我有一个简单的任务，已经使用 pubsub 在 javascript 中完成了，这里是任务:

我有对象让说 A 和另外 2 个对某些元素感兴趣的对象(在这种情况下为字符串)，让说 Foo 对元素 m 感兴趣, n 和 Bar 对元素 n, o, p 感兴趣。兴趣可以相交。

A 对象有添加/删除元素的方法，当该对象包含 Foo 感兴趣的 m, n 元素时，该对​​象存储在 Foo 这是使用 pubsub 的 javascript 伪代码

var A = {};

var Foo = {
    interests: ['m', 'n'],
    storedObj: {},
    tempObj: {}
};

// Bar same as Foo with different interest ['n', 'o', 'p']

// somewhere in Foo and Bar constructor
// Foo and Bar subscribe too each interests element
// for each interests when add
subscribe('add'+interest, function(obj) {
    // store this obj in tempObj and increment until satisfy all 
    // interest
    tempObj[obj]++;

    // if this obj satisfy all interest then store it in array of obj
    if(tempObj[obj] === len(interests)) {
        storedObj[obj] = true;
    }
});

// for each interests when remove
subscribe('remove'+interest, function(obj) {
    // remove from storedObj
    delete storedObj[obj];

    // decrement tempObj so it can be used for later if the interest 
    // is adding again
    tempObj[obj]--;
});

// inside A prototype
prototype.add = function(interest) {
    publish('add'+interest, this);
    return this;
}
prototype.remove = function(interest) {
    publish('remove'+interest, this);
    return this;
}

// implementation
A.add('m')
 .add('n')
 .add('o')

// then A is stored inside Foo but not in Bar because A doesn't have 
// `p`, but it still stored Bar.tempObj and have value 2 and waiting 
// for `p` to be add

A.remove('m')
 .add('p')

// then A is removed from Foo and stored in Bar

我想将此任务移植到 golang 中，但我不想使用 pubsub，我想要更符合 golang 习惯的方式。注意:我也已经在 golang 中使用 pubsub。

你能告诉我如何用 golang 做吗？我正在考虑使用 channel ，但找不到解决方案。

最佳答案

只是给你一个想法，不一定是你的真实用例。

package main

import (
    "fmt"
    "time"
)

type Publisher struct {
    subscription map[string]chan string
}

func (p *Publisher)Subscribe(interest string) chan string{
    if p.subscription == nil {
        p.subscription = make(map[string]chan string)
    }
    p.subscription[interest] = make(chan string)
    return p.subscription[interest]
}

func (p *Publisher) Add(val string) {
    if p.subscription[val] != nil {
        fmt.Println("Adding " + val)
        p.subscription[val] <- "added " + val
    }
}
func (p *Publisher) Remove(val string) {
    if p.subscription[val] != nil {
        p.subscription[val] <- "removed " + val
    }
}

type Subscriber struct {
    subscriptions [] chan string
    publisher *Publisher
}

func (s *Subscriber) RegisterInterest(interest string){
    s.subscriptions = append(s.subscriptions, s.publisher.Subscribe(interest))
}
func (s *Subscriber) run(channel chan string) {
    for  {
        fmt.Println("Waiting for message")
        m := <- channel
        fmt.Println("Got message : " + m)
    }
}
func (s *Subscriber) Listen()  {
    for _, elem := range s.subscriptions {
        go s.run(elem)
    }

}
func main() {
    pub := Publisher{}
    sub := &Subscriber{publisher: &pub}
    sub.RegisterInterest("m")
    sub.RegisterInterest("n")
    sub.Listen()
    pub.Add("m")
    pub.Add("n")
    pub.Remove("m")
    pub.Remove("n")
    time.Sleep(time.Second * 10)
}

关于javascript - golang 中的 pubsub 替代方案，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44000215/