51 lines
905 B
Go
51 lines
905 B
Go
package main
|
||
|
||
import (
|
||
"fmt"
|
||
"math/rand"
|
||
"time"
|
||
)
|
||
|
||
// 有两个服务会产生数据
|
||
// 希望有数据就处理
|
||
// 希望不要交替执行，谁有就处理谁
|
||
|
||
// 方案1: 起一个函数，这两个channel有数据，就把这个数据给到另外一个channel，然后读取另外一个channel
|
||
func main() {
|
||
m1, m2 := genMsg("service1"), genMsg("service2")
|
||
m := merge(m1, m2)
|
||
for{
|
||
fmt.Println(<-m)
|
||
}
|
||
}
|
||
|
||
func genMsg(name string) chan string {
|
||
c := make(chan string)
|
||
go func() {
|
||
i := 0
|
||
for {
|
||
time.Sleep(time.Duration(rand.Intn(2000)) * time.Millisecond)
|
||
c <- fmt.Sprintf("the service name is %v,the message is %v", name, i)
|
||
i++
|
||
}
|
||
}()
|
||
return c
|
||
}
|
||
|
||
// 将两个channel里的数据送到另外一个channel里
|
||
func merge(c1, c2 chan string) chan string {
|
||
c := make(chan string)
|
||
go func() {
|
||
for {
|
||
c <- <-c1
|
||
}
|
||
}()
|
||
|
||
go func() {
|
||
for {
|
||
c <- <-c2
|
||
}
|
||
}()
|
||
return c
|
||
}
|