Question

为什么RSVP Deferred会在两次调用promise时产生错误？

似乎deferred.promise.then().finally()和deferred.promise.then(); deferred.promise.finally()之间存在差异。为什么呢？

RSVP.on('error', function(reason) {
  console.log('Error: ' + reason);
});

var deferred = RSVP.defer();
var deferred2 = RSVP.defer();
var deferred3 = RSVP.defer();
var promise3 = deferred3.promise;

deferred.promise.then(function() {
  console.log('Resolved');
}, function() {
  console.log('Rejected');
}).finally(function() {
  console.log('Finally');
});

deferred2.promise.then(function() {
  console.log('Resolved2');
}, function() {
  console.log('Rejected2');
});

deferred2.promise.finally(function() {
  console.log('Finally2');
});

promise3 = promise3.then(function() {
  console.log('Resolved3');
}, function() {
  console.log('Rejected');
});

promise3.finally(function() {
  console.log('Finally3');
});


deferred.reject('Reject!');
deferred2.reject('Reject2!');
deferred3.reject('Reject3!');

<script src="https://cdnjs.cloudflare.com/ajax/libs/rsvp/4.8.1/rsvp.js"></script>

编辑：我发现了如何解决这个问题。请参阅代码中的Deferred3。

Answer 1

我发现promise.then()（和其他方法）会返回修改后的承诺，因此您必须将then，finally，catch ...方法链接起来每次都保存promise。

RSVP.on('error', function(reason) {
  console.log('Error: ' + reason);
});

var deferred = RSVP.defer();
var promise = deferred.promise;

promise = promise.then(function() {
  console.log('Resolved');
}, function() {
  console.log('Rejected');
});

promise.finally(function() {
  console.log('Finally');
});


deferred.reject('Reject!');

<script src="https://cdnjs.cloudflare.com/ajax/libs/rsvp/4.8.1/rsvp.js"></script>

两次调用promise时，为什么RSVP Deferred会产生错误

1 个答案: