'该函数转自互联网
function urldecode(encodestr)
newstr=""
havechar=false
lastchar=""
for i=1 to len(encodestr)
char_c=mid(encodestr,i,1)
if char_c="+" then
newstr=newstr & " "
elseif char_c="%" then
next_1_c=mid(encodestr,i+1,2)
next_1_num=cint("&H" & next_1_c)
if havechar then
havechar=false
newstr=newstr & chr(cint("&H" & lastchar & next_1_c))
else
if abs(next_1_num)<=127 then
newstr=newstr & chr(next_1_num)
else
havechar=true
lastchar=next_1_c
end if
end if
i=i+2
else
newstr=newstr & char_c
end if
next
urldecode=newstr
end function