go - 使用 time.AfterFunc 在 golang 中按时间间隔执行重复性任务，只是一个示例

Question

我只想在 Go 中做重复的后台任务，使用 time.AfterFunc，但似乎逻辑有问题。输出只是:间隔调用间隔调用

但如果一切正常，至少要调用该函数 5 次。

package main

import (
    "fmt"
    "time"
    "os"
    "os/signal"
)

type Timer struct {
    Queue chan *TimeCall
}

func NewTimer(l int) *Timer {
    timer := new(Timer)
    timer.Queue = make(chan *TimeCall,l)
    return timer
}

type TimeCall struct {
    timer    *time.Timer
    callback func()
}

func (this *TimeCall) CallBack() {
    defer func() { recover() }()
    if this.callback != nil {
        this.callback()
    }
}

func (this *Timer) AfterFunc(d time.Duration, callback func()) *TimeCall {
    call := new(TimeCall)
    call.callback = callback
    call.timer = time.AfterFunc(d, func() {
        this.Queue <- call
    })
    return call
}



type PipeService struct {
    TimeCall *Timer
}

func (this *PipeService) AfterFunc(delay time.Duration, callback func()) *TimeCall {
    return this.TimeCall.AfterFunc(delay, callback)
}

func (this *PipeService) IntervalCall(interval time.Duration, callback func()) {
    this.TimeCall.AfterFunc(interval,func(){
        if callback != nil {
            callback()
        }
        this.AfterFunc(interval,callback)
    })
}

func (this *PipeService) Run(closeSig chan bool) {
    for {
        select {
        case <-closeSig:
            return
        case call := <-this.TimeCall.Queue:
            call.CallBack()
        }
    }
}

func main() {
    var closeChan chan bool
    InsPipeService := &PipeService{TimeCall: NewTimer(10)}
    InsPipeService.IntervalCall(2*time.Second,func(){
        fmt.Println("interval call")
    })

    c := make(chan os.Signal, 1)
    signal.Notify(c, os.Interrupt, os.Kill)

    go func(){
        InsPipeService.Run(closeChan)
    }()

    time.Sleep(10*time.Second)
}

Run Code

Answer 1

time.AfterFunc()返回 *time.Timer ，引用其文档:

The Timer type represents a single event. When the Timer expires, the current time will be sent on C, unless the Timer was created by AfterFunc.

time.AfterFunc() 返回的time.Timer 不会重复，所以您看到的是完全正常的:在您的 PipeService.IntervalCall() 立即执行回调，它会在超时后执行。

另请注意，您将 2 作为 PipeService.IntervalCall() 方法的间隔传递。此 interval 参数的类型为 time.Duraion .因此，当您传递 2 时，这不会是 2 秒(但实际上是 2 纳秒)。您应该传递一个由 time 包中的常量构造的值，例如:

InsPipeService.IntervalCall(2 * time.Second, func(){
    fmt.Println("interval call")
})

如果你想重复，使用time.Ticker .例如，以下代码每 2 秒打印一条消息:

t := time.NewTicker(2 * time.Second)
for now := range t.C {
    fmt.Println("tick", now)
}

或者如果您不需要 Ticker 并且不想将其关闭:

c := time.Tick(2 * time.Second)
for now := range c {
        fmt.Println("tick", now)
}