当我从数据库 select 一个文件时，下载 link 没有生成
Download link doesn't generate when I select a file from database
我尝试生成 link 以便从我的数据库下载文件。我首先使用 select 下拉列表中的一个文件，然后它会将您定向到另一个页面，并在该页面中创建您的 link 但是当我想这样做时，如果我单击下载文件 returns 我上一页蚂蚁它没有创建 link。
这是我的 download.php:
<html>
<body>
<title>Download your friend's uploads</title>
<form action="download2.php" method="post">
<?php
session_start();
$username =$_SESSION["uname"];
?>
<?php
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql= mysqli_query($con, "SELECT imagesnotes FROM userdata where sharedpeople='$username'");
echo "File or Image";
echo'<select name="imagesnotes">';
echo'<option value="" selected="selected">Select a File</option>';
while($row = mysqli_fetch_array($sql))
{
echo'<option value="' . $row['imagesnotes'] . '">'. $row['imagesnotes'] .'</option>';
}
echo'</select></p><p>';
mysqli_close($con);
?>
<td width="80"><input name="download" type="submit" class="box" id="download" value=" download "></td>
</form>
</body>
</html>
从那里用户 select 一张图片并点击提交按钮。
这是我的 download2.php:
<?php
session_start();
$username =$_SESSION["uname"];
?>
<?php
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$imagesnotes = $_POST['imagesnotes'];
$query = "SELECT imagesnotes From userdata where imagesnotes='$imagesnotes' and sharedpeople='$username'";
$res = mysqli_query($con,$query);
$res or die('Error, query failed');
if(mysqli_num_rows($res) == 0){
echo "<br>Database is empty </br>";
}
else{
while(list($id) = mysqli_fetch_array($res)){
?>
<br><a href="download.php?id=<?php=$id;?>">download your file</a></br>
<?php
}
}
mysqli_close($con);
?>
在这里，它必须创建一个link来下载文件。我做错了什么？你能帮帮我吗？谢谢。
更新：
所以我的uploaded2.php最终版本是这样的：
<?php
session_start();
$username =$_SESSION["uname"];
?>
<?php
if(isset($_GET['imagesnotes']))
{
// if id is set then get the file with the id from database
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$id = $_GET['imagesnotes'];
$query = "SELECT imagesnotes, type, size, content FROM datas WHERE imagesnotes = '$id'";
$result = mysqli_query($query) or die('Error, query failed');
list($name, $type, $size, $content) = mysqli_fetch_array($result);
header("Content-length: $size");
header("Content-type: $type");
header("Content-Disposition: attachment; filename=$name");
echo $content;
mysqli_close($con);
}
?>
<?php
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$imagesnotes = $_POST['imagesnotes'];
$query = "SELECT imagesnotes From userdata where imagesnotes='$imagesnotes' and sharedpeople='$username'";
$res = mysqli_query($con,$query);
$res or die('Error, query failed');
if(mysqli_num_rows($res) == 0){
echo "<br>Database is empty </br>";
}
else{
while(list($id) = mysqli_fetch_array($res)){
?>
<br><a href="download.php?id=<?php echo $id;?>">download your file</a></br>
<?php
}
}
mysqli_close($con);
?>
但它仍将我带到上一页。为什么会这样？
更新 2：所以当我下载文件时，我下载了损坏的文件。这是我的 uploadfile.php:
<html>
<body>
<?php
session_start();
$username =$_SESSION["uname"];
?>
<title>Uploading Page</title>
<b>Hello again,<?php echo $username ?>. From here, you can select your image or file and upload our database.</b>
<form method="post" action ="uploaded.php" enctype="multipart/form-data">
<table width="350" border="0" cellpadding="1" cellspacing="1" class="box">
<tr>
<td width="246">
<input type="hidden" name="MAX_FILE_SIZE" value="2000000">
<input name="userfile" type="file" id="userfile">
</td>
<td width="80"><input name="upload" type="submit" class="box" id="upload" value=" Upload "></td>
</tr>
</table>
</form>
</body>
</html>
这是我的 uploaded.php 来确认更新操作：
<html>
<body>
<title>Uploading Page</title>
<?php
session_start();
$username =$_SESSION["uname"];
?>
<?php
$con=mysqli_connect("localhost","root","","webpage");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0)
{
$fileName = $_FILES['userfile']['name'];
$tmpName = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
$fp = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);
if(!get_magic_quotes_gpc())
{
$fileName = addslashes($fileName);
}
$query = "INSERT INTO datas (username,imagesnotes,size,type,content) ".
