Question

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

<li ng-repeat="shop in shops | filter:search">
search.locations.city_id = 22

我只显示父元素，但希望按两者进行过滤，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
  ]
]

Answer 1

您也可以像这样过滤（版本1.2.13 +）

<li ng-repeat="shop in shops | filter: { locations: [{ city_id: search.locations.city_id }] }">

Answer 2

是的，如果我理解你的例子，你可以。

根据您的集合的大小，最好计算您在ng-repeat中迭代的集合，以便在模型更改时过滤器不会持续执行此操作。

http://jsfiddle.net/suCWn/

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;
};

Answer 3

更新了“像Jared这样的单词”的答案，使用正则表达式来检查它是否包含searchterm。这种方式在您键入1个数字时开始过滤，因此您不必匹配整个单词

JSfiddle

    angular.forEach(shop.locations, function (location) {          
        if (checknum(location.city_id)) {
            found = true;
        }
    });

    function checknum(num){
        var regx = new RegExp($scope.selectedCityId);
        return regx.test(num);
    };

Angularjs过滤嵌套对象

3 个答案: