Question

我一直在网上搜索几个小时，在整个“外部js文件”-jungle中找不到答案。我希望你们能帮忙！

简而言之：我的外部javascript文件似乎没有得到我在main.php文件中定义的变量..

在 main.php 上我定义了php变量并将它们“转换”为 javascript变量

<head>...
<script type="text/javascript">
var phpmain_img = <?php echo json_encode($main_img); ?>;
var phpvar1_large = <?php echo json_encode($var1_large); ?>;
var phpvar2_large = <?php echo json_encode($var2_large); ?>;
var phpvar3_large = <?php echo json_encode($var3_large); ?>;
var phpvar4_large = <?php echo json_encode($var4_large); ?>;
</script>
...
<script language="javascript" type="text/javascript" src="/wshop/ext.js"></script>
</head>

在我的 ext.js 文件中，我想处理这些变量。在里面 ext.js文件我定义了将要使用的函数swapImage（）回到主PHP：

        var imgArray = new Array(
        phpmain_img,
        phpvar1_large,
        phpvar2_large,
        phpvar3_large
        );

    function swapImage(imgID) {
        var theImage = document.getElementById('theImage');
        var newImg;
        newImg = imgArray[imgID];
        theImage.src = newImg;
    }

    function preloadImages() {      
        for(var i = 0; i < imgArray.length; i++) {
            var tmpImg = new Image;
            tmpImg.src = imgArray[i];
        }
    }

结果： main.php 中的swapImage（）无法正常工作

<div id="image">
    <img id="theImage" src="<?=$main_img; ?>" alt="" />
</div>

<div id="thumbs">
<?php

        echo "<img src=\"<$main_img_small\" alt=\"\" onmouseover=\"swapImage(0);\">";
        echo "<img src=\"$var1_small\" alt=\"\" onmouseover=\"swapImage(1);\">";
        echo "<img src=\"$var2_small\" alt=\"\" onmouseover=\"swapImage(2);\">";
        echo "<img src=\"$var3_small\" alt=\"\" onmouseover=\"swapImage(3);\">";

        ?>


    <br />
</div>

非常感谢任何帮助！

更新

我没有收到特定错误，swapImage函数在mouseover时不起作用。但是，我试图用例如输出变量。 document.write(phpimg_main)但似乎没有任何东西让我相信交换变量有问题......

这是源代码浏览器输出

<html>
    <head>
        <link href="../demo.css" rel="stylesheet" type="text/css" />
        <style type="text/css">
            ....
</style>

        <script type="text/javascript">
        var phpmain_img = {"0":"http:\/\/path\/to\/main\/image.jpg"};
        var phpvar1_large = {"0":"http:\/\/path\/to\/image1.jpg"};
        var phpvar2_large = {"0":"http:\/\/path\/to\/image2.jpg"};
        var phpvar3_large = null;
        var phpvar4_large = null;
        </script>   

        <script language="javascript" type="text/javascript" src="/wshop/ext.js"></script>
</head>




<body onload="preloadImages()">

    <div id="image">
        <img id="theImage" src="http://path-to-main-image.jpg" alt="" />
    </div>

    <div id="thumbs">
    <img src="http://path-to-main-image.jpg" alt="" onmouseover="swapImage(0);"><img src="http://path-to-image1.jpg" alt="" onmouseover="swapImage(1);"><img src="http://path-to-image2.jpg" alt="" onmouseover="swapImage(2);">

        <br />
    </div>



</body>

`

更新2：

感谢您的意见和建议！当然，你是对的，我需要一个字符串而不是一个对象，所以编码是一个很好的提示。

然而问题仍然没有用[0]解决。即使我像下面那样硬编码，第二个javascript块（我之前尝试将其外包为外部js文件）也不会获得第一个javascript块中定义的变量。

<script type="text/javascript">
        var phpmain_img = "http://www.abc.de/path-img_main.jpg";
        var phpvar1_large = "http://www.abc.de/path-img1.jpg";
        var phpvar2_large = "http://www.abc.de/path-img2.jpg";
        var phpvar3_large = "http://www.abc.de/path-img3.jpg";
        var phpvar4_large = "http://www.abc.de/path-img4.jpg";
        </script>

<script language="javascript" type="text/javascript">


                   var imgArray = new Array(
                phpmain_img,
                phpvar1_large,
                phpvar2_large,
                phpvar3_large,
                phpvar4_large
            );



            function swapImage(imgID) {
                var theImage = document.getElementById('theImage');
                var newImg;
                newImg = imgArray[imgID];
                theImage.src = newImg;
            }

            function preloadImages() {      
                for(var i = 0; i < imgArray.length; i++) {
                    var tmpImg = new Image;
                    tmpImg.src = imgArray[i];
                }
            }



            </script>

浏览器来源视图输出：

<script type="text/javascript">
        var phpmain_img = "http://www.abc.de/path-img_main.jpg";
        var phpvar1_large = "http://www.abc.de/path-img1.jpg";
        var phpvar2_large = "http://www.abc.de/path-img2.jpg";
        var phpvar3_large = "http://www.abc.de/path-img3.jpg";
        var phpvar4_large = "http://www.abc.de/path-img4.jpg";
        </script>

        <script language="javascript" type="text/javascript"> //this i actually wanted to outsource into an external js-file

        var imgArray = new Array(
            phpmain_img,
            phpvar1_large,
            phpvar2_large,
            phpvar3_large,
            phpvar4_large
        );



        function swapImage(imgID) {
            var theImage = document.getElementById('theImage');
            var newImg;
            newImg = imgArray[imgID];
            theImage.src = newImg;
        }

        function preloadImages() {      
            for(var i = 0; i < imgArray.length; i++) {
                var tmpImg = new Image;
                tmpImg.src = imgArray[i];
            }
        }



        </script>

对不起，这里有一个长话题。我希望你能效仿！我正在尽力学习！

Answer 1

虽然我不是PHP专家，但你的意思是

var phpmain_img = "<?php echo urlencode($main_img); ?>";

而不是

var phpmain_img = <?php echo json_encode($main_img); ?>;

如果查看渲染标记，很明显图像变量不包含您的预期：

var phpmain_img = {"0":"http:\/\/path\/to\/main\/image.jpg"};

而不是字符串，它是一个对象。

外部Javascript文件无法从头部的javascript块接收变量

1 个答案: