请问这代码应当怎写????????
图片上传时,获取上传的文件?
提交表单程序:
<TABLE><FORM action=Add_Product_Ok.php method=post enctype="multipart/form-data" >
<tr >
<TD >ID:</TD>
<TD 5>
<input name="ID" type="text" id="ID2" style="font-size: 14px" size="10" >
<INPUT TYPE = "hidden" NAME = "MAX_FILE_SIZE" VALUE ="1000000"></TD>
</tr>
<tr >
<TD c>产品编号:</TD>
<TD ><input name="productname" type="text" id="productname" style="font-size: 14px" size="40" ></TD>
</tr>
<tr>
<TD >图片:</TD>
<TD > <input name="UploadFile" type="file" id="UploadFile"></TD>
</tr>
<TR >
<TD >
<INPUT class=main type=submit size=3 value=' 完成 ' name=Submit2>
</TD>
</TR>
</form>
</TABLE>
下面是处理的:
<?
$UploadFile=$HTTP_POST_VARS['$UploadFile'];
If(($UploadFile != "none")&&($UploadFile != ""))
{
$UploadPath = AddSlashes(dirname($PATH_TRANSLATED))."\\upload\\";
$FileName = $UploadPath.$UploadFile_name; //上载文件名
if($UploadFile_size <1024) //上载文件大小
{
$FileSize = (string)$UploadFile_size . "字节";
}
elseif($UploadFile_size <(1024 * 1024))
{
$FileSize = number_format((double)($UploadFile_size / 1024), 1) . " KB";
}
else
{
$FileSize = number_format((double)($UploadFile_size/(1024*1024)),1)."MB";
}
if(!file_exists($FileName))
{
if(copy($UploadFile,$FileName))
{
echo "文件 $UploadFile_name ($FileSize)上载成功!";
}
else
{
echo "文件 $UploadFile_name上载失败!";
}
unlink($UploadFile);
}
else
{
echo "文件 $UploadFile_name已经存在!";
}
}
else
{
echo "你没有选择任何文件上载!";
}
?>
运行错误提示:
Notice: Undefined index: $UploadFile in E:\PHPWEB\万豪实业\admin\Add_Product_Ok.php on line 7
你没有选择任何文件上载!
没这语句:”$UploadFile=$HTTP_POST_VARS['$UploadFile'];“也是一样的错?
请问这代码应当怎写?