將一系列有序操作合並為一個Observable

Question

var state = [];
var operation1 = function() {
    return Rx.Observable.fromPromise(new Promise((resolve, reject) => {
        state.push(1, 2);
        setTimeout(resolve, 300, state);
    }));
};
var operation2 = function() {
    return Rx.Observable.fromPromise(new Promise((resolve, reject) => {
        state = state.map(x => x * 2);
        setTimeout(resolve, 200, state);
    }));
};
var operation3 = function() {
    return Rx.Observable.fromPromise(new Promise((resolve, reject) => {
        state = state.reduce( (prev, next) => prev + next );
        setTimeout(resolve, 100, state);
    }));
};
var operations = [operation1, operation2, operation3];

給定上面的代碼，我試圖將operations合並為一個Observable ，以發出每個操作的狀態。 因此， Observable需要執行以下任一操作：

• 發出3次： [1, 2], [2, 4], 6
• 發出1次： [[1, 2], [2, 4], 6]

Answer 1

你可以嘗試（ jsbin ）

var state = [];
var operation1 = Rx.Observable.defer(function() {
    return Rx.Observable.fromPromise(new Promise((resolve, reject) => {
        state.push(1, 2);
        setTimeout(resolve, 300, state);
    }));
});
var operation2 = Rx.Observable.defer(function() {
    return Rx.Observable.fromPromise(new Promise((resolve, reject) => {
        state = state.map(x => x * 2);
        setTimeout(resolve, 200, state);
    }));
});
var operation3 = Rx.Observable.defer(function() {
    return Rx.Observable.fromPromise(new Promise((resolve, reject) => {
        state = state.reduce( (prev, next) => prev + next );
        setTimeout(resolve, 100, state);
    }));
});
var operations = Rx.Observable.from([operation1, operation2, operation3]).merge(1);
operations.subscribe(function(x){console.log(x)})

請檢查是否可以解決問題，稍后我將詳細說明其工作原理。

Answer 2

您可以使用Rx.Observable.concat但我認為如果沒有rx.java和Promise.all，這種簡單的情況會更容易

 let slow = new Promise((resolve) => { setTimeout(resolve, 200, 'slow'); }); let instant = new Promise((resolve) => { setTimeout(resolve, 0, 'instant'); }); let quick = new Promise((resolve) => { setTimeout(resolve, 50, 'quick'); }); var operation1 = function() { return Rx.Observable.fromPromise(slow); } var operation2 = function() { return Rx.Observable.fromPromise(instant); } var operation3 = function() { return Rx.Observable.fromPromise(quick); } var operations = [operation1(), operation2(), operation3()]; var source = Rx.Observable.concat(operations); var subscription = source.subscribe( function(x) { console.log('Next: ' + x); }, function(err) { console.log('Error: ' + err); }, function(a) { console.log('Completed', a); }); // Or with promises operations = [slow, instant, quick] Promise.all(operations) .then(console.log.bind(console, "Promise all")) 

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