"VALUES ('$username','$fileName', '$fileSize', '$fileType', '$content')";
$res=mysqli_query($con,$query);
$res or die('Error, query failed');
echo "<br>File $fileName uploaded<br> <a href = 'default.php'>Return</a> ";
}
mysqli_close($con);
?>
</body>
</html>
我做错了什么？谢谢。
更新 3：现在我将文件上传到服务器而不是数据库。但现在我遇到了一个问题，我无法下载该文件 link。我怎样才能做到这一点？
这是我的 downloadyours.php:
<html>
<body>
<title>Download your uploads</title>
<?php
session_start();
$username =$_SESSION["uname"];
?>
<form action="downloadyours2.php" method="post">
<?php
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql= mysqli_query($con, "SELECT imagesnotes FROM datas where username='$username'");
echo "File or Image";
echo'<select name="imagesnotes">';
echo'<option value="" selected="selected">Select a File</option>';
while($row = mysqli_fetch_array($sql))
{
echo'<option value="' . $row['imagesnotes'] . '">'. $row['imagesnotes'] .'</option>';
}
echo'</select></p><p>';
mysqli_close($con);
?>
<td width="80"><input name="download" type="submit" class="box" id="download" value=" download "></td>
</form>
</body>
</html>
这是我的 downloadyours2.php:
<!DOCTYPE html>
<html>
<head>
<title>Your download link is ready</title>
</head>
<body>
<?php
// No need to query the database again, you get the filename from POST data
echo '<br /><a href="downloadedyours.php?imagesnotes=' . $_POST['imagesnotes'] . '">Download your file</a><br />';
?>
</body>
</html>
这是我的 downloadedyours.php:
<?php
session_start();
$username =$_SESSION["uname"];
?>
<?php
if(isset($_GET['imagesnotes']))
{
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$imagesnotes = $_GET['imagesnotes'];
// Set database name
$db = "webpage";
// If $_GET['imagesnotes'] is set then get the file with that filename from database
// Expecting a single result row: let's use mysqli_fetch_array(query(...), MYSQLI_ASSOC) function
$sql = mysqli_fetch_array(mysqli_query($con, "SELECT type, size FROM datas WHERE imagesnotes='$imagesnotes'"), MYSQLI_ASSOC);
header("Content-length:" . $sql['size']);
header("Content-Disposition: attachment; filename=" . $imagesnotes);
header("Content-Transfer-Encoding: Binary");
header("$target_file");
echo $sql['content'];
}
mysqli_close($con);
?>
</body>
</html>
<?php=$id;?>
应该是
<?php echo $id; ?>
就这样：
header('Location: uploaddir/'.$_GET['id']);
例如，如果文件是：document.docx
header('Location: uploaddir/document.docx');
会起作用
如果 .htcaccess 配置正确，该文件将自动下载。
因此，如果这对您还不起作用：
How to configure .htcaccess
我在你的 download.php（第一个文件）中没有看到 $_GET['id'] 的管理并且你在 [=118= 中创建了一个 link 到 download.php（第一个文件） ]（第二个文件）。所以它只打印 download.php 页面内容。
您应该重定向到管理 $_GET['id'] 的另一个页面（download3.php？）以重定向到真实文件 link，或者首先在 download.php 中执行并打印 download.php 如果没有提供 $_GET['id']。
如果您还没有这样做，请像这样在您的 MySQL table 中分配和 ID：
id | filename
-----------------
1 | myfile.zip
2 | myfile2.rar
然后将其放在 download.php 之上或新的 download3.php 中（不要忘记在 download2.php 中上传 link）：
<?php
if ($_GET['id'] != "") {
$id = $_GET['id'];
// Query your database to get the filename corresponding to $id and save it in a $filename variable
header('Location: uploads/'.$filename);
// Dont forget to change "uploads" to your real uploads folder
}
?>
还有linkid打印在download2.php是错误的，使用：<?php echo $id; ?>
编辑：根据您更新的问题。请阅读我写的代码里面的注释。
文件名: "mysql.php"
目的：压缩 mysqli 函数，避免在更改数据库时编辑每个文件user/password。
注释: 编辑 define() 变量来设置你的数据库 user/password.
