該代碼向主機發送并行請求，如果我發送有效的主機，該程序工作正常，但是如果我在執行10次后發送無效主機，則該程序將停止工作。但是這個程序已經在結構中返回錯誤。和main函數處理不正確，我無法找出為什么這個程序在返回10個無效結果后停止。它不會返回 100 個結果。import ( "fmt" "net/http" "sort" "time" )// a struct to hold the result from each request including an index// which will be used for sorting the results after they come intype result struct { index int res http.Response err error }// boundedParallelGet sends requests in parallel but only up to a certain// limit, and furthermore it's only parallel up to the amount of CPUs but// is always concurrent up to the concurrency limitfunc boundedParallelGet(urls []string, concurrencyLimit int) []result { // this buffered channel will block at the concurrency limit semaphoreChan := make(chan struct{}, concurrencyLimit) // this channel will not block and collect the http request results resultsChan := make(chan *result) // make sure we close these channels when we're done with them defer func() { close(semaphoreChan) close(resultsChan) }() // keen an index and loop through every url we will send a request to for i, url := range urls { // start a go routine with the index and url in a closure go func(i int, url string) { // this sends an empty struct into the semaphoreChan which // is basically saying add one to the limit, but when the // limit has been reached block until there is room semaphoreChan <- struct{}{} // send the request and put the response in a result struct // along with the index so we can sort them later along with // any error that might have occoured res, err := http.Get(url) if err != nil { fmt.Println(err) return }
1 回答
慕蓋茨4494581
TA貢獻1850條經驗 獲得超11個贊
當 觸發錯誤時，您將立即返回，而不會從通道中刪除阻塞項。Get
您可以在此處刪除阻止項，也可以刪除并實際在結構中存儲指向 的指針（這樣就可以從失敗的請求中存儲）。returnhttp.Responsenil
下面是使用第二種方法的 for 循環的代碼，請注意，狀態的打印現在處于一個中，因為它可能為 nil。else
// a struct to hold a pointer to the result from each request
// including an index which will be used for sorting the
// results after they come in
type result struct {
index int
res *http.Response
err error
}
// ...
// keen an index and loop through every url we will send a request to
for i, url := range urls {
// start a go routine with the index and url in a closure
go func(i int, url string) {
// this sends an empty struct into the semaphoreChan which
// is basically saying add one to the limit, but when the
// limit has been reached block until there is room
semaphoreChan <- struct{}{}
// send the request and put the response in a result struct
// along with the index so we can sort them later along with
// any error that might have occoured
res, err := http.Get(url)
if err != nil {
fmt.Println(err)
} else {
fmt.Println(res.Status)
}
result := &result{i, res, err}
// now we can send the result struct through the resultsChan
resultsChan <- result
// once we're done it's we read from the semaphoreChan which
// has the effect of removing one from the limit and allowing
// another goroutine to start
<-semaphoreChan
}(i, url)
}
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