dongpinyao2203 2017-02-16 09:19
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jquery最接近特定div内的图像

I used the following HTML in my page

<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 script:-

$( ".edge-img-sec" ).click(function(){
    alert('aa');
    $(this).closest('img').find('.edge-icon1-right').hide();
});

By clicking the edge-img-sec div i want to hide the corresponding edge-icon1-right image.

I used the above code. But it is not working. What I am doing wrong here. please help me.

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  • dongmuzhan4705 2017-02-16 09:26
    关注

    You can do it like below (using children()):-

    $( ".edge-img-sec" ).click(function(){
        $(this).children('.edge-icon1-right').hide();
    });
    <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>

    Or using find():-

    $( ".edge-img-sec" ).click(function(){
        $(this).find('.edge-icon1-right').hide();
    });
    <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>

    </div>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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