Question

$scope.property = new Property();
$scope.property.propertyType = {};

$scope.propertyTypes = [
    { value: 'ResidentialPlot', name: 'Residential Plot' },
    { value: 'CommercialPlot', name: 'Commercial Plot' },
    { value: 'Apartment', name: 'Apartment/Flat' },
    { value: 'Townhouse', name: 'Townhouse' },
    { value: 'House', name: 'Single Family House' },
    { value: 'Commercial', name: 'Commercial Property' }
];

<label for="ptype" class="col-sm-2 control-label">Property Type</label>
<p>Populated: {{property.propertyType}}</p>
<ui-select ng-model="property.propertyType" id="ptype" theme="selectize" ng-disabled="disabled" style="width: 300px;" title="Choose Property Type">
    <ui-select-match placeholder="Select a Property Type">{{$select.selected.value}}</ui-select-match>
    <ui-select-choices repeat="propType in propertyTypes">
        <span ng-bind-html="propType.name"></span>
        <small ng-bind-html="propType.value"></small>    
</ui-select-choices>

这给了我：

$scope.PropertyType = {"value":"Apartment","name":"Apartment/Flat"}

我的架构中的PropertyType只是一个字符串，所以我想绑定选定的值而不是选择的JSON项。

$scope.PropertyType = "Apartment"

我应该将什么绑定到我的ng模型才能得到它？

Answer 1

你不需要观看。

<ui-select ng-model="property.propertyType" id="ptype" theme="selectize" ng-disabled="disabled" style="width: 300px;" title="Choose Property Type">
    <ui-select-match placeholder="Select a Property Type">{{$select.selected.value}}</ui-select-match>
    <ui-select-choices repeat="propType.value as propType in propertyTypes track by $index | filter: $select.search">
        <div ng-bind-html="propType.value | highlight: $select.search"></div>    
</ui-select-choices>

Answer 2

您需要在选择输入中将ng-model属性更改为selected_propertyType并在其更改时进行观察，然后提取值并将其分配给propertyType

$scope.property = new Property();
$scope.property.propertyType = {};

$scope.propertyTypes = [
    { value: 'ResidentialPlot', name: 'Residential Plot' },
    { value: 'CommercialPlot', name: 'Commercial Plot' },
    { value: 'Apartment', name: 'Apartment/Flat' },
    { value: 'Townhouse', name: 'Townhouse' },
    { value: 'House', name: 'Single Family House' },
    { value: 'Commercial', name: 'Commercial Property' }
];

$scope.$watch('selected_propertyType',function(newValue,oldValue){
      if (newValue && newValue!=oldValue){
           $scope.propertyType = $scope.selected_propertyType.value;

      }

})


<label for="ptype" class="col-sm-2 control-label">Property Type</label>
<p>Populated: {{property.selected_propertyType}}</p>
<ui-select ng-model="property.selected_propertyType" id="ptype" theme="selectize" ng-disabled="disabled" style="width: 300px;" title="Choose Property Type">
    <ui-select-match placeholder="Select a Property Type">{{$select.selected.value}}</ui-select-match>
    <ui-select-choices repeat="propType in propertyTypes">
        <span ng-bind-html="propType.name"></span>
        <small ng-bind-html="propType.value"></small>    
</ui-select-choices>

Answer 3

我遇到了同样的问题。我查阅了文档：

https://github.com/angular-ui/ui-select/wiki/ui-select-choices

最好的方法是：

<ui-select ng-model="animal.names"> <ui-select-match>{{$select.selected.name}}</ui-select-match> <ui-select-choices group-by="groupFind" repeat="value.id as value in animals | filter: $select.search"> <div ng-bind-html="value.name | highlight: $select.search"></div> </ui-select-choices> </ui-select>

请注意我们如何在重复中指定value.id，同时仍然使用value.name来显示组合框中显示的内容。这将适用于将ng-model设置为value.id（无论保存什么）。

我确认这对我有用。在此发布，因为Google会将人们带到此页面。

Answer 4

您可以使用select as表示法：

repeat="propType as propType.value for propType in propertyTypes"

Angular ui-select：如何仅将选定值绑定到ng-model

4 个答案: