Question

我正在建立一个流动的网站。如果您使用其中的一些图像，那将是一个地狱。 IE有问题，或FF或Chrome。

我又遇到了一个问题。基本上我是在运行时通过javascript构建网站。我将宽度放在首位。之后，我正在设置主容器的高度。（否则图像会出现突破问题）。

现在我的问题是：如果我在本地执行网页，它可以在所有3个浏览器中使用。如果我在线执行它，它不会设置主容器的高度（包含所有内容）。（var wrapperHeight = objLogo.offsetHeight; - ＆gt;返回0）

如果我刷新网页，一切看起来都正常，上面一行返回有效高度...... 我正在使用它和一个简单的jquery脚本来翻转div（但是这是在我的简单脚本之后）。

注意：我也在身体上使用相当大的背景图像，这是在css中设置的.. 第二个注意事项：它只发生在Chrome ...

代码：

<head>
    <title></title>
    <link rel="stylesheet" href="css/default.css" type="text/css" />
    <script type="text/javascript" src="scripts/jquery-1.3.2.js"></script>
    <script type="text/javascript" src="scripts/jquery.quickflip.js"></script>
    <script type="text/javascript">
        $('document').ready(function(){
                            //getting inner width/height
            var windowWidth = window.innerWidth;
            var windowHeight = window.innerHeight;

                            //getting width for the main container
            var wrapperWidth = schermBreedte*0.641;

                            //getting width for left and right div in main container
            var totalWrapperWidth = wrapperWidth - 40;
            var widthLeft = totalWrapperWidth*0.345;
            var widthRight = totalWrapperWidth*0.655;

                            //getting elements
            var objOrange = document.getElementById('orange');
            var objGreen = document.getElementById('green');
            var objFliptabs = document.getElementById('flip-tabs');
            var objLeft = document.getElementById('left');
            var objRight = document.getElementById('right');
            var objContent = document.getElementById('content');
            var objLogo = document.getElementById('main_logo');
            var objV_line = document.getElementById('v_line');

                            //setting main container (objOrange & ObjGreen are 2 main containers used for the flip jquery script, they are actually placed above eachother (see jquery ref above and script beneath)
            objOrange.style.width = wrapperWidth + "px";
            objGreen.style.width = wrapperWidth + "px";
            objFliptabs.style.width = wrapperWidth + "px";

                            //setting the left & right div
            objLeft.style.width = widthLeft + "px";
            objRight.style.width = widthRight + "px";

                            //this logo is placed in the left div. The actual main container height is determined by getting the height of this logo after setting the width
            objLogo.style.width = (widthLeft-20)+"px";

                            //the right div is splitted into two pieces: a top which contains 6 images and a bottom which contains 1 image, the objectContent refers to the bottom as the top height is set dynamically by setting the width of the 6 images in %
            objContent.style.width = widthRight  + "px";

                            //getting the height for the main container by getting the height of the image (we've set the width of the image earlier - see above - so that it is scaled proportionally)
            var wrapperHeight =  objLogo.offsetHeight;
                            //setting the height of the main containers
            objOrange.style.height = wrapperHeight + "px";
            objGreen.style.height = wrapperHeight + "px";

                           //setting the height of a 1px line next to the left div
            objV_line.style.height = 100 + "%"

                            //getting the height of the content div (=2nd and bottom div of the right div)
            var contentHeight = wrapperHeight*0.445;

                            //setting the height
            objContent.style.height = contentHeight + "px";

                            //another line
            var length = parseInt(document.getElementsByTagName("div")["right"].offsetWidth) - 20;
            var line = document.getElementById('hor_lijn');
            line.style.width = lengte + "px";   

            $('#flip-container').quickFlip();
            $('#flip-navigation li a').each(function(){
                $(this).click(function(){
                    $('#flip-navigation li').each(function(){
                        $(this).removeClass('selected');
                    });
                    $(this).parent().addClass('selected');
                    var flipid=$(this).attr('id').substr(4);
                    $('#flip-container').quickFlipper({ }, flipid, 1);
                    return false;
                });
            });


        });
    </script>
</head>

Answer 1

看起来您需要等待运行该代码，直到加载图像以便知道它们的大小，因此可以正确计算高度。

尝试使用此jQuery插件：

https://github.com/desandro/imagesloaded

$(document).ready(function() {
    $('#my-image-container').imagesLoaded(function() {
        /* your code here */
    });
});

Answer 2

当触发document.ready时，浏览器不一定在内存中有图像来确定它们的高度/宽度。

一种选择是将调整大小逻辑放在$（窗口）.ready（）而不是$（document）.ready（）中。两者之间的区别在于window.ready将等待加载所有内容。

Javascript / jQuery运行得太快 - 适用于重新加载页面

2 个答案: