有个js的数组问题(初级问题)

Question

惨了,基础太不扎实.
刚查到
ie5.5
array.shift
array.unshift

Answer 1

array.splice(startIndex,deletecount) ie5.5

Answer 2

呵呵，顺手给IE5也实现了一个：
<SCRIPT LANGUAGE="JavaScript">
<!--
if (Array.splice==null)
Array.prototype.splice = Splice;
function Splice(a,b){
a = parseInt(a);b=parseInt(b);
if (isNaN(a)||isNaN(b)||a>this.length||a<0||b<0) return null;
if (a+b>this.length) b=this.length-a;
var ar=[],i,l=this.length-b;
for (i=a;i<a+b;i++) ar[i-a]=this[i];
for (i=a;i<l;i++) this[i]=this[i+b];
this.length=l;
return ar;
}
var a=["a","b","c","d","e"];
alert(a.splice(2,3));
alert(a)
//-->
</SCRIPT>

Answer 3

更严谨一点的版本：<SCRIPT LANGUAGE="JavaScript">
<!--
if (Array.splice==null)
Array.prototype.splice = Splice;
function Splice(a,b){
a = parseInt(a);b=parseInt(b);
if (isNaN(a)||isNaN(b)||a>this.length||a<0||b<0) return null;
if (a+b>this.length) b=this.length-a;
var ar=[],rs=[],i,l=this.length-b+arguments.length-2;
for (i=a;i<a+b;i++) rs[i-a]=this[i];
for (var i=2;i<arguments.length;i++) ar[ar.length]=arguments[i];
for (var i=a+b;i<this.length;i++) ar[ar.length]=this[i];
for (var i=0;i<ar.length;i++) this[a+i]=ar[i];
this.length=l;
return rs;
}
var a=["a","b","c","d","e","f"];
alert(a.splice(2,2,"i","j","k"));
alert(a)
//-->
</SCRIPT>