您可以使用sync.WaitGroup。引用链接的示例：

package main

import (
        "net/http"
        "sync"
)

func main() {
        var wg sync.WaitGroup
        var urls = []string{
                "http://www.golang.org/",
                "http://www.google.com/",
                "http://www.somestupidname.com/",
        }
        for _, url := range urls {
                // Increment the WaitGroup counter.
                wg.Add(1)
                // Launch a goroutine to fetch the URL.
                go func(url string) {
                        // Decrement the counter when the goroutine completes.
                        defer wg.Done()
                        // Fetch the URL.
                        http.Get(url)
                }(url)
        }
        // Wait for all HTTP fetches to complete.
        wg.Wait()
}

WaitGroups绝对是做这件事的标准方法。不过，为了完整起见，下面是在引入WaitGroups之前常用的解决方案。其基本思想是使用一个通道来表示“我完成了”，并让主goroutine等待，直到每个派生的例程都报告了它的完成。

func main() {
    c := make(chan struct{}) // We don't need any data to be passed, so use an empty struct
    for i := 0; i < 100; i++ {
        go func() {
            doSomething()
            c <- struct{}{} // signal that the routine has completed
        }()
    }

    // Since we spawned 100 routines, receive 100 messages.
    for i := 0; i < 100; i++ {
        <- c
    }
}

同步。WaitGroup可以在此为您提供帮助。

package main

import (
    "fmt"
    "sync"
    "time"
)

func wait(seconds int, wg * sync.WaitGroup) {
    defer wg.Done()

    time.Sleep(time.Duration(seconds) * time.Second)
    fmt.Println("Slept ", seconds, " seconds ..")
}

func main() {
    var wg sync.WaitGroup

    for i := 0; i <= 5; i++ {
        wg.Add(1)   
        go wait(i, &wg)
    }
    wg.Wait()
}

虽然sync.waitGroup（wg）是规范的前进方向，但它确实要求您在wg.Wait之前至少执行一些wg.Add调用，以便完成所有调用。这对于像web crawler这样简单的东西来说可能不可行，因为您事先不知道递归调用的数量，并且需要一段时间来检索驱动wg.Add调用的数据。毕竟，在知道第一批子页面的大小之前，您需要加载和解析第一个页面。
我写了一个使用通道的解决方案，在我的解决方案Tour of Go - web crawler练习中避免了waitGroup。每次启动一个或多个go-routines时，您将数字发送到children通道。每次一个go例程即将完成时，您就向done通道发送一个1。当孩子的总和等于完成的总和时，我们就完成了。
我唯一关心的是results通道的硬编码大小，但这是（当前）Go的限制。

// recursionController is a data structure with three channels to control our Crawl recursion.
// Tried to use sync.waitGroup in a previous version, but I was unhappy with the mandatory sleep.
// The idea is to have three channels, counting the outstanding calls (children), completed calls 
// (done) and results (results).  Once outstanding calls == completed calls we are done (if you are
// sufficiently careful to signal any new children before closing your current one, as you may be the last one).
//
type recursionController struct {
    results  chan string
    children chan int
    done     chan int
}

// instead of instantiating one instance, as we did above, use a more idiomatic Go solution
func NewRecursionController() recursionController {
    // we buffer results to 1000, so we cannot crawl more pages than that.  
    return recursionController{make(chan string, 1000), make(chan int), make(chan int)}
}

// recursionController.Add: convenience function to add children to controller (similar to waitGroup)
func (rc recursionController) Add(children int) {
    rc.children <- children
}

// recursionController.Done: convenience function to remove a child from controller (similar to waitGroup)
func (rc recursionController) Done() {
    rc.done <- 1
}

// recursionController.Wait will wait until all children are done
func (rc recursionController) Wait() {
    fmt.Println("Controller waiting...")
    var children, done int
    for {
        select {
        case childrenDelta := <-rc.children:
            children += childrenDelta
            // fmt.Printf("children found %v total %v\n", childrenDelta, children)
        case <-rc.done:
            done += 1
            // fmt.Println("done found", done)
        default:
            if done > 0 && children == done {
                fmt.Printf("Controller exiting, done = %v, children =  %v\n", done, children)
                close(rc.results)
                return
            }
        }
    }
}

Full source code for the solution

下面是一个使用WaitGroup的解决方案。
首先，定义2个实用程序方法：

package util

import (
    "sync"
)

var allNodesWaitGroup sync.WaitGroup

func GoNode(f func()) {
    allNodesWaitGroup.Add(1)
    go func() {
        defer allNodesWaitGroup.Done()
        f()
    }()
}

func WaitForAllNodes() {
    allNodesWaitGroup.Wait()
}

然后，替换callback的调用：

go callback(fileName)

通过调用你的工具函数：

util.GoNode(func() { callback(fileName) })

最后一步，将这一行添加到main的末尾，而不是sleep。这将确保主线程正在等待所有例程完成，然后程序才能停止。

func main() {
  // ...
  util.WaitForAllNodes()
}