Question

<table class="col-md-12">
    <th>values</th>
    <tr>
        <td>
            <span ng-repeat="item in items">
                <span ng-repeat="subitem in item.allowitems" ng-if "item.allowid = 1">
                    <span ng-repeat="point in subitem.points">{{point.value}}</span>
                </span>
            </span>
        </td>
    </tr>
    <tr>
        <td>
            <span ng-repeat="item in items">
                <span ng-repeat="subitem in item.allowitems" ng-if "item.allowid = 3">
                    <span ng-repeat="point in subitem.points">{{point.value}}</span>
                </span>
            </span>
        </td>
    </tr>
    <tr>
        <td>
            <span ng-repeat="item in items">
                <span ng-repeat="subitem in item.allowitems" ng-if "item.allowid = 5">
                    <span ng-repeat="point in subitem.points">{{point.value}}</span>
                </span>
            </span>
        </td>
    </tr>
</table>

如果point.value为null或者item.allowid为null，我想隐藏相应的行trs。但是我如何在父对象tr中检查ng-if或ng-hide

Answer 1

可能有更优雅的方法，但一个选项是使用指令隐藏父.closest()。由于jqLite没有.hide()或.directive('hideParent', function(){ return { restrict: 'E', link: function(scope, element, attrs){ element.closest('tr').hide(); } } })，因此您必须包含对jQuery的引用才能生效。

<tr>

然后，当您想要隐藏父<span ng-repeat="item in items"> <hide-parent ng-if="item.allowid === null"></hide-parent> <span ng-repeat="subitem in item.allowitems" ng-if="item.allowid === 1"> <span ng-repeat="point in subitem.points"> {{point.value}} <hide-parent ng-if="point.value === null"></hide-parent> </span> </span> </span>：

时，您可以使用该指令

ng-repeat

有些人可能会认为这是解决问题的黑客或肮脏方法，因为除了操作DOM之外，你包含的元素没有任何意义。您可以随时将其更改为包含hideParent的元素的属性，但是您必须移动确定是否将父项隐藏到<nav> <a href="">HTML</a> <a href="">CSS</a> <a href="">JS</a> </nav>指令中的逻辑。

根据角度ng-repeat中的子跨度值隐藏父tr

1 个答案: