Angularjs 和 UI-Select:如何从代码中选择一个选项

in angularjs 我有一个ui-select：

<ui-select ng-model="itemSelected.selected" theme="selectize" ng-disabled="disabled">
   <ui-select-match placeholder="Select an item...">{{$select.selected.Name}}</ui-select-match>
   <ui-select-choices repeat="item in itemsList">
      <span ng-bind-html="item.Name"></span>
   </ui-select-choices>
</ui-select>

加载页面时如何从代码中选择项目？当我在控制器中加载页面时，我得到$scope.itemsList：如何从controller中选择特定项目？

谢谢

您可以在控制器加载本身上设置它

标记

<body ng-controller="DemoCtrl">
  <p>Selected: {{item.selected}}</p>
  <ui-select ng-model="item.selected" theme="select2" ng-disabled="disabled" style="min-width: 300px;">
    <ui-select-match placeholder="Select a item in the list">{{$select.selected.name}}</ui-select-match>
    <ui-select-choices repeat="item in items | propsFilter: {name: $select.search, age: $select.search}">
      <div ng-bind-html="item.Code | highlight: $select.search"></div>
    </ui-select-choices>
  </ui-select>
</body>

法典

app.controller('DemoCtrl', function($scope, $http) {
  $scope.disabled = undefined;
  $scope.clear = function() {
    $scope.item.selected = undefined;
  };
  $scope.item = {};
  $scope.items = [
    { name: 'Item1', Code: 'Code1', },
    { name: 'Item2', Code: 'Code3'},
    { name: 'Item3',  Code: 'Code4'},
    { name: 'Item4',  Code: 'Code4' },
    { name: 'Item5', Code: 'Code5' },
  ];
  $scope.item.selected = $scope.items[0] //here you can set the item selected
});

工作普伦克

尔

@Pankaj Parkar 的 plunkr 不再用于预期用途，所以我分叉了它并让它在这里工作：

http://plnkr.co/edit/Y6cdgJQ3YPq7Ncb3bT4A?p=preview

关键更改涉及在控制器中实际设置所选项目：

$scope.address = {};
$scope.refreshAddresses = function(address) {
    var params = {address: address, sensor: false};
    return $http.get(
    'http://maps.googleapis.com/maps/api/geocode/json',
    {params: params}
    ).then(function(response) {
    $scope.addresses = response.data.results
    $scope.address.selected = $scope.addresses[0];
    $scope.$apply();
    });
};
$scope.refreshAddresses('New York, NY');

关键的补充是$scope.address.selected = $scope.addresses[0];和$scope.refreshAddresses('New York, NY');让它继续下去。

我还更新了其他选择以预填充。

相关内容

最新更新

热门标签：