Question

我正在努力让以下工作。我可以使用:has而不是三个来执行2个ID。我希望只有在span_element的孙子时才能将CSS应用于disabling_div。

的jQuery

$("[id^=span_element]:has[id^=another_id]:has([id^=DisablingDiv])").css({ "cursor": "not-allowed" });

或

$("[id^=span_element].find([id^=DisablingDiv])").css({ "cursor": "not-allowed" });

HTML

<div id="span_element_1"> <!--apply style to this, only when DisablingDiv_1 is below-->
    <div id="another_id">
        <div id="DisablingDiv_1">

             Content

        </div>
    </div>
</div>

Answer 1

使用查找和长度

if( $('[id^=span_element]').find('[id^=DisablingDiv]').length!==0 ){
   alert("Yes! #span_element_1 contains a #DisablingDiv_1");
   // And here your magix, like showing the span.
}

这会尝试找到#span_element_1，然后它会尝试查找#DisablingDiv_1作为其中一个后代。如果它找到多于零的匹配，则它存在。

Answer 2

使用>作为直接子选择器

$("[id^=span_element] > > [id^=DisablingDiv]").css({
  "color": "red"
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="span_element_1">not red
  <div id="another_id"> not red
    <div id="DisablingDiv_1">

      red

    </div>
  </div>
</div>

Answer 3

在阅读完您的编辑和一些评论后，我认为您会搜索此内容：

$("[id^=DisablingDiv]").parent().css(); // That's #another_id
$("[id^=DisablingDiv]").parent().parent().css(); // That's #span_element_1

这样你就走到了底部＆gt;顶部，您不必验证#span_element或another_div是否包含DisablingDiv。

Answer 4

$("[id^=span_element]").find("[id^=DisablingDiv]").css({ "cursor": "not-allowed" });

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="span_element_1"> <!--apply style to this, only when DisablingDiv_1 is below-->
    <div id="another_id">
        <div id="DisablingDiv_1">

             Content

        </div>
    </div>
</div>

Answer 5

<div class="wrapper">
  <div id="span_element_1"> <!--apply style to this, only when DisablingDiv_1 is below-->
    <div id="another_id">
        <div id="DisablingDiv_1" class="active">
           Content
        </div>
    </div>
   </div>
</div>
if ( $("[id^=span_element].parents('.wrapper').find('#another_id').hasClass("active)") {
  $("[id^=span_element].css(); // Do css if active class present
} else {
  $("[id^=span_element].css(); // Do css if active class not present
}

Jquery。查找id的id

5 个答案: