我有以下代码：
<table class="table">
<tr>
<th>Name___</th>
</tr>
<tr ng-repeat="app in apps"
ng-click="go('/editApp/' + plugin.name);">
<td>
<span>{{app.name}}</span>
</td>
<td style="width: 100px;">
<i class="glyphicon glyphicon-pencil"
ng-click="openPopup(app)"></i>
</td>
</tr>
</table>
当我点击OpenPopup,还有go()方法触发,我该怎么做,如果我点击弹出窗口,弹出窗口会触发？
解决方法
由于您的< td>嵌套在< tr>中,然后单击首先触发openPopup()
然后点火go().您可以使用$event.stopPropagation()来停止事件传播到< tr>.
尝试
<table class="table">
<tr>
<th>Name___</th>
</tr>
<tr ng-repeat="app in apps"
ng-click="go('/editApp/' + plugin.name);">
<td>
<span>{{app.name}}</span>
</td>
<td style="width: 100px;">
<i class="glyphicon glyphicon-pencil"
ng-click="openPopup(app);$event.stopPropagation()"></i>
</td>
</tr>
</table>
