AngularJS承诺链 [英] AngularJS promise chain

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本文介绍了AngularJS承诺链的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有我的应用程序应该打开一个弹出的用户请确认,然后做出一个Ajax CAL和关闭弹出。结果
我试图用链做的的(我已经用它,我记得它应该以这种方式工作),但似乎在调用后,阻止 reservationService.confirm($ scope.object); 。现在,它是实现了一个假的服务的setTimeout $ Q 只返回一个承诺(在未来它将使Ajax调用)。这是一个有效的code或没undestood承诺是如何工作的?结果
对于我选择AngularUI并在弹出的code是这样的:

I have my application that should open a popup ask a confirmation at the user, then make an ajax cal and close the popup.
I tried to do it using a chain of promise (I've already used it, and I remember that it should work in this way), but it seems to block after the call toreservationService.confirm($scope.object);. Now it is a fake service implemented with a setTimeout and $q just to return a promise (in future it will make the ajax call). Is this a valid code or I didn't undestood how promise works?
For the popup I choose AngularUI and the code is this:

 reservationService.book($scope.object, day)
        .then(function(){
            var dialogOpts = {/* dialog options omitted*/}
            return $dialog.dialog(dialogOpts).open();

        })
        .then(function(result){
            console.log('confirmed? ', result);
            if (result){
                //After this line it doesn't do nothing, also if the promise is resolved 
                return reservationService.confirm($scope.object);
            }
        })
        .then(function(){
            //this function is never executed
            $scope.$emit('object:detail',{object: $scope.object});
        });

reservationService:

reservationService:

function confirm(){
     var deferred = $q.defer();
     setTimeout(function(){
          console.log('Confirming');
          deferred.resolve(true)
     }, 500);
     return deferred.promise;
}

解决
变化的setTimeout $超时角度的服务。

推荐答案

用于 $超时而不是的setTimeout 导致它的工作原理togheter的角度范围,迫使消化阶段(或使用 $ scope.apply()里面的的setTimeout )。

Used $timeout instead of setTimeout 'cause it works togheter at the angular scope, forcing the digest phase (or use $scope.apply() inside the setTimeout).

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