看看什么原因,通过此asp上传的文件,收到的是二进制数,如何收到文件?
<%
response.expires=0
idate=request.binaryread(request.totalbytes)
dim str1,str2,str3
dim st,st2,st3
for i=1 to len("filename")
str1=str1 & chrb(ascb(mid("filename",i,1)))
next
str2=str2 & chrb(ascb(chr(10))) & chrb(ascb(chr(13)))
st=instrb(1,idate,str1)
st2=instrb(st,idate,str2)
st2=st2+3
str3=str3 & chrb(ascb("-")) & chrb(ascb("-"))
st3=instrb(st2,idate,str3)
st3=st3-2
dim filepath,strpath
dim fi,fi2
fi=instrb(st,idate,chrb(ascb(chr(34))))
fi2=instrb(fi+1,idate,chrb(ascb(chr(34))))
filepath=midb(idate,fi+1,fi2-fi-1)
for i=1 to lenb(filepath)
sss=midb(filepath,i,1)
strpath=strpath & chrw(ascb(sss))
next
fi=instrrev(strpath,"\",-1,1)
strpath=right(strpath,len(strpath)-fi)
'response.write(strpath)
idate=midb(idate,st2,st3-st2)
'set f=createobject("myupload2.upload")
'f.spath="f:\" & strpath
'f.idate=idate
'f.uploadfile
'response.binarywrite(idate)
'set f=createobject("scripting.filesystemobject")
'set fc=f.createtextfile("f:\" & strpath,true)
'for i=1 to lenb(idate)
'fc.write chr(ascb(midb(idate,i,1)))
'next
'fc.close
'response.write(st & "<br>")
'response.write(st2 & "<br>")
'response.write(st3 & "<br>")
'response.write(str1 & "<br>")
'response.write(str2 & "<br>")
'response.write(str3 & "<br>")
'response.write(lenb(idate) & "<br>")
%>
该如何做上传呀?