點咩可以整到個xanga有"雪花飄飄"???

script language="JavaScript"
N = 200;
Y = new Array();
X = new Array();
S = new Array();
A = new Array();
B = new Array();
M = new Array();
V = (document.layers)?1:0;
iH=(document.layers)?window.innerHeight:window.document.body.clientHeight;
iW=(document.layers)?window.innerWidth:window.document.body.clientWidth;
for (i=0; i < N; i++){
Y[i]=Math.round(Math.random()*iH);
X[i]=Math.round(Math.random()*iW);
S[i]=Math.round(Math.random()*5+2);
A[i]=0;
B[i]=Math.random()*0.1+0.1;
M[i]=Math.round(Math.random()*1+1);
}if (V){for (i = 0; i < N; i++)
{document.write(" ")}
}else{document.write(&#39;<div style="">&#39;);
document.write(&#39;<div style="">&#39;);
for (i = 0; i &lt; N; i++)
{document.write(&#39;<div style="background:#fffff0;"></div>&#39;)}
document.write(&#39;</div></div>&#39;);}
function snow(){var H=(document.layers)?window.innerHeight:window.document.body.clientHeight;
var W=(document.layers)?window.innerWidth:window.document.body.clientWidth;
var T=(document.layers)?window.pageYOffset:document.body.scrollTop;
var L=(document.layers)?window.pageXOffset:document.body.scrollLeft;
for (i=0; i &lt; N; i++){sy=S[i]*Math.sin(90*Math.PI/180);
sx=S[i]*Math.cos(A[i]);
Y[i]+=sy;
X[i]+=sx;
if (Y[i] &gt; H){Y[i]=-10;
X[i]=Math.round(Math.random()*W);
M[i]=Math.round(Math.random()*1+1);
S[i]=Math.round(Math.random()*5+2);}
if (V){document.layers[&#39;sn&#39;+i].left=X[i];document.layers[&#39;sn&#39;+i].top=Y[i]+T}
else{si[i].style.pixelLeft=X[i];si[i].style.pixelTop=Y[i]+T}
A[i]+=B[i];}setTimeout(&#39;snow()&#39;,10);}window.onLoad=snow()
/script

這語法不需用圖片做飄雪

語法中　N = 200;　200 是設定速度及數量

數字越小速度越快、雪粒相對亦越少

語法中　background: #fffff0　是飄雪的顏色、可修改

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2007-02-09 22:46:20 補充：
因為yahoo ge顯示問題你要o係頭同尾gescript language=&quot;JavaScript&quot;同/script加番個&lt;&gt;即係好似咁&lt; /script &gt;不過唔好有空格

2007-02-16 13:47:46 補充：
http://kkx3.net/或者你直接去e度丫^^