决心和q.All()问题 [英] Resolve and q.All() issue

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本文介绍了决心和q.All()问题的处理方法,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

我试图解决类似下面使用的方法 UI路由器

I am trying to resolve a method like below using ui-router

$stateProvider
        .state('abc', {
            url: 'xyz',
            templateUrl: 'templateURL',
            controller: ctrl,
            resolve:{
                data: function(someService){
                    data = someService.init();
                    return data;
                }
            }
        })

和我的服务code看起来像这样

And my service code looks like this

var someObject = {
  data1: ...,
  data2: ....,
  ...
}
return{
   init: function(){
   promise1 = ...
   promise2 = ...
   promise3 = $http.get('someurl').then(function(){
   ...
   ...
   //do some manipulation with someObj
   return someObject;
   });
   $q.all(promise1 , promise2 ).then(promise3);
   }
}

当我调试code,它是来行返回someObject 但它没有解决。

When I debug the code, it is coming to line return someObject but then it is not resolving.

我在做什么错了?

推荐答案

如果你想 promise3 promise1 promise2 然后尝试

return $q.all([promise1, promise2])
    .then(function(arrayOfData) {
        return promise3;
    });

下面是说明差的示例:

    var promise1 = $timeout(function() {
        return 'promise1Data';
    }, 1000); 
    var promise2 = $timeout(function() {
        return 'promise2Data';
    }, 2000);
    var promise3 = $timeout(function() {
        return 'promise3Data';
    }, 5000);

    // This is what you're essentially doing when your route resolves
    $q.all([promise1, promise2])
        .then(promise3)
        .then(function(data) {
            console.log(data); // You get ["promise1Data", "promise2Data"]
        });

    // This is what I think you want
    $q.all([promise1, promise2])
        .then(function(arrayOfResolvedData) {
            return promise3;
        })
        .then(function(data) {
            console.log(data); // You get promise3Data
        });

这篇关于决心和q.All()问题的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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