我有两个列表，我想删除所有li标记，但那些包含-all作为其id的输入元素的li标记除外。这是我在jQuery中使用选择器可以实现的吗？可能不使用选择器？
这就是我目前所拥有的：$("#search-filter-column ul li").remove();
<ul>
<li>
<input id="filter-types-all" name="filter-types-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-types-suv" name="filter-types-suv" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-types-special" name="filter-types-special" type="checkbox" checked="checked">
</li>
</ul>
<ul>
<li>
<input id="filter-company-all" name="filter-company-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-company-1" name="filter-company-1" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-company-2" name="filter-company-2" type="checkbox" checked="checked">
</li>
</ul>
<ul>
<li>
<input id="filter-location-1" name="filter-location-1" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-location-2" name="filter-location-2" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-location-3" name="filter-location-3" type="checkbox" checked="checked">
</li>
</ul>
答案 0 :(得分：9)
您可以测试input元素内的li元素，然后使用 {{3}，而不是搜索input元素然后测试其内容和 :not() CSS pseudo-class 以及 "ends-with" attribute selector ($=) 排除任何不相关的li元素。然后，您可以使用 "starts-with" (^=) attribute selector 方法删除所有匹配元素的父// Remove all input elements that are descendants of an <li> that don't have
// an id that ends with ($=) "all" or an id that starts with "filter-location-".
$("#search-filter-column ul > li input:not([id$='-all']):not([id^='filter-location-'])").parent().remove();。
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="search-filter-column">
<ul>
<li>
<input id="filter-types-all" name="filter-types-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-types-suv" name="filter-types-suv" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-types-special" name="filter-types-special" type="checkbox" checked="checked">
</li>
</ul>
<ul>
<li>
<input id="filter-company-all" name="filter-company-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-company-1" name="filter-company-1" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-company-2" name="filter-company-2" type="checkbox" checked="checked">
</li>
</ul>
<ul>
<li>
<input id="filter-location-1" name="filter-location-1" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-location-2" name="filter-location-2" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-location-3" name="filter-location-3" type="checkbox" checked="checked">
</li>
</ul>
</div>must_not
答案 1 :(得分：3)
是的，您可以使用:not，:has和attribute-ends-with selector：
$("#search-filter-column ul li:not(:has(input[id$=-all]))").remove();
匹配li ul中#search-filter_column个input元素，这些元素不包含id以-all结尾的:not
在其他地方的评论中你说过
我也有包含-location-的li元素。我也不想删除这些li元素。
您可以拥有多个li:not(this):not(that)条件，*=。它们和CSS选择器的其他方面一样。所以你要attribute contains selector（$("#search-filter-column ul li:not(:has(input[id$=-all])):not(:has(input[id*=-location-]))").remove();
）这样做：
$("#search-filter-column ul li:not(:has(input[id$=-all])):not(:has(input[id*=-location-]))").remove();直播示例：
<div id="search-filter-column">
Types:
<ul>
<li>
<input id="filter-types-all" name="filter-types-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-types-suv" name="filter-types-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-types-special" name="filter-types-all" type="checkbox" checked="checked">
</li>
</ul>
Companies:
<ul>
<li>
<input id="filter-company-all" name="filter-company-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-company-1" name="filter-company-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-company-2" name="filter-company-all" type="checkbox" checked="checked">
</li>
</ul>
Locations:
<ul>
<li>
<input id="filter-location-1" name="filter-location-1" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-location-2" name="filter-location-2" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-location-3" name="filter-location-3" type="checkbox" checked="checked">
</li>
</ul>
</div>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script><android.support.constraint.ConstraintLayout
android:id="@+id/layout_charts_list"
tools:context="apppath.fragments.ChartListFragment">
<TextView
android:id="@+id/loading_list"
android:text="@string/loading_my_charts" />
<TextView
android:id="@+id/empty_list"
android:text="@string/my_charts_empty"
android:visibility="gone"/>
<android.support.v7.widget.RecyclerView
android:id="@+id/charts_list" />
</android.support.constraint.ConstraintLayout>
答案 2 :(得分：0)
您可以使用jQuery的filter函数来过滤掉您不想删除的元素：
$("#search-filter-column ul li").filter(function() {
return $(this).find('input[id$=-all]').length;
}).remove();<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<div id="search-filter-column">
<ul>
<li>
<input id="filter-types-all" name="filter-types-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-types-suv" name="filter-types-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-types-special" name="filter-types-all" type="checkbox" checked="checked">
</li>
</ul>
<ul>
<li>
<input id="filter-company-all" name="filter-company-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-company-1" name="filter-company-all" type="checkbox" checked="checked">
</li>
<li>
<input id="filter-company-2" name="filter" type="checkbox" checked="checked">
</li>
</ul>
</div>
在您的示例中，我更改了一些过滤器，以便在ID中没有-all，以显示它实际上会删除它们。
答案 3 :(得分：-4)
您尝试按ID选择，获取父级并删除父级。 例如：
$('#ID_REMOVE').parent().remove()