[英]Can I use $q.all when one of the functions in the array does not return a promise?
[英]Can I use $q.all in AngularJS with a function that does not return a .promise?
如果我有以下功能:
doTask1: function ($scope) {
var defer = $q.defer();
$http.get('/abc')
.success(function (data) {
defer.resolve();
})
.error(function () {
defer.reject();
});
return defer.promise;
},
doTask2: function ($scope) {
var defer = $q.defer();
var x = 99;
return defer.promise;
},
我被告知我可以等待這兩個承諾:
$q.all([
doTask1($scope),
doTask2($scope)
])
.then(function (results) {
});
如果任務2沒有返回承諾怎么樣? 我在AngularJS的$ q文檔中看到有“when”。 但是我不知道如何使用它並且沒有例子。
是不是我必須讓doTask2通過兩行來返回一個承諾:
var defer = q.defer()
return defer.promise
或者有更簡單的方法嗎?
下面的示例/ plunker顯示了一個方法,其結果在$ q.all中使用,並且每次調用時都返回不同類型的對象(int或promise):
app.controller('MainController', function($scope, $q, $http) {
var count = 0;
function doTask1() {
var defer = $q.defer();
$http.get('abc.json')
.success(function (data) {
defer.resolve(data);
})
.error(function () {
defer.reject();
});
return defer.promise;
}
/**
* This method will return different type of object
* every time it's called. Just an example of an unknown method result.
**/
function doTask2() {
count++;
var x = 99;
if(count % 2){
console.log('Returning', x);
return x;
} else {
var defer = $q.defer();
defer.resolve(x);
console.log('Returning', defer.promise);
return defer.promise;
}
}
$scope.fetchData = function(){
// At this point we don't know if doTask2 is returning 99 or promise.
// Hence we wrap it in $q.when because $q.all expects
// all array members to be promises
$q.all([
$q.when(doTask1()),
$q.when(doTask2())
])
.then(function(results){
$scope.results = results;
});
};
});
<body ng-app="myApp" ng-controller='MainController'>
<button ng-click="fetchData()">Run</button>
<pre>{{results|json}}</pre>
</body>
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