我如何实现这种情况:
1.我有LoginHandler接收一些用户数据 - email和signedXml:
func LoginHandler(c *gin.Context) {
var (
err error
data LoginPost
)
if err = c.BindJSON(&data); err != nil {
c.JSON(http.StatusBadRequest, gin.H{"status": "error"})
return
}
...
c.JSON(http.StatusOK, gin.H{"status": "ok"})
}
2.我需要通过websocket
将signedXml发送到另一台服务器3.保存结果(成功或错误)
4.关闭连接
每个HTTP请求都会打开连接,发送1条消息,得到1条结果,最后关闭套接字。我在尝试使用频道,但没有成功。这有可能实现我的案例吗?
更新
package main
import (
"log"
"net/url"
"github.com/gorilla/mux"
"github.com/gorilla/websocket"
"net/http"
)
func indexHandler(w http.ResponseWriter, r *http.Request) {
message := r.FormValue("message")
w.Write([]byte(message))
}
func postHandler(w http.ResponseWriter, r *http.Request) {
var (
message = r.FormValue("message")
u = url.URL{Scheme: "ws", Host: "echo.websocket.org", Path: "/"}
err error
out []byte
conn *websocket.Conn
)
log.Printf("message: %s\n", message)
log.Printf("connecting to %s\n", u.String())
conn, _, err = websocket.DefaultDialer.Dial(u.String(), nil)
if err != nil {
log.Fatal("dial:", err)
}
log.Println("write")
if err = conn.WriteMessage(websocket.TextMessage, []byte(message)); err != nil {
log.Fatal("write:", err)
}
log.Println("read")
if _, out, err = conn.ReadMessage(); err != nil {
log.Fatal("read:", err)
}
w.Write(out)
log.Println("close")
conn.Close()
}
func main() {
r := mux.NewRouter()
r.HandleFunc("/", indexHandler).Methods("GET")
r.HandleFunc("/post", postHandler).Methods("POST")
http.Handle("/", r)
http.ListenAndServe(":8080", nil)
}
答案 0 :(得分:1)
依次致电Dial,WriteMessage,ReadMessage和Close。
c, _, err := websocket.DefaultDialer.Dial(url, nil)
if err != nil {
// handle error
}
err := c.WriteMessage(websocket.TextMessage, signedXML)
if err != nil {
// handle error
}
_, p, err := c.ReadMessage()
if err != nil {
// handle error
}
c.Close()
// p is a []byte with the first received message.