篩選ng選項

Question

我有2選擇ng選項。我希望第二個選擇重複選擇name的values字段。

我發現的唯一解決方案是選擇對象本身，而不僅僅是第一個選擇中的名稱字段（類似d as d.name for d in data） 但爲了一致性我寧願不要。

我可以使用過濾器或類似的東西來解決這個問題嗎？

angular.module("App", []).controller("Ctrl", function($scope) { 
 
    $scope.data = [{ 
 
    name : "john", 
 
    values : ["Hello", "World"] 
 
    }, { 
 
    name : "peter", 
 
    values : ["Bonjour", "le", "monde"] 
 
    }]; 
 
});

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script> 
 
<div ng-app="App" ng-controller="Ctrl"> 
 
<label>Name :</label> 
 
<select ng-model="name" ng-options="d.name as d.name for d in data"></select> 
 
<label>Value :</label> 
 
<select ng-model="value" ng-options="d as d for d in data"></select> 
 
</div>

Answer 1

喜歡這個？

angular.module("App", []).controller("Ctrl", function($scope) { 
 
    $scope.data = [{ 
 
    name : "john", 
 
    values : ["Hello", "World"] 
 
    }, { 
 
    name : "peter", 
 
    values : ["Bonjour", "le", "monde"] 
 
    }]; 
 
});

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script> 
 
<div ng-app="App" ng-controller="Ctrl"> 
 
<label>Name :</label> 
 
<select ng-model="name" ng-options="d.name for d in data"></select> 
 
<label>Value : </label> 
 
<select ng-model="value" ng-options="d as d for d in name.values"></select> 
 
</div>

Answer 2

我不知道是否有更好的方法，但這是一種工作方式：現在

angular.module("App", []).controller("Ctrl", function($scope) { 
 
    $scope.data = [{ 
 
    name : "john", 
 
    values : ["Hello", "World"] 
 
    }, { 
 
    name : "peter", 
 
    values : ["Bonjour", "le", "monde"] 
 
    }]; 
 
});

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script> 
 
<div ng-app="App" ng-controller="Ctrl"> 
 
<label>Name :</label> 
 
<select ng-model="name" ng-options="d.name as d.name for d in data"></select> 
 
<label>Value :</label> 
 
<select ng-model="value" ng-repeat="d in data" ng-if="name === d.name" ng-options="val for val in d.values"></select> 
 
</div>