考虑下面以串行/顺序方式读取文件数组的代码。readFiles返回一个承诺,只有在顺序读取所有文件后才会解析这个承诺。
var readFile = function(file) {
... // Returns a promise.
};
var readFiles = function(files) {
return new Promise((resolve, reject) => {
var readSequential = function(index) {
if (index >= files.length) {
resolve();
} else {
readFile(files[index]).then(function() {
readSequential(index + 1);
}).catch(reject);
}
};
readSequential(0); // Start with the first file!
});
};
上面的代码可以工作,但是我不喜欢为了使事情按顺序发生而进行递归。是否有一种更简单的方法可以重写这段代码,这样我就不必使用奇怪的readSequential函数了?
最初我尝试使用Promise。但是这会导致所有的readFile调用并发发生,这不是我想要的:
var readFiles = function(files) {
return Promise.all(files.map(function(file) {
return readFile(file);
}));
};
我真的很喜欢@joelnet的回答,但对我来说,这种编码风格有点难以消化,所以我花了几天时间试图弄清楚如何以更可读的方式表达相同的解决方案,这就是我的想法,只是使用了不同的语法和一些注释。
// first take your work
const urls = ['/url1', '/url2', '/url3', '/url4']
// next convert each item to a function that returns a promise
const functions = urls.map((url) => {
// For every url we return a new function
return () => {
return new Promise((resolve) => {
// random wait in milliseconds
const randomWait = parseInt((Math.random() * 1000),10)
console.log('waiting to resolve in ms', randomWait)
setTimeout(()=>resolve({randomWait, url}),randomWait)
})
}
})
const promiseReduce = (acc, next) => {
// we wait for the accumulator to resolve it's promise
return acc.then((accResult) => {
// and then we return a new promise that will become
// the new value for the accumulator
return next().then((nextResult) => {
// that eventually will resolve to a new array containing
// the value of the two promises
return accResult.concat(nextResult)
})
})
};
// the accumulator will always be a promise that resolves to an array
const accumulator = Promise.resolve([])
// we call reduce with the reduce function and the accumulator initial value
functions.reduce(promiseReduce, accumulator)
.then((result) => {
// let's display the final value here
console.log('=== The final result ===')
console.log(result)
})
这是上面另一个答案的轻微变化。使用原生承诺:
function inSequence(tasks) {
return tasks.reduce((p, task) => p.then(task), Promise.resolve())
}
解释
如果你有这些任务[t1, t2, t3],那么上面的等价于Promise.resolve().then(t1).then(t2).then(t3)。这是约简的行为。
如何使用
首先你需要构造一个任务列表!任务是不接受实参的函数。如果需要将参数传递给函数,则使用bind或其他方法创建任务。例如:
var tasks = files.map(file => processFile.bind(null, file))
inSequence(tasks).then(...)
我不得不运行大量的顺序任务,并使用这些答案来伪造一个函数,将照顾处理任何顺序任务…
function one_by_one(objects_array, iterator, callback) {
var start_promise = objects_array.reduce(function (prom, object) {
return prom.then(function () {
return iterator(object);
});
}, Promise.resolve()); // initial
if(callback){
start_promise.then(callback);
}else{
return start_promise;
}
}
该函数接受2个参数+ 1个可选参数。第一个参数是我们将要处理的数组。第二个参数是任务本身,一个返回承诺的函数,只有当这个承诺解决时,下一个任务才会开始。第三个参数是在所有任务完成后运行的回调。如果没有传递回调,则函数返回它创建的promise,以便我们可以处理结束。
下面是一个用法示例:
var filenames = ['1.jpg','2.jpg','3.jpg'];
var resize_task = function(filename){
//return promise of async resizing with filename
};
one_by_one(filenames,resize_task );
希望它能节省一些时间…
我真的很喜欢@joelnet的回答,但对我来说,这种编码风格有点难以消化,所以我花了几天时间试图弄清楚如何以更可读的方式表达相同的解决方案,这就是我的想法,只是使用了不同的语法和一些注释。
// first take your work
const urls = ['/url1', '/url2', '/url3', '/url4']
// next convert each item to a function that returns a promise
const functions = urls.map((url) => {
// For every url we return a new function
return () => {
return new Promise((resolve) => {
// random wait in milliseconds
const randomWait = parseInt((Math.random() * 1000),10)
console.log('waiting to resolve in ms', randomWait)
setTimeout(()=>resolve({randomWait, url}),randomWait)
})
}
})
const promiseReduce = (acc, next) => {
// we wait for the accumulator to resolve it's promise
return acc.then((accResult) => {
// and then we return a new promise that will become
// the new value for the accumulator
return next().then((nextResult) => {
// that eventually will resolve to a new array containing
// the value of the two promises
return accResult.concat(nextResult)
})
})
};
// the accumulator will always be a promise that resolves to an array
const accumulator = Promise.resolve([])
// we call reduce with the reduce function and the accumulator initial value
functions.reduce(promiseReduce, accumulator)
.then((result) => {
// let's display the final value here
console.log('=== The final result ===')
console.log(result)
})
我的答案基于https://stackoverflow.com/a/31070150/7542429。
Promise.series = function series(arrayOfPromises) {
var results = [];
return arrayOfPromises.reduce(function(seriesPromise, promise) {
return seriesPromise.then(function() {
return promise
.then(function(result) {
results.push(result);
});
});
}, Promise.resolve())
.then(function() {
return results;
});
};
该解决方案以Promise.all()等数组的形式返回结果。
用法:
Promise.series([array of promises])
.then(function(results) {
// do stuff with results here
});
如果其他人在执行CRUD操作时需要一种有保证的严格顺序的方法来解析promise,您也可以使用以下代码作为基础。
只要你在调用每个函数之前添加'return',描述一个Promise,并使用这个例子作为基础,下一个.then()函数调用将在前一个函数完成后consistent启动:
getRidOfOlderShoutsPromise = () => {
return readShoutsPromise('BEFORE')
.then(() => {
return deleteOlderShoutsPromise();
})
.then(() => {
return readShoutsPromise('AFTER')
})
.catch(err => console.log(err.message));
}
deleteOlderShoutsPromise = () => {
return new Promise ( (resolve, reject) => {
console.log("in deleteOlderShouts");
let d = new Date();
let TwoMinuteAgo = d - 1000 * 90 ;
All_Shouts.deleteMany({ dateTime: {$lt: TwoMinuteAgo}}, function(err) {
if (err) reject();
console.log("DELETED OLDs at "+d);
resolve();
});
});
}
readShoutsPromise = (tex) => {
return new Promise( (resolve, reject) => {
console.log("in readShoutsPromise -"+tex);
All_Shouts
.find({})
.sort([['dateTime', 'ascending']])
.exec(function (err, data){
if (err) reject();
let d = new Date();
console.log("shouts "+tex+" delete PROMISE = "+data.length +"; date ="+d);
resolve(data);
});
});
}