如何限制http.HandleFunc并发数量?

leenzhu · · 4252 次点击
楼主的那个效果只能改源码
#12
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``` package main import ( "fmt" "net/http" "net/http/httptest" "time" "golang.org/x/time/rate" ) type LimiterOption struct { lm *rate.Limiter } func WithLimiter(duration time.Duration, count int) func(o *LimiterOption) { return func(o *LimiterOption) { o.lm = rate.NewLimiter(rate.Every(duration), count) } } // LimiterWrap 每个 handler 单独限制 func LimiterWrap(f http.HandlerFunc, opts ...func(o *LimiterOption)) http.HandlerFunc { o := &LimiterOption{ lm: rate.NewLimiter(rate.Every(100*time.Millisecond), 10), } for _, opt := range opts { opt(o) } return func(w http.ResponseWriter, r *http.Request) { if !o.lm.Allow() { w.WriteHeader(http.StatusInternalServerError) return } f(w, r) } } func xHandle(w http.ResponseWriter, r *http.Request) { w.Write([]byte("Hello\n")) } func main() { // 单独控制一个接口 http.HandleFunc("/", LimiterWrap(xHandle, WithLimiter(100*time.Millisecond, 1))) // 所有接口都走同一个限流 // http.ListenAndServe(":8080", LimiterWrap(http.DefaultServeMux.ServeHTTP)) for i := 0; i < 10; i++ { w := httptest.NewRecorder() req, _ := http.NewRequest(http.MethodGet, "/", nil) http.DefaultServeMux.ServeHTTP(w, req) fmt.Println(i, w.Code) time.Sleep(time.Millisecond * 20) } //output //0 200 //1 500 //2 500 //3 500 //4 500 //5 200 //6 500 //7 500 //8 500 //9 500 } ```
#1
```go package main func success() {} func fail() {} func main() { c := make(chan struct{}, 16) select { case c <- struct{}{}: success() <-c default: fail() } } ```
#2