<?php
define("MYSQL_SERVER", "localhost");
define("MYSQL_USER", "root");
define("MYSQL_PASS", "YourPassword");
// Query function to use when expecting single result row or no result
// USAGE when needing query result:
// $result = mysqli_fetch_array(query("db_name", "SELECT column1, column2 FROM table_name WHERE column3='$something'"), MYSQLI_ASSOC);
// or
// USAGE when NOT needing query result:
// query("db_name", "INSERT INTO table_name (column1, column2) VALUES ('$column1_value','$column2_value') ");
function query($db, $arg) {
$con = mysqli_connect(MYSQL_SERVER, MYSQL_USER, MYSQL_PASS, $db)
or die("Cannot connect: " . mysqli_error());
$query = sprintf($arg);
$result = mysqli_query($con, $query);
mysqli_close($con);
}
// Creates a multidimensional array from query result, to use when expecting multiple result rows
// $type is the array type, accepted values are MYSQLI_ASSOC or MYSQLI_NUM or MYSQLI_BOTH
// usually MYSQLI_ASSOC is good
// USAGE:
// mysqli_fetch_alls("db_name", "SELECT column1, column2 FROM table_name WHERE column3='$something'", MYSQLI_ASSOC);
function mysqli_fetch_alls($db, $arg, $type) {
$data = query($db, $arg);
$result = array();
$i = 0;
while($row = mysqli_fetch_array($data, $type))
{
$result[$i] = $row;
$i++;
}
return $result;
}
?>
文件名: "download.php"
目的：创建一个包含所有可下载文件列表的表单。
注意事项: 检查mysqli_fetch_alls()并确保数据库table名称正确。
<!DOCTYPE html>
<html>
<head>
<title>Download your friend's uploads</title>
</head>
<body>
<form action="download2.php" method="post">
<?php
// Let's load mysql.php where are defined MySQL database parameters and query functions
if (!defined('ABSPATH'))
define('ABSPATH', dirname(__FILE__) . '/');
require_once(ABSPATH . 'mysql.php');
session_start();
$username = $_SESSION["uname"];
// Set database name
$db = "webpage";
// Query the database. Expecting multiple result rows: let's use mysqli_fetch_alls() function
$sql = mysqli_fetch_alls($db, "SELECT imagesnotes FROM userdata WHERE sharedpeople='$username'", MYSQLI_ASSOC);
echo "File or Image";
echo'<select name="imagesnotes">';
echo'<option value="" selected>Select a File</option>';
// Print results
for ($i = 0; $i < count($sql); $i++)
echo'<option value="' . $sql[$i]['imagesnotes'] . '">'. $sql[$i]['imagesnotes'] .'</option>';
echo'</select>';
?>
<input name="download" type="submit" class="box" id="download" value="download">
</form>
</body>
</html>
文件名: "download2.php"
目的：打印link下载文件。
备注：没什么好说的。
<!DOCTYPE html>
<html>
<head>
<title>Your download link is ready</title>
</head>
<body>
<?php
// No need to query the database again, you get the filename from POST data
echo '<br /><a href="download3.php?imagesnotes=' . $_POST['imagesnotes'] . '">Download your file</a><br />';
?>
</body>
</html>
文件名: "download3.php"
用途：打印文件内容。
注意事项: 检查query()并确保数据库table名称正确。
<!DOCTYPE html>
<html>
<head>
<title>Download your friend's uploads</title>
</head>
<body>
<?php
// Let's load mysql.php where are defined MySQL database parameters and query functions
if (!defined('ABSPATH'))
define('ABSPATH', dirname(__FILE__) . '/');
require_once(ABSPATH . 'mysql.php');
session_start();
$username = $_SESSION["uname"];
// $_GET['imagesnotes'] is the GET parameter name printed in download2.php
// as "download3.php?imagesnotes=filename"
// If it was "download3.php?somethingelse=filename"
// you should have looked for $_GET['somethingelse']
if(isset($_GET['imagesnotes']))
{
$imagesnotes = $_GET['imagesnotes'];
// Set database name
$db = "webpage";
// If $_GET['imagesnotes'] is set then get the file with that filename from database
// Expecting a single result row: let's use mysqli_fetch_array(query(...), MYSQLI_ASSOC) function
$sql = mysqli_fetch_array(query($db, "SELECT type, size, content FROM datas WHERE imagesnotes='$imagesnotes'"), MYSQLI_ASSOC);
