我在页面中使用了以下HTML
<div class="edge-img-sec">
<img src="img1" alt="edge-icon" class="edge-icon1">
<img src="img2" alt="right-icon" class="edge-icon1-right">
</div>
<div class="edge-img-sec">
<img src="img3" alt="edge-icon" class="edge-icon1">
<img src="img4" alt="right-icon" class="edge-icon1-right">
</div>
Jquery脚本: -
$( ".edge-img-sec" ).click(function(){
alert('aa');
$(this).closest('img').find('.edge-icon1-right').hide();
});
点击edge-img-sec
div我要隐藏相应的edge-icon1-right
图像。
我使用了上面的代码。但它没有用。我在这做错了什么。请帮我。
答案 0 :(得分:4)
您可以像下面这样(使用children()
): -
$( ".edge-img-sec" ).click(function(){
$(this).children('.edge-icon1-right').hide();
});
&#13;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="edge-img-sec">
<img src="img1" alt="edge-icon" class="edge-icon1"><br>
<img src="img2" alt="right-icon" class="edge-icon1-right"><br>
</div>
<br>
<div class="edge-img-sec">
<img src="img3" alt="edge-icon" class="edge-icon1"><br>
<img src="img4" alt="right-icon" class="edge-icon1-right"><br>
</div>
&#13;
或使用find()
: -
$( ".edge-img-sec" ).click(function(){
$(this).find('.edge-icon1-right').hide();
});
&#13;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="edge-img-sec">
<img src="img1" alt="edge-icon" class="edge-icon1"><br>
<img src="img2" alt="right-icon" class="edge-icon1-right"><br>
</div>
<br>
<div class="edge-img-sec">
<img src="img3" alt="edge-icon" class="edge-icon1"><br>
<img src="img4" alt="right-icon" class="edge-icon1-right"><br>
</div>
&#13;
答案 1 :(得分:3)
试试这个:
$( ".edge-img-sec" ).click(function(){
$(this).find('img.edge-icon1-right').hide();
});