 AddInstanceMethods 或 AddAllMethods 这两个方法。AddNetRPCMethods 这个方法是用来发布为 net/rpc 编写的 RPC 服务的。你看一下文档就知道了。你用错了方法。

[fromiuan]

think小马哥耐心解答，点赞。