我試圖創建一個頁面,用於顯示和刷新網絡攝像頭的圖像(camimg.jpg),但如果圖像最近未更新,則會顯示靜態圖像(notlive.jpg)。我已經找到了AJAXCam腳本,該腳本用於在設定的間隔重新加載「活」形象的目的(15秒),但與沒有使用Javascript的經驗,我無法弄清楚如何確定當圖像上次更新以決定是否顯示camimg.jpg或notlive.jpg。確定在Javascript中的圖像的創建日期?
我與編程有限的經驗我想它應該是沿着線的東西:
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代碼(最初由<body onLoad="holdUp()">
的稱呼)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>
是什麼我試圖完成甚至可能?如果是這樣,我將如何去實施它?先謝謝您的幫助!
對於那些誰可能會發現這個問題,並面臨着類似的問題,我發現cavemonkey50的[網絡攝像頭更新腳本](http://cavemonkey50.com/code/webcam_update_script/)。將他的PHP和AJAXCam的腳本結合起來,實現了我原本打算的結果。 – bobby 2010-10-11 20:05:59