我正在使用Angularjs,并且收到了这样的对象数组:
{
"_id": 310243,
"name": "Five Star Babies: Inside the Portland Hospital",
"airsDayOfWeek": "Wednesday",
"airsTime": "21:00",
"firstAired": "2016-04-13T00:00:00.000Z",
"network": "BBC Two",
"overview": "An exclusive glimpse inside the UK's only private maternity hospital.\r\n",
"rating": null,
"ratingCount": 0,
"status": "Continuing",
"poster": "/someImage/path",
"subscribers": []
}
实际上,该对象来自此函数:
$scope.shows = Show.query(); // returns an array
_.forEach($scope.shows, function(item) {
if (item.rating === null) {
console.log(angular.toJson(item, 'pretty'));
}
})
如您所见,我需要如果
null
中包含“ rating”或“ poster”属性,则不要在视图中显示该元素,这是合适的情况,如您所见,rating来自因此,我需要隐藏整个元素。视图:
<div ng-repeat="show in shows | filter:query | orderBy:'rating':true">
<a href="/shows/{{show._id}}">
<img ng-src="{{show.poster}}" width="100%"/>
</a>
<div>
<a href="/shows/{{show._id}}">{{show.name}}</a>
<p>
Episodes: {{show.episodes.length}} <br>
<span>Rating: {{show.rating}}</span>
</p>
</div>
</div>
有什么建议么?
最佳答案
似乎您可以只过滤所需值的数组。
$scope.shows = Show.query().filter(function(val) {
if (val.rating !== null && val.poster !== null) return val;
});
这将为您提供没有评级/海报的节目,而没有这些节目。