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)i==0,index==-1 i>index 往下index=0fn=f1Promise.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 1next() 其实就是调用dispatch(1) bind的功劳递归开始dispatch(1)i==1,index==0 i>index 往下index=1fn=f2Promise.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 2next() 其实就是调用dispatch(2)接着递归dispatch(2)i==2,index==1 i>index 往下index=2fn=f3Promise.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 3next() 其实就是调用dispatch(3)接着递归dispatch(3)i==3,index==2 i>index 往下index=3i === middleware.lengthfn=nextPromise.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 4next() 其实就是调用dispatch(4)接着递归dispatch(4)i==4,index==3 i>index 往下index=4fn=middleware[4]fn=undefinedreuturn Promise.resolve(‘fn is undefined’) 回到next执行上下文console.log(‘middleware 4’);输出middleware 4return ‘middleware 4 return’Promise.resolve(‘middleware 4 return’)回到f3执行上下文console.log(‘middleware 3’);输出middleware 3return ‘middleware 3 return’Promise.resolve(‘middleware 3 return’)回到f2执行上下文console.log(‘middleware 2’);输出middleware 2return ‘middleware 2 return’Promise.resolve(‘middleware 2 return’)回到f1执行上下文console.log(‘middleware 1’);输出middleware 1return ‘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。