取数据库文件名

Question

本帖最后由 u013650821 于 2014-03-11 18:06:54 编辑

Answer 1

用超链接只能下载浏览器不认识的扩展名，对于html/txt 等浏览器认识的扩展名需要用header来下载。

Answer 2

echo "$row[0],$row[1],<a href=\"download.php?filename=".urlencode($row[2])."\">(Download)</a><br><br>";download.php$filename=$_GET['filename'];
header('content-type:application/octet-stream');  
header('content-disposition:attachment; filename='.$filename); 
echo readfile("upload/".$filename);

Answer 3

echo readfile("upload/".$filename); 改 readfile("upload/".$filename);

Answer 4

取不到文件下载呀! 下载的是download.php

Answer 5

改成了如下，可是老是下载download.pdf，不是数据库里的文件名
<?php
$filename=$_GET['filename'];
header('content-type:application/pdf');  
header('content-disposition:attachment; filename='.$filename); 
readfile("upload/".$filename);
?>echo "$row[0],$row[1],<a href=\"download.php?filename=".urlencode($row[2])."\">(Download)</a><br><br>";

Answer 6

产生的连接是
www.mydoamain.com/download.php?filename=

Answer 7

原用是download.asp, 只是想改成用 php，为何php反而复杂了?sqlString = "SELECT * FROM upload ORDER BY uploaddate DESC"
set rs = conn.execute(sqlString)

Do While Not rs.EOF
Response.Write("<p><font size=""2"" face=""Times new roman"">")
Response.Write(rs("uploaddate"))
Response.Write(", ")
Response.Write(rs("topic"))
Response.Write(" <a href=""upload/" & rs("filename") & """ target=""_blank"">(Download)</a></font></p>" )
Response.Write(vbCrLf)
rs.movenext()
Loop

Answer 8

echo "$row[0],$row[1],<a href=\"download.php?filename=".urlencode($row[2])."\">(Download)</a><br><br>";
产生的是
www.mydoamain.com/download.php?filename=path/name
这样的连接而
header('Content-disposition:attachment; filename='.$filename); 
中的 $filename 是不能带路径的，如果带有路径，则提示的文件名是 download.php