Question

在我的模板中，我有：

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在我的组件中

<div ng-if="$ctrl.show()">
  <input class="form-control" type="text">
</div>

我希望angular只在单击属性show() { if (angular.isDefined(this.parking.parkingType)) { return this.parking.parkingType.labelKey === 'parking_type.air' } }的选择输入（ui-select）时处理该函数：

on-select="$ctrl.show()"

此案例可能类似于以下示例：仅在单击<ui-select ng-model="$ctrl.parking.parkingType" on-select="$ctrl.show()"> <ui-select-match allow-clear="true"> <span>{{ $select.selected.label }}</span> </ui-select-match> <ui-select-choices repeat="item in $ctrl.parkingType | filter: { label: $select.search }"> <span ng-bind-html="item.label"></span> </ui-select-choices> </ui-select>时启动该功能

Answer 1

将您的ng-show更改为变量并按原样保留on-select="$ctrl.show()"

在您看来：

<div ng-if="$ctrl.shouldShow">
  <input class="form-control" type="text">
</div>

在您的组件中：

$ctrl.show = function() {
  if (angular.isDefined(this.parking.parkingType)) {
    $ctrl.shouldShow = (this.parking.parkingType.labelKey === 'parking_type.air')
  }
}

不要在ng-if，ng-show和ng-hide中使用功能，这是一个很好的做法，因为它会影响性能

Angular 1 / ng-如果仅在条件下进行一次性绑定

1 个答案: