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。
本文来自极简博客,作者:柠檬味的夏天,转载请注明原文链接:解读koa中间件处理模块koa-compose