window.open参数的问题
<script   Language= "JavaScript ">
function   processForm(form){
var   type;
if   (xf> 70)   type   = "A ";
if   (xf <60   &&   xf> 50)   type   = "B ";
if   (xf <40   &&   xf> 30)   type   = "C ";
if   (xf <20   &&   xf> 10)   type   = "D ";
if   (xf <10)   type   = "E ";
window.open( "frank.asp?Type= "+   type   , "aaa ", " ")
} </script>

为什么TYPE参数传过去结果是undefined？

------解决方案--------------------
因为xf是undefined,if里面的判断都是false,所以type也是undefined