我试图创建一个页面,用于显示和刷新网络摄像头的图像(camimg.jpg),但如果图像最近未更新,则会显示静态图像(notlive.jpg)。我已经找到了AJAXCam脚本,该脚本用于在设定的间隔重新加载“活”形象的目的(15秒),但与没有使用Javascript的经验,我无法弄清楚如何确定当图像上次更新以决定是否显示camimg.jpg或notlive.jpg。确定在Javascript中的图像的创建日期?
我与编程有限的经验我想它应该是沿着线的东西:
pageLoaded = time page loaded in browser
imageUpdated = time the image was uploaded to server
if imageUpdated is not within 20 seconds of pageLoaded:
display notlive.jpg
else:
run AJAXCam
的AJAXCam代码(最初由<body onLoad="holdUp()">
的称呼)如下:
<script type="text/javascript">
<!--
//
//AJAXCam v0.8b (c) 2005 Douglas Turecek http://www.ajaxcam.com
//
function holdUp()
{
//
// This function is first called either by an onLoad event in the <body> tag
// or some other event, like onClick.
//
// Set the value of refreshFreq to how often (in seconds)
// you want the picture to refresh
//
refreshFreq=15;
//
//after the refresh time elapses, go and update the picture
//
setTimeout("freshPic()", refreshFreq*1000);
}
function freshPic()
{
//
// Get the source value for the picture
// e.g. http://www.mysite.com/doug.jpg and
//
var currentPath=document.campic.src;
//
// Declare a new array to put the trimmed part of the source
// value in
//
var trimmedPath=new Array();
//
// Take everything before a question mark in the source value
// and put it in the array we created e.g. doug.jpg?0.32234 becomes
// doug.jpg (with the 0.32234 going into the second array spot
//
trimmedPath=currentPath.split("?");
//
// Take the source value and tack a qustion mark followed by a random number
// This makes the browser treat it as a new image and not use the cached copy
//
document.campic.src = trimmedPath[0] + "?" + Math.random();
//
// Go back and wait again.
holdUp();
}
// -->
</script>
是什么我试图完成甚至可能?如果是这样,我将如何去实施它?先谢谢您的帮助!
对于那些谁可能会发现这个问题,并面临着类似的问题,我发现cavemonkey50的[网络摄像头更新脚本](http://cavemonkey50.com/code/webcam_update_script/)。将他的PHP和AJAXCam的脚本结合起来,实现了我原本打算的结果。 – bobby 2010-10-11 20:05:59