同步示例:
type job struct {
Id int
Message string
}
for {
// getJob() blocks until job is received
job := getJob()
doSomethingWithJob(job)
}
我希望处理来自getJob
doSomethingWithJob
的工作。例如getJob可以是从MessagingQueue接收的有效负载,例如RabbitMQ / Beanstalkd或处理HTTP请求。
我不想阻止getJob
,而我doSomethingWithJob
&反之亦然。但是我想控制/缓冲作业的数量,这样我就不会使系统过载。例如最大并发度为5.
go例程的概念目前让我很困惑,所以任何正确方向的指针都会非常感激,以帮助我学习。
更新:感谢@JimB的帮助。为什么工人5总是接受这份工作?
jobCh := make(chan *job)
// Max 5 Workers
for i := 0; i < 5; i++ {
go func() {
for job := range jobCh {
time.Sleep(time.Second * time.Duration(rand.Intn(3)))
log.Println(i, string(job.Message))
}
}()
}
for {
job, err := getJob()
if err != nil {
log.Println("Closing Channel")
close(jobCh)
break
}
jobCh <- job
}
log.Println("Complete")
示例输出
2016/06/09 22:19:57 5 {"id":10692,"name":"Test Message"}
2016/06/09 22:19:57 5 {"id":10687,"name":"Test Message"}
2016/06/09 22:19:57 5 {"id":10699,"name":"Test Message"}
2016/06/09 22:19:57 5 {"id":10701,"name":"Test Message"}
2016/06/09 22:19:57 5 {"id":10703,"name":"Test Message"}
2016/06/09 22:19:57 5 {"id":10704,"name":"Test Message"}
答案 0 :(得分:4)
您可以从频道开始播放5个goroutines来呼叫doSomethingWithJob
。这样,同时处理的作业永远不会超过5个。
jobCh := make(chan *job)
// start 5 workers to process jobs
for i := 0; i < 5; i++ {
go func() {
for job := range jobCh {
doSomethingWithJob(job)
}
}()
}
// send jobs to workers as fast as we can
for {
jobCh <- getJob()
}