如何限制每分鐘 20 個(gè)請求？import "golang.org/x/time/rate"limiter := rate.NewLimiter(rate.Every(1*time.Minute), 20)for { limiter.Wait() //more code}這是行不通的。它的作用是，它允許前 20 個(gè)請求，然后每分鐘只允許 1 個(gè)請求。預(yù)期的是第一分鐘有 20 個(gè)請求（不需要像每 3 秒 1 個(gè)一樣均勻分布），然后在第二分鐘有 20 個(gè)請求。在任何 1 分鐘的時(shí)間間隔內(nèi)，發(fā)送的請求不能超過 20 個(gè)。
4 回答
天涯盡頭無女友
TA貢獻(xiàn)1831條經(jīng)驗(yàn) 獲得超9個(gè)贊
設(shè)置您期望的費(fèi)率：
limiter := rate.NewLimiter(rate.Every(1*time.Minute/20), 20)
for {
limiter.Wait()
//more code
}
操場：https ://go.dev/play/p/ZpxpHj0vK7P
您似乎在尋找允許“最多 20 次爆發(fā)，每分鐘重置一次”的東西。這是一個(gè)嘗試：
type Limiter struct {
maxCount int
count int
ticker *time.Ticker
ch chan struct{}
}
func (l *Limiter) run() {
for {
// if counter has reached 0: block until next tick
if l.count <= 0 {
<-l.ticker.C
l.count = l.maxCount
}
// otherwise:
// decrement 'count' each time a message is sent on channel,
// reset 'count' to 'maxCount' when ticker says so
select {
case l.ch <- struct{}{}:
l.count--
case <-l.ticker.C:
l.count = l.maxCount
}
}
}
func (l *Limiter) Wait() {
<-l.ch
}
func NewLimiter(d time.Duration, count int) *Limiter {
l := &Limiter{
maxCount: count,
count: count,
ticker: time.NewTicker(d),
ch: make(chan struct{}),
}
go l.run()
return l
}
https://go.dev/play/p/5WiOJL5nqCy
慕標(biāo)琳琳
TA貢獻(xiàn)1830條經(jīng)驗(yàn) 獲得超9個(gè)贊
我的解決方案
import (
"sync"
"time"
)
type Limiter interface {
Wait()
}
type limiter struct {
tick time.Duration
count uint
entries []time.Time
index uint
mutex sync.Mutex
}
func NewLimiter(tick time.Duration, count uint) Limiter {
l := limiter{
tick: tick,
count: count,
index: 0,
}
l.entries = make([]time.Time, count)
before := time.Now().Add(-2 * tick)
for i, _ := range l.entries {
l.entries[i] = before
}
return &l
}
func (l *limiter) Wait() {
l.mutex.Lock()
defer l.mutex.Unlock()
last := &l.entries[l.index]
next := last.Add(l.tick)
now := time.Now()
if now.Before(next) {
time.Sleep(next.Sub(now))
}
*last = time.Now()
l.index = l.index + 1
if l.index == l.count {
l.index = 0
}
}
慕森卡
TA貢獻(xiàn)1806條經(jīng)驗(yàn) 獲得超8個(gè)贊
試試這個(gè)
import (
"fmt"
"time"
)
func main() {
limiter := time.Tick(3 * time.Minute) //changed from 20 to 3 because i realized you wanted 20 requests per minute, not 1 request per 20 minutes
for true {
<-limiter
fmt.Println(time.Now())
}
}
我實(shí)際上并沒有嘗試過20 * time.Minute作為價(jià)值，但我嘗試過200 * time.Milisecond并且它有效。這是我的回應(yīng)
2022-05-31 00:06:13.447108 -0600 MDT m=+0.200889915
2022-05-31 00:06:13.651373 -0600 MDT m=+0.405148283
2022-05-31 00:06:13.851522 -0600 MDT m=+0.605291066
2022-05-31 00:06:14.051481 -0600 MDT m=+0.805244205
2022-05-31 00:06:14.250144 -0600 MDT m=+1.003900790
2022-05-31 00:06:14.450952 -0600 MDT m=+1.204703429
2022-05-31 00:06:14.648365 -0600 MDT m=+1.402110595
2022-05-31 00:06:14.848223 -0600 MDT m=+1.601961982
2022-05-31 00:06:15.04909 -0600 MDT m=+1.802823232
2022-05-31 00:06:15.250164 -0600 MDT m=+2.003891217
精慕HU
TA貢獻(xiàn)1845條經(jīng)驗(yàn) 獲得超8個(gè)贊
你可以像time.Tick這樣使用for-select：
package main
import (
"fmt"
"time"
)
func main() {
max := 20
limiter := time.Tick(1 * time.Minute)
exit := make(chan struct{})
go func() {
count := 0
exit2 := 0
defer func() {
exit <- struct{}{}
}()
for {
select {
case <-limiter:
count = 0
fmt.Println("exit2: ", exit2)
if exit2 == 3 {
return
}
exit2++
default:
if count == max {
continue
}
fmt.Println("accepting request", count)
count++
}
}
}()
<-exit
}
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