1

我试图复制一个图像交换，但是当我只希望它在最后一个时，鼠标移出最终在三个交换中的每一个上都是图像 3。任何帮助我弄清楚我如何使交换彼此不同，因此他们不会调用相同的图像，我将不胜感激，谢谢！

//---imageswap1

if(document.images) {
    cars1 = new Array();
    cars1[1] = new Image();
    cars1[1].src = "car4.png";
    cars1[2] = new Image();
    cars1[2].src = "car1.png";
}

function swapping_pics(picture_name, value_2) {
    document.images[picture_name].src = cars1[value_2].src;
}

//---imageswap2

if(document.images) {
    cars2 = new Array();
    cars2[1] = new Image();
    cars2[1].src = "car5.png";
    cars2[2] = new Image();
    cars2[2].src = "car2.png";
}

//---imageswap3

function swapping_pics(picture_name, value_2) {
    document.images[picture_name].src = cars2[value_2].src;
}

if(document.images) {
    cars3 = new Array();
    cars3[1] = new Image();
    cars3[1].src = "car6.png";
    cars3[2] = new Image();
    cars3[2].src = "car3.png";
}

function swapping_pics(picture_name, value_2) {
    document.images[picture_name].src = cars3[value_2].src;
}


<div id="imageswap1" onMouseOver="swapping_pics('car1',1)" onMouseOut="swapping_pics('car1',2)" href="javascript:void">
    <img name="car1" border=”0” src="car1.png" alt="car1">
</div>

<div id="imageswap2" onMouseOver="swapping_pics('car2',1)" onMouseOut="swapping_pics('car2',2)" href="javascript:void">
    <img name="car2" border=”0” src="car2.png" alt="car2">
</div>

<div id="imageswap3" onMouseOver="swapping_pics('car3',1)" onMouseOut="swapping_pics('car3',2)" href="javascript:void">
    <img name="car3" border=”0” src="car3.png" alt="car3">
</div>
4

1 回答 1

1

您不能有多个具有相同名称的函数：swapping_pics，为了解决您的问题，您可以为每个函数添加一个 id，例如：swapping_pics_01、swapping_pics_02、swapping_pics_03。

但这并不能解决你所拥有的混乱，而不是所有的代码，CSS 可以以更好的方式做到这一点......

HTML：

<div id="imageswap1" class="swap"></div>
<div id="imageswap2" class="swap"></div>
<div id="imageswap3" class="swap"></div>

CSS：

// This class "swap" is general to all the divs
.swap {
    width: 500px; // This is the image size
    height: 400px;
}

#imageswap1       { background-image: url("car01.png"); }
#imageswap1:hover { background-image: url("car04.png"); } // Mouse over 1

#imageswap2       { background-image: url("car02.png"); }
#imageswap2:hover { background-image: url("car05.png"); } // Mouse over 2

#imageswap3       { background-image: url("car03.png"); }
#imageswap3:hover { background-image: url("car06.png"); } // Mouse over 3

摆弄这个例子：http: //jsfiddle.net/qnw6j/

于 2013-02-09T03:21:28.153 回答