Question

我有img普通班级

<div debug-id="slide-0" style="width: 800px; height: 356px; top: 0px; left: -800px; position: absolute; overflow: hidden;">
<img class="commonclass" u="image" id="imgid_1" src="upload/img_1.jpg" border="0" style="width: 800px; height: 356px; top: 0px; left: 0px; position: absolute;"></div>

img类名是完全相同但div debug-id不同所以如何调用样式的父div

<div debug-id="slide-1" style="width: 800px; height: 356px; top: 0px; left: 0px; position: absolute; overflow: hidden;">
<img class="commonclass" u="image" id="imgid_2" src="upload/img_2.jpg" border="0" style="width: 800px; height: 356px; top: 0px; left: 0px; position: absolute;"></div>

以及更多

<div debug-id="slide-2" style="width: 800px; height: 356px; top: 0px; left: 0px; position: absolute; overflow: hidden;">
<img class="commonclass" u="image" id="imgid_3" src="upload/img_2.jpg" border="0" style="width: 800px; height: 356px; top: 0px; left: 0px; position: absolute;">

Answer 1

如果您尝试选择父div，尽管它没有可用的ID和类选择器，只需选择图像并使用如下的parent（）：

$('img.commonclass').parent();

Answer 2

如果您希望从属性SELECT p.*, CONCAT(IFNULL(GROUP_CONCAT(c_parent...), ''), ',', IFNULL(GROUP_CONCAT(c_direct_parent...), '')), GROUP_CONCAT(c_child...), FROM `places` AS p LEFT JOIN `places_categories_rel` AS rel ON rel.`place_id` = p.`id` LEFT JOIN `categories` AS c_direct_parent ON c_direct_parent.`id` = rel.`category_id` AND c_direct_parent.`parent_id` IS NULL LEFT JOIN `categories` AS c_child ON c_child.`id` = rel.`category_id` AND c_child.`parent_id` IS NOT NULL LEFT JOIN `categories` AS c_parent ON c_parent.`id` = c_child.`parent_id` GROUP BY p.`id`获取父div，

debug-id

如何调用父div标签但没有id和class？

2 个答案: