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鼠標(biāo)離開時返回原始src

CSS
錢斌斌1年前8瀏覽0評論

上下文:我試圖在鼠標(biāo)經(jīng)過時替換一個圖像，在鼠標(biāo)離開時返回原始圖像。當(dāng)我使用圖像src的原始鏈接時，它是有效的。但我的問題是，我將有多個部分。我知道我可以通過在每個原始src上制作一個腳本并在那里輸入圖像鏈接來使它工作，但是有沒有一種方法可以使它返回到原始img src而不添加img的硬鏈接呢？

<div class="col-12 text-center">
    <h3>TITLE</h3>
    <div class="name-wrapper">
        <span id="name1" class="some-name opacity-0">Name 1</span>
        <span id="name2" class="some-name opacity-0">Name 2</span>
    </div>
</div>
<div class="col-12">
    <div class="img-set">
        <img src="images/2023/some-img.jpg" alt="Some Image" />
    </div>
    <div class="hotspots">
        <a href="#" data-src="images/2023/some-img2.jpg" data-target="name1">test1</a>
        <a href="#" data-src="images/2023/some-img3.jpg" data-target="name2">test2</a>
    </div>
</div>

$(".hotspots a").mouseover( function() {

        var value = $(this).attr('data-src');
        $(".img-set img").attr("src", value);

        var target = $(this).data("target");

        var targetName = document.getElementById(target);

        if($(targetName).attr('id') == target) {
            $("#" + target).removeClass("opacity-0");
          }
    })

    $(".hotspots a").mouseout( function() {

        $( '.img-set img' ).attr("src","images/2023/some-img.jpg");

        var target = $(this).data("target");
        var targetName = document.getElementById(target);

        if($(targetName).attr('id') == target) {
            $("#" + target).addClass("opacity-0");
          }
    });

我怎樣才能使它不是放入實際的src，而是原始的src呢？

所以我試著用這個，但這一切都不起作用了

$(document).ready(function() {

    var original = document.getElementsByClassName(".img-set img").getAttribute("src");

    $(".hotspots a").mouseover( function() {

        var value = $(this).attr('data-src');
        $(".img-set img").attr("src", value);

        var target = $(this).data("target");

        var targetName = document.getElementById(target);

        if($(targetName).attr('id') == target) {
            $("#" + target).removeClass("opacity-0");
          }
    })

    $(".hotspots a").mouseout( function() {

        $( '.img-set img' ).attr("src", original);

        var target = $(this).data("target");
        var targetName = document.getElementById(target);

        if($(targetName).attr('id') == target) {
            $("#" + target).addClass("opacity-0");
          }
    });
    
})

問題是你打了電話

document.getElementsByClassName(".img-set img")

這與您實際需要的元素不匹配。(您得到空值)

請改用

document.querySelector(".img-set img")

或者

$(".img-set img")

有點偏左，但是...有一種方法只使用CSS:has()偽類就可以做到這一點。將兩幅圖像放入你的標(biāo)記中，然后在:懸停選擇器上切換顯示模式。

唯一的警告是，目前firefox或一些舊瀏覽器不支持這一功能。

這里有一個凱文·鮑威爾關(guān)于:has()的很好的視頻，我希望它能解釋一些事情。

下面標(biāo)記的代碼:

#image2 {
  display:none;
}

/*on the body element, if the a tags in the div with class hotspot is hovered then on the element with ID image2 set display to intial */
body:has(.hotspots a:hover) #image2 {
  display:initial;
}

/*similarly for the other one, display none */
body:has(.hotspots a:hover) #image1 {
  display:none;
}

<div class="col-12 text-center">
  <h3>TITLE</h3>
  <div class="name-wrapper">
    <span id="name1" class="some-name opacity-0">Name 1</span>
    <span id="name2" class="some-name opacity-0">Name 2</span>
  </div>
</div>
<div class="col-12">
  <div class="img-set">
    <!-- added both images but we hide the 2nd one intially -->
    <img id='image1' src="https://picsum.photos/id/237/200/200">
    <img id='image2' src="https://picsum.photos/id/238/200/200">
  </div>
  <div class="hotspots">
    <a href="#" id='hover1'>test1</a>
    <a href="#" id='hover2'>test2</a>
  </div>
</div>

只需在第一次懸停時保存原始src，并將其設(shè)置為事件mouseleave:

$(document).on('mouseenter', '.hotspots a', function() {
  const t = $(this);
  const img = t.closest('.hotspots-wrapper').find('.img-set img');
  if ( !img.data('src') ) {
    img.data( 'src', img.attr('src') );
  }
  img.attr( 'src', t.data('src') );
})

$(document).on('mouseleave', '.hotspots a', function() {
  const img = $(this).closest('.hotspots-wrapper').find('.img-set img');
  img.attr( 'src', img.data('src') );
})

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="hotspots-wrapper">
  <div class="img-set">
    <img src="https://picsum.photos/id/230/200/300" alt="Some Image">
  </div>
  <div class="hotspots">
    <a href="#" data-src="https://picsum.photos/id/231/200/300">test1</a>
    <a href="#" data-src="https://picsum.photos/id/232/200/300">test2</a>
  </div>
</div>

你可以只使用jQuery而不使用javascript來選擇原始圖像src。

$(document).ready(function() {

    var original = $(".img-set img").attr("src");

    $(".hotspots a").mouseover( function() {

        var value = $(this).attr('data-src');
        $(".img-set img").attr("src", value);

        var target = $(this).data("target");

        var targetName = document.getElementById(target);

        if($(targetName).attr('id') == target) {
            $("#" + target).removeClass("opacity-0");
          }
    })

    $(".hotspots a").mouseout( function() {

        $( '.img-set img' ).attr("src", original);

        var target = $(this).data("target");
        var targetName = document.getElementById(target);

        if($(targetName).attr('id') == target) {
            $("#" + target).addClass("opacity-0");
          }
    });
    
})

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="col-12 text-center">
    <h3>TITLE</h3>
    <div class="name-wrapper">
        <span id="name1" class="some-name opacity-0">Name 1</span>
        <span id="name2" class="some-name opacity-0">Name 2</span>
    </div>
</div>
<div class="col-12">
    <div class="img-set">
        <img src="https://picsum.photos/id/230/200/300" alt="Some Image" />
    </div>
    <div class="hotspots">
        <a href="#" data-src="https://picsum.photos/id/231/200/300" data-target="name1">test1</a>
        <a href="#" data-src="https://picsum.photos/id/232/200/300" data-target="name2">test2</a>
    </div>
</div>