AngularJS处理被拒绝的资源的$q.all。

I'm trying to handle errors with my resources, and then handle rejection of resources in my $q.all().

我尝试使用我的资源处理错误，然后处理$q.all()中的资源拒绝。

This is my code:

这是我的代码:

var user = User.get({id: 1}, function() {
    // Success
}, function(response) {
    // Error
    return $q.reject(response);
});

var promiseList = [user];
$q.all(promiseList).then(function(){
  // Success <-- this seems to run all the time
  }, function(response) {
    // Error <-- this never seems to run but I want it to
});

When my User resource receives a 404, the error callback handles it and returns a $q.reject.

当我的用户资源收到404时，错误回调处理它并返回$q.reject。

However the success callback in my $q.all gets called, not my error callback. I would've thought because I am rejecting my promise the $q.all error callback would be fired?

但是成功回调在我的$q。所有的都被调用，而不是我的错误回调。我可能会想，因为我拒绝了我的承诺。所有的错误回调将被触发?

I appreciate I only have 1 item in my promiseList but that shouldn't make a difference should it?

我很感激我的承诺中只有一项，但这不应该有什么不同，是吗?

1 个解决方案

#1

According to Michael in the comments

根据Michael的评论

I changed

我改变了

var promiseList = [user];

To this:

var promiseList = [user.$promise];

And now the $q.reject() is being picked up by the $q.all().

现在$q.reject()被$q.all()接收。

Awesome, thanks for the advice.

太棒了，谢谢你的建议。

#1