ui.bootstrap 弹出框点击关闭 [英] ui.bootstrap popover close on click

问题描述

我有这个带有模板的弹出框

I have this popover with template

<i class="fa fa-link" popover-placement="right" uib-popover-template="'newReferenceTemplate.html'" popover-title="New link"> Add new external reference </i>

因此，当我单击该链接图标时，会打开一个带有此模板的弹出窗口

So when I click on that link icon, a popover opens witht this tamplate

<script type="text/ng-template" id="newReferenceTemplate.html">
  <label>Title</label> <br>
  <input ng-model="link.Title"> <br>
  <label>Url</label> <br>
  <input ng-model="link.Url"><br>
  <i class="fa fa-floppy-o" > Save </i>
</script>

当我按下那个软盘"图标时，我想关闭弹出窗口.有没有办法做到这一点?

When I press that 'floppy' icon, I'd like to close the popover. Are there any ways of doing this?

我能找到的文档就是popover-is-open 值，但我不知道我是否可以以某种方式使用它，有什么想法吗?

All I can find on documentation is the popover-is-open value, but I don't know if I can use this somehow, any thoughts?

推荐答案

第 1 步: 将 popover-is-open="isOpen" 添加到触发链接.

Step 1 : Add popover-is-open="isOpen" to the trigger link.

<i class="fa fa-link add-link" 
        popover-placement="right" 
        uib-popover-template="'newReferenceTemplate.html'" 
        popover-is-open="isOpen"
        popover-title="New link"> Add new external reference </i>

第 2 步:当您单击弹出窗口内的软盘图标时，将 isOpen 设置为 false:

Step 2 : When you click the floppy icon inside the popover, set isOpen to false:

这是弹出框的保存图标:

This is the save icon of the popover:

<i class="fa fa-floppy-o" ng-click="save()"> Save </i>

这是在控制器中:

$scope.save = function () {
  $scope.isOpen = false;  
};

参见plunker

See plunker

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