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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=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)
i==1,index==0 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)
i==2,index==1 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)
i==3,index==2 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)
i==4,index==3 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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