小编Har*_*iss的帖子

我有一些带有图像的div,当我点击这些图像时,我想要用我点击的图像打开另一个div.

JS

$('.examples img').click(function() {
    var loc = $(this).attr("src");
    $('#image-zoom').attr("src",loc);
});

HTML

<div class="container examples"  >
    <div id="image-zoom">
        <img class="img-thumbnail zoom" src="" alt="dental">
    </div>
    <div class="row">
        <div class="col-sm-12">
            <img id="zoom" class="img-thumbnail zoom" src="images/01.png" alt="dental">
            <img class="img-thumbnail zoom" src="images/02.png" alt="dental">
            <img class="img-thumbnail zoom" src="images/03.png" alt="dental">
        </div>
    </div>

    <div class="row">
        <div class="col-sm-12">
            <img class="img-thumbnail zoom" src="images/04.png" alt="dental">
            <img class="img-thumbnail zoom" src="images/05.png" alt="dental">
            <img class="img-thumbnail zoom" src="images/06.png" alt="dental">
        </div>
    </div>
</div>

当我试图隐藏div工作时,我认为我的语法有错误