javascript - For Each 循环中的 Promise 在 AngularJS 中不起作用

Question

传入的“types”参数的格式为["hourAverage","hourMedian","dayAverage","dayMedian","weekAverage","weekMedian","monthAverage","monthMedian"]

来源:

$scope.buildCharts = function(types){
    var deferred = $q.defer();
    var promises = [];

    types.forEach(function(type){
                $scope.prepareData(type);
                deferred.resolve([
                {
                    title: type,
                    curveType: 'function',
                    legend: { position: 'bottom' }
                },type]);
                promises.push(deferred.promise);
    });
    return $q.all(promises);
};

输出:

[
  [
    {
      "title": "hourAverage",
      "curveType": "function",
      "legend": {
        "position": "bottom"
      }
    },
    "hourAverage"
  ],
  [
    {
      "title": "hourAverage",
      "curveType": "function",
      "legend": {
        "position": "bottom"
      }
    },
    "hourAverage"
  ],
  [
    {
      "title": "hourAverage",
      "curveType": "function",
      "legend": {
        "position": "bottom"
      }
    },
    "hourAverage"
  ],
  [
    {
      "title": "hourAverage",
      "curveType": "function",
      "legend": {
        "position": "bottom"
      }
    },
    "hourAverage"
  ],
  [
    {
      "title": "hourAverage",
      "curveType": "function",
      "legend": {
        "position": "bottom"
      }
    },
    "hourAverage"
  ],
  [
    {
      "title": "hourAverage",
      "curveType": "function",
      "legend": {
        "position": "bottom"
      }
    },
    "hourAverage"
  ],
  [
    {
      "title": "hourAverage",
      "curveType": "function",
      "legend": {
        "position": "bottom"
      }
    },
    "hourAverage"
  ],
  [
    {
      "title": "hourAverage",
      "curveType": "function",
      "legend": {
        "position": "bottom"
      }
    },
    "hourAverage"
  ]
]

Answer 1

这说明 Promise 只能以一个值解析。对同一个延迟对象进一步调用 resolve 都不会产生任何效果。这就是为什么你一遍又一遍地获取第一个对象作为值。这是 Promises/A+ specifications, point 2.1.2 所要求的行为(强调我的):

2.1.2 When fulfilled, a promise:

     2.1.2.1 must not transition to any other state.

     2.1.2.2 must have a value, which must not change.

当然，如果允许更改 Promise 的值，那么您仍然不会得到所需的结果，因为结果会重复 < em>最后对象值。

要解决此问题，请在循环内创建一个新的延迟对象，以便始终解决新的 Promise:

移动这个:

var deferred = $q.defer();

进入forEach回调:

types.forEach(function(type){
    var deferred = $q.defer();
    // etc...