angularjs - $q promise 错误回调链

Question

在以下代码片段 error 1 中和 success 2将被记录。如果原始延迟被拒绝，我如何传播被调用的错误回调而不是成功回调被调用。

angular.module("Foo", []);
angular
.module("Foo")
.controller("Bar", function ($q) {
    var deferred = $q.defer();
      deferred.reject();

      deferred.promise
          .then(
              /*success*/function () { console.log("success 1"); },
              /*error*/function () { console.log("error 1"); })
          .then(
              /*success*/function () { console.log("success 2"); },
              /*error*/function () { console.log("error 2"); });
});

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>
<div ng-app="Foo">
    <div ng-controller="Bar"></div>
</div>

Answer 1

错误通过返回 $q.reject 传播在错误回调中

    var deferred = $q.defer();
      deferred.reject();

      deferred.promise
          .then(
              /*success*/function () { console.log("success 1"); },
              /*error*/function () { console.log("error 1"); return $q.reject('error 1')})
          .then(
              /*success*/function () { console.log("success 2"); },
              /*error*/function () { console.log("error 2"); });
});