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  • 傻瓜式解读koa中间件处理模块koa-compose

    最近需要单独使用到koa-compose这个模块,虽然使用koa的时候大致知道中间件的执行流程,但是没仔细研究过源码用起来还是不放心(主要是这个模块代码少,多的话也没兴趣去研究了)。

    koa-compose看起来代码少,但是确实绕。闭包,递归,Promise。。。看了一遍脑子里绕不清楚。看了网上几篇解读文章,都是针对单行代码做解释,还是绕不清楚。最后只好采取一种傻瓜的方式:

    koa-compose去掉一些注释,类型校验后,源码如下:

    function compose (middleware) {
      return function (context, next) {
        // last called middleware #
        let index = -1
        return dispatch(0)
        function dispatch (i) {
          if (i <= index) return Promise.reject(new Error('next() called multiple times'))
          index = i
          let fn = middleware[i]
          if (i === middleware.length) fn = next
          if (!fn) return Promise.resolve()
          try {
            return Promise.resolve(fn(context, dispatch.bind(null, i + 1)));
          } catch (err) {
            return Promise.reject(err)
          }
        }
      }
    }
    

    写出如下代码:

    var index = -1;
    function compose() {
        return dispatch(0)
    }
    function dispatch (i) {
          if (i <= index) return Promise.reject(new Error('next() called multiple times'))
          index = i
          var fn = middleware[i]
          if (i === middleware.length) fn = next
          if (!fn) return Promise.resolve('fn is undefined')
          try {
            return Promise.resolve(fn(context, dispatch.bind(null, i + 1)));
          } catch (err) {
            return Promise.reject(err)
          }
     }
     
     function f1(context,next){
        console.log('middleware 1');
        next().then(data=>console.log(data));
        console.log('middleware 1');
        return 'middleware 1 return';
      }
      function f2(context,next){
        console.log('middleware 2');
        next().then(data=>console.log(data));
        console.log('middleware 2');
        return 'middleware 2 return';
      }
      function f3(context,next){
        console.log('middleware 3');
        next().then(data=>console.log(data));
        console.log('middleware 3');
        return 'middleware 3 return';
      }
    var middleware=[
      f1,f2,f3
    ]
    
    var context={};
    var next=function(context,next){
        console.log('middleware 4');
        next().then(data=>console.log(data));
        console.log('middleware 4');
        return 'middleware 4 return';
    };
    compose().then(data=>console.log(data));
    

    直接运行结果如下:

    "middleware 1"

    "middleware 2"

    "middleware 3"

    "middleware 4"

    "middleware 4"

    "middleware 3"

    "middleware 2"

    "middleware 1"

    "fn is undefined"

    "middleware 4 return"

    "middleware 3 return"

    "middleware 2 return"

    "middleware 1 return"

    按着代码运行流程一步步分析:

    dispatch(0)

    i0,index-1 i>index 往下

    index=0

    fn=f1

    Promise.resolve(f1(context, dispatch.bind(null, 0 + 1)))

    这就会执行

    f1(context, dispatch.bind(null, 0 + 1))

    进入到f1执行上下文

    console.log('middleware 1');

    输出middleware 1

    next()

    其实就是调用dispatch(1) bind的功劳

    递归开始

    dispatch(1)

    i1,index0 i>index 往下

    index=1

    fn=f2

    Promise.resolve(f2(context, dispatch.bind(null, 1 + 1)))

    这就会执行

    f2(context, dispatch.bind(null, 1 + 1))

    进入到f2执行上下文

    console.log('middleware 2');

    输出middleware 2

    next()

    其实就是调用dispatch(2)

    接着递归

    dispatch(2)

    i2,index1 i>index 往下

    index=2

    fn=f3

    Promise.resolve(f3(context, dispatch.bind(null, 2 + 1)))

    这就会执行

    f3(context, dispatch.bind(null, 2 + 1))

    进入到f3执行上下文

    console.log('middleware 3');

    输出middleware 3

    next()

    其实就是调用dispatch(3)

    接着递归

    dispatch(3)

    i3,index2 i>index 往下

    index=3

    i === middleware.length

    fn=next

    Promise.resolve(next(context, dispatch.bind(null, 3 + 1)))

    这就会执行

    next(context, dispatch.bind(null, 3 + 1))

    进入到next执行上下文

    console.log('middleware 4');

    输出middleware 4

    next()

    其实就是调用dispatch(4)

    接着递归

    dispatch(4)

    i4,index3 i>index 往下

    index=4

    fn=middleware[4]

    fn=undefined

    reuturn Promise.resolve('fn is undefined')

    回到next执行上下文

    console.log('middleware 4');

    输出middleware 4

    return 'middleware 4 return'

    Promise.resolve('middleware 4 return')

    回到f3执行上下文

    console.log('middleware 3');

    输出middleware 3

    return 'middleware 3 return'

    Promise.resolve('middleware 3 return')

    回到f2执行上下文

    console.log('middleware 2');

    输出middleware 2

    return 'middleware 2 return'

    Promise.resolve('middleware 2 return')

    回到f1执行上下文

    console.log('middleware 1');

    输出middleware 1

    return 'middleware 1 return'

    Promise.resolve('middleware 1 return')

    回到全局上下文

    至此已经输出

    "middleware 1"

    "middleware 2"

    "middleware 3"

    "middleware 4"

    "middleware 4"

    "middleware 3"

    "middleware 2"

    "middleware 1"

    那么

    "fn is undefined"

    "middleware 4 return"

    "middleware 3 return"

    "middleware 2 return"

    "middleware 1 return"

    怎么来的呢

    回头看一下,每个中间件里都有

    next().then(data=>console.log(data));

    按照之前的分析,then里最先拿到结果的应该是next中间件的,而且结果就是Promise.resolve('fn is undefined')的结果,然后分别是f4,f3,f2,f1。那么为什么都是最后才输出呢?

    Promise.resolve('fn is undefined').then(data=>console.log(data));
    console.log('middleware 4');
    

    运行一下就清楚了

    或者

    setTimeout(()=>console.log('fn is undefined'),0);
    console.log('middleware 4');
    

    整个调用过程还可以看成是这样的:

    function composeDetail(){
      return Promise.resolve(
        f1(context,function(){
          return Promise.resolve(
            f2(context,function(){
              return Promise.resolve(
                f3(context,function(){
                  return Promise.resolve(
                    next(context,function(){
                      return Promise.resolve('fn is undefined')
                    })
                  )
                })
              )
            })
          )
        })
      )
    }
    composeDetail().then(data=>console.log(data));
    

    方法虽蠢,但是compose的作用不言而喻了

    最后,if (i <= index) return Promise.reject(new Error('next() called multiple times'))这句代码何时回其作用呢?

    一个中间件里调用两次next(),按照上面的套路走,相信很快就明白了。

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  • 原文地址:https://www.cnblogs.com/jaycewu/p/9873980.html
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