我有像这样的角度嵌套对象.
有没有办法过滤它的嵌套属性
有没有办法过滤它的嵌套属性
<li ng-repeat="shop in shops | filter:search"> search.locations.city_id = 22
我只显示父元素,但想要同时过滤它们,例如:
search = category_id: 2 locations: city_id: 368 [ name: "xxx" category_id: 1 locations: [ city_id: 368 region_id: 4,city_id: 368 region_id: 4,city_id: 368 region_id: 4 ],name: "xxx" category_id: 2 locations: [ city_id: 30 region_id: 4,city_id: 22 region_id: 2 ] ]
是的,如果我理解你的例子,你可以.
原文链接:https://www.f2er.com/angularjs/141853.html根据集合的大小,最好计算在ng-repeat中迭代的集合,以便过滤器在模型更改时不会持续执行.
如果我理解正确的话,基本上你会做这样的事情:
$scope.search = function (shop) { if ($scope.selectedCityId === undefined || $scope.selectedCityId.length === 0) { return true; } var found = false; angular.forEach(shop.locations,function (location) { if (location.city_id === parseInt($scope.selectedCityId)) { found = true; } }); return found; };