Question

当我只声明了一个data-*属性时，该函数运行良好，但如果我的产品（product-image）有多个data-*属性，则该函数不起作用，换句话说，产品图片未显示，会生成错误的.src。

总共我有10种图像组合（2种颜色，每种颜色有5种不同的形状）。

我是javascript / jquery的新手。所以非常感谢一些帮助。

<div class="container">

    <h2>Product Page</h2>

    <div class="product-view">
        <img id="myimg" src="img/square_orange_img.jpg" width="372" height="511">
    </div>

    <div class="options">
        <p>Choose a Color:</p>
        <ul>
            <li><img class="colorbox" data-color="orange" style="background-color:orange;" onclick="changecolor(this)" alt=""></li>
            <li><img class="colorbox" data-color="purple" style="background-color:purple;" onclick="changecolor(this)" alt=""></li>
        </ul>

        <p>Select Shape:</p>
        <ul>
            <li><img class="shapes" data-shape="square" src="img/square-vertical.png" width="50" height="69" onclick="changeshape(this)" alt=""></li>
            <li><img class="shapes" data-shape="round" src="img/round-vertical.png" width="50" height="69" onclick="changeshape(this)" alt=""></li>
            <li><img class="shapes" data-shape="classic" src="img/classic-vertical.png" width="50" height="69" onclick="changeshape(this)" alt=""></li>
            <li><img class="shapes" data-shape="curved" src="img/curved-vertical.png" width="50" height="69" onclick="changeshape(this)" alt=""></li>
            <li><img class="shapes" data-shape="scallop" src="img/scallop-vertical.png" width="50" height="69" onclick="changeshape(this)" alt=""></li>
        </ul>
    </div>

</div>

<script>
function changecolor(elem) {
    var color = elem.getAttribute("data-color");
    var shape = elem.getAttribute("data-shape");
    var newImageSrc = "img/" + shape + "_" + color + "_" + "img.jpg";
    document.getElementById("myimg").src = newImageSrc;
}
function changeshape(elem){
    var color = elem.getAttribute("data-color");
    var shape = elem.getAttribute("data-shape");
    var newImageSrc = "img/" + shape + "_" + color + "_" + "img.jpg";
    document.getElementById("myimg").src = newImageSrc;
}
</script>

Answer 1

你可以将json传递给data属性，不要使用内联javascript，尝试将逻辑与标记分开

data-json='{"color": "blue", "shape": "circle"}'

或使用jquery简单访问DOM

data-color="blue"
data-shape="circle" 

$("selector").data("color")
$("selector").data("shape")


<a href="#" id="elem" data-color="blue" data-shape="circle">click me!</a>

<script>

    $(function(){

        var el = $("#elem"),
            color, shape;

        el.click(function(event){
            event.preventDefault();

            color = $(this).data("color");
            shape = $(this).data("shape");

            changeSomething(color, shape);
        });

        function changeSomething(color, shape){

            console.log(color + " " + shape);
        }

    });

</script>

或

<a href="#" id="elem" data-param='{"color": "blue", "shape": "circle"}'>click me!</a>

<script>

    $(function(){

        var el = $("#elem"),
            param = {};

        el.click(function(event){
            event.preventDefault();

            param = $(this).data("param");

            changeSomething(param);
        });

        function changeSomething(param){

            console.log(param);
        }

    });

</script>

Answer 2

您需要一些变量来保持当前color和shape。然后，每次更改颜色或形状时，您都将更新此变量，然后相应地加载图像。

尽量保留当前代码，您可以将color和shape变量移到函数之外。然后changecolor更改当前颜色，changeshape更改当前形状。您可以将color和shape变量初始化为初始值。在你的情况下橙色和方形。

var color = 'orange';
var shape = 'square';
function changecolor(elem) {
    color = elem.getAttribute("data-color");
    var newImageSrc = "img/" + shape + "_" + color + "_" + "img.jpg";
    document.getElementById("myimg").src = newImageSrc;
}
function changeshape(elem){
    shape = elem.getAttribute("data-shape");
    var newImageSrc = "img/" + shape + "_" + color + "_" + "img.jpg";
    document.getElementById("myimg").src = newImageSrc;
}

查看demo。这不起作用，因为图像不可用但应该在您的环境中工作。

多个html data- *属性和js功能不起作用

2 个答案: