我遇到過一種情況，我想跟蹤某個 goroutine 以在特定點上進行同步，例如在獲取所有 url 時。然后，我們可以將它們全部放置并按特定順序顯示。我認為這是障礙進來的。它是在go與sync.WaitGroup. 但是，在實際情況下，我們無法確保所有的 fetch 操作都會在短時間內成功。所以，我想wait在獲取操作時引入超時。我是新手Golang，所以有人可以給我一些建議嗎？我正在尋找的是這樣的： wg := &sync.WaigGroup{} select { case <-wg.Wait(): // All done! case <-time.After(500 * time.Millisecond): // Hit timeout. }我知道Wait不支持Channel。
3 回答
慕桂英4014372
TA貢獻1871條經驗 獲得超13個贊
如果你想要的只是你的簡潔選擇，你可以通過生成一個調用方法并在完成后關閉/發送通道的例程來輕松地將阻塞功能轉換為通道。
done := make(chan struct{})
go func() {
wg.Wait()
close(done)
}()
select {
case <-done:
// All done!
case <-time.After(500 * time.Millisecond):
// Hit timeout.
}
慕尼黑的夜晚無繁華
TA貢獻1864條經驗 獲得超6個贊
將您的結果發送到一個緩沖通道，足以獲取所有結果，而不會阻塞，并在主線程的 for-select 循環中讀取它們：
func work(msg string, d time.Duration, ret chan<- string) {
time.Sleep(d) // Work emulation.
select {
case ret <- msg:
default:
}
}
// ...
const N = 2
ch := make(chan string, N)
go work("printed", 100*time.Millisecond, ch)
go work("not printed", 1000*time.Millisecond, ch)
timeout := time.After(500 * time.Millisecond)
loop:
for received := 0; received < N; received++ {
select {
case msg := <-ch:
fmt.Println(msg)
case <-timeout:
fmt.Println("timeout!")
break loop
}
}
游樂場：http : //play.golang.org/p/PxeEEJo2dz。
慕標5832272
TA貢獻1966條經驗 獲得超4個贊
另一種方法是在內部對其進行監控，您的問題是有限的，但我假設您正在通過循環啟動 goroutines，即使您不是，您也可以重構它來為您工作，但您可以這樣做這兩個示例之一，第一個將超時每個請求單獨超時，第二個將超時整批請求，如果時間過長則繼續
var wg sync.WaitGroup
wg.Add(1)
go func() {
success := make(chan struct{}, 1)
go func() {
// send your request and wait for a response
// pretend response was received
time.Sleep(5 * time.Second)
success <- struct{}{}
// goroutine will close gracefully after return
fmt.Println("Returned Gracefully")
}()
select {
case <-success:
break
case <-time.After(1 * time.Second):
break
}
wg.Done()
// everything should be garbage collected and no longer take up space
}()
wg.Wait()
// do whatever with what you got
fmt.Println("Done")
time.Sleep(10 * time.Second)
fmt.Println("Checking to make sure nothing throws errors after limbo goroutine is done")
或者，如果您只是想要一個通用的簡單方法來超時所有請求，您可以執行以下操作
var wg sync.WaitGroup
waiter := make(chan int)
wg.Add(1)
go func() {
success := make(chan struct{}, 1)
go func() {
// send your request and wait for a response
// pretend response was received
time.Sleep(5 * time.Second)
success <- struct{}{}
// goroutine will close gracefully after return
fmt.Println("Returned Gracefully")
}()
select {
case <-success:
break
case <-time.After(1 * time.Second):
// control the timeouts for each request individually to make sure that wg.Done gets called and will let the goroutine holding the .Wait close
break
}
wg.Done()
// everything should be garbage collected and no longer take up space
}()
completed := false
go func(completed *bool) {
// Unblock with either wait
wg.Wait()
if !*completed {
waiter <- 1
*completed = true
}
fmt.Println("Returned Two")
}(&completed)
go func(completed *bool) {
// wait however long
time.Sleep(time.Second * 5)
if !*completed {
waiter <- 1
*completed = true
}
fmt.Println("Returned One")
}(&completed)
// block until it either times out or .Wait stops blocking
<-waiter
// do whatever with what you got
fmt.Println("Done")
time.Sleep(10 * time.Second)
fmt.Println("Checking to make sure nothing throws errors after limbo goroutine is done")
這樣你的 WaitGroup 將保持同步，你不會有任何 goroutines 處于不確定狀態
http://play.golang.org/p/g0J_qJ1BUT在這里試試你可以改變周圍的變量以查看它的工作方式不同
編輯：我在移動設備上如果有人可以修復格式，那將是非常感謝。
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