为什么在两次调用 promise 时 RSVP Deferred 会产生错误

Question

为什么在两次调用 promise 时 RSVP Deferred 会产生错误？

deferred.promise.then().finally() 和deferred.promise.then(); deferred.promise.finally() 似乎有区别。为什么？

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。查看代码中的 Deferred3。

Answer 1

我发现promise.then()（和其他方法）返回一个修改过的承诺，所以你必须链接then、finally、catch...方法，否则你必须每次都保存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>