角度承诺解决了内部功能,但不在外部

Question

我有一个递归函数检查每半秒左右的一些数据.该函数返回一个promise.一旦找到数据,我想解决承诺并将数据作为分辨率传递.问题是,承诺不会在函数外部调用.then().这是小提琴:http://jsfiddle.net/btyg1u0g/1/.

这是小提琴代码:

服务:

myApp.factory('myService', function($q, $timeout) {

    var checkStartTime = false;
    var checkTimeout = 30000;

    function checkForContent() {

        var deferred = $q.defer();

        // simulating an $http request here
        $timeout(function () {

            console.log("Checking...");

            if (!checkStartTime) checkStartTime = new Date();

            // this would normally be 'if (data)'
            if ((new Date()) - checkStartTime > 3000) {
                deferred.resolve("Finished check");
                checkStartTime = false; // reset the timeout start time
            } else {
                // if we didn't find data, wait a bit and check again
                $timeout(function () {
                    checkForContent();
                }, 400);
            }

        }, 5);

        // then is called in here when we find data
        deferred.promise.then(function(message) {
             console.log("Resolved inside checkForContent");
        });

        return deferred.promise;

    }

    return {
        runCheck: function() {
            return checkForContent()
        }
    }
});

控制器:

myApp.controller('MyCtrl', function ($scope, myService) {
    $scope.name = 'Superhero';

    // then is never called
    myService.runCheck()
    .then(function (message) {
        console.log("Resolved outside checkForContent");
    });

});

Answer 1

看看这个小提琴。

外部$timeout函数已命名，因此可以从其内部调用它。

$timeout(function inner() {
  // ...

然后像这样递归调用：

$timeout(function () {
  inner();
}, 400);