打开网站提示是否收藏JS代码

• <P>最近客户要打开网站 提示用户是否收藏，本来的想法的 改IE的文件。但是这样容易被认为是木马。</P> <P>所有就有了下面的代码。测试不错！</P>只会运行一次，不管用户点击是否，都会把cookie记录下来。下次来就不会出现了。 这样不会对 用户有多大的印象<br/> <script language="JavaScript"> <!-- var caution = false function setCookie(name, value, expires, path, domain, secure) { var curCookie = name + "=" + escape(value) + ((expires) ? "; expires=" + expires.toGMTString() : "") + ((path) ? "; path=" + path : "") + ((domain) ? "; domain=" + domain : "") + ((secure) ? "; secure" : "") if (!caution || (name + "=" + escape(value)).length <= 4000) document.cookie = curCookie else if (confirm("Cookie exceeds 4KB and will be cut!")) document.cookie = curCookie } function getCookie(name) { var prefix = name + "=" var cookieStartIndex = document.cookie.indexOf(prefix) if (cookieStartIndex == -1) return null var cookieEndIndex = document.cookie.indexOf(";", cookieStartIndex + prefix.length) if (cookieEndIndex == -1) cookieEndIndex = document.cookie.length return unescape(document.cookie.substring(cookieStartIndex + prefix.length, cookieEndIndex)) } function deleteCookie(name, path, domain) { if (getCookie(name)) { document.cookie = name + "=" + ((path) ? "; path=" + path : "") + ((domain) ? "; domain=" + domain : "") + "; expires=Thu, 01-Jan-70 00:00:01 GMT" } } function fixDate(date) { var base = new Date(0) var skew = base.getTime() if (skew > 0) date.setTime(date.getTime() - skew) } var now = new Date() fixDate(now) now.setTime(now.getTime() + 365 * 24 * 60 * 60 * 1000) var visits = getCookie("counter") if (!visits) visits = 1 else visits = parseInt(visits) + 1 setCookie("counter", visits, now) if(visits==1) { window.external.AddFavorite('http://www.phpxd.com/', '小戴编程文章'); } // --> </script>
  
  
• [时间：2010-05-25 21:05:36][点击:]

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