Question

我迭代一个范围并为每个索引填充新的选择选项。这些选项与我迭代的范围无关，但是是不同类型的选项。代码如下：

<div ng-repeat="i in range(booking.numberOfRooms) track by $index">
    <p>Room {{$index + 1}}</p>
    <select ng-model="booking.roomSelection[$index]" ng-options="obj.roomType as obj.roomType for obj in roomTypes" ng-init="booking.roomSelection[$index] = { id: $index + 1, roomType: 'Double' }"> </select>
</div>

如何将对象数组分配给ng-model（与ng-init中的对象一样）？例如。对于两个房间，ng-model的结果应该类似于：

booking.roomSelection = [{id: 1, roomSelection: 'Double'}, {id: 2, roomSelection: 'Double'}]

Answer 1

只需绑定到您的booking.roomSelection并移除ng-init。如果您选择两个选项，其值pushed将booking.roomSelection和booking.roomSelection将是一个数组

<select multiple ng-model="booking.roomSelection" ng-options="obj.roomType as obj.roomType for obj in roomTypes"></select>

Answer 2

原来必须将ng-model属性更改为：booking.roomSelection[$index].roomType

对象数组也必须在控制器中声明，例如$scope.booking.roomSelection = [];

完整的声明是：

<div ng-repeat="i in range(booking.numberOfRooms) track by $index">
    <p>Room {{$index + 1}}</p>
    <select ng-model="booking.roomSelection[$index].roomType" ng-options="obj.roomType as obj.roomType for obj in roomTypes" ng-init="booking.roomSelection[$index] = { id: $index + 1, roomType: 'Double' }"></select>
</div>

为选择中的对象分配ng-model

2 个答案: