如何在AngularJS中的ui-select中不允许复制粘贴值

Question

我有一个dropdown list和一个ui-select.在下拉值的基础上ui-select绑定.但如果我直接粘贴其中的特定值ui-select显示为已选中.我们如何防止ui-select复制粘贴值？

示例如下所示.

选择查询

                    <div class="dropdown-Finding">
                        <ui-select class="form-control dropdown-reviwerFinding-select" id="searchBarArea" onkeypress="return false;" multiple tagging tagging-label="false" ng-model="QiReviewerFindingType.selectedItems" theme="bootstrap">
                            <ui-select-match placeholder="Select Reviewer Findings">{{$item.ShortDescription}}</ui-select-match>
                            <ui-select-choices repeat="qiQueryFinding in qiQueryFindings | filter:$select.search">
                                {{qiQueryFinding.ShortDescription}}
                            </ui-select-choices>
                        </ui-select>
                        <span class="carat" open-menu-by-click="searchBarArea"></span>
                    </div>

Answer 1

http://plnkr.co/edit/BVaXpviCACi5sd3aw9oX?p=​​preview

使用这些类型的代码。

<ui-select ng-model="ctrl.person.selected" theme="select2" on-select="ctrl.onSelectCallback($item, $model)" ng-disabled="ctrl.disabled" style="min-width: 300px;" title="Choose a person">
<ui-select-match placeholder="Select a person in the list or search his name/age...">{{$select.selected.name}}</ui-select-match>
<ui-select-choices repeat="person.email as person in ctrl.people | propsFilter: {name: $select.search, age: $select.search}">
  <div ng-bind-html="person.name | highlight: $select.search"></div>
  <small>
    email: {{person.email}}
    age: <span ng-bind-html="''+person.age | highlight: $select.search"></span>
  </small>
</ui-select-choices>

我试图找出确切的差异，但我做不到。但上面的示例可能会对您有所帮助。