header("Content-length:" . $sql['size']);
header("Content-type:" . $sql['type']);
header("Content-Disposition: attachment; filename=" . $imagesnotes);
echo $sql['content'];
}
?>
</body>
</html>
回复您的更新 3
首先确保你的上传功能正常。这意味着您应该通过上传 php 脚本上传文件并通过 FTP 客户端下载文件，以查看文件是否已正确上传。如果是这样，那么请着手修复您的下载功能。
您的下载功能有几个错误。
$_FILES["fileToUpload"]["name"]
你为什么期望这会重新运行某些内容？ - 参见 PHP $_FILES
header("$target_file");
没有任何意义 - 请参阅 PHP header() function
echo $sql['content'];
没有打印任何内容，因为您没有存储任何内容。
下载功能应该如下所示。无需调用数据库，您将文件名作为 $_GET 参数传递。
<?php
session_start();
$username = $_SESSION["uname"];
if(isset($_GET['imagesnotes'])) {
$filename = $_GET['imagesnotes'];
$dir = "uploads/";
$file = $dir . $filename;
if (file_exists($filename)) {
header('Content-Description: File Transfer');
header('Content-Disposition: attachment; filename=' . $filename);
header('Content-Transfer-Encoding: Binary');
header('Expires: 0');
header('Cache-Control: must-revalidate');
header('Pragma: public');
header('Content-Length: ' . filesize($file));
ob_clean();
flush();
readfile($file);
}
else
echo "File not found!";
exit;
}
?>
我尝试生成 link 以便从我的数据库下载文件。我首先使用 select 下拉列表中的一个文件，然后它会将您定向到另一个页面，并在该页面中创建您的 link 但是当我想这样做时，如果我单击下载文件 returns 我上一页蚂蚁它没有创建 link。
这是我的 download.php:
<html>
<body>
<title>Download your friend's uploads</title>
<form action="download2.php" method="post">
<?php
session_start();
$username =$_SESSION["uname"];
?>
<?php
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql= mysqli_query($con, "SELECT imagesnotes FROM userdata where sharedpeople='$username'");
echo "File or Image";
echo'<select name="imagesnotes">';
echo'<option value="" selected="selected">Select a File</option>';
while($row = mysqli_fetch_array($sql))
{
echo'<option value="' . $row['imagesnotes'] . '">'. $row['imagesnotes'] .'</option>';
}
echo'</select></p><p>';
mysqli_close($con);
?>
<td width="80"><input name="download" type="submit" class="box" id="download" value=" download "></td>
</form>
</body>
</html>
从那里用户 select 一张图片并点击提交按钮。
这是我的 download2.php:
<?php
session_start();
$username =$_SESSION["uname"];
?>
<?php
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$imagesnotes = $_POST['imagesnotes'];
$query = "SELECT imagesnotes From userdata where imagesnotes='$imagesnotes' and sharedpeople='$username'";
$res = mysqli_query($con,$query);
$res or die('Error, query failed');
if(mysqli_num_rows($res) == 0){
echo "<br>Database is empty </br>";
}
else{
while(list($id) = mysqli_fetch_array($res)){
?>
<br><a href="download.php?id=<?php=$id;?>">download your file</a></br>
<?php
}
}
mysqli_close($con);
?>
在这里，它必须创建一个link来下载文件。我做错了什么？你能帮帮我吗？谢谢。
更新：
所以我的uploaded2.php最终版本是这样的：
<?php
session_start();
$username =$_SESSION["uname"];
?>
<?php
if(isset($_GET['imagesnotes']))
{
// if id is set then get the file with the id from database
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$id = $_GET['imagesnotes'];
$query = "SELECT imagesnotes, type, size, content FROM datas WHERE imagesnotes = '$id'";
$result = mysqli_query($query) or die('Error, query failed');
list($name, $type, $size, $content) = mysqli_fetch_array($result);
header("Content-length: $size");
header("Content-type: $type");
header("Content-Disposition: attachment; filename=$name");
echo $content;
mysqli_close($con);
}
?>
<?php
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$imagesnotes = $_POST['imagesnotes'];
$query = "SELECT imagesnotes From userdata where imagesnotes='$imagesnotes' and sharedpeople='$username'";
$res = mysqli_query($con,$query);
$res or die('Error, query failed');
if(mysqli_num_rows($res) == 0){
echo "<br>Database is empty </br>";
}
else{
while(list($id) = mysqli_fetch_array($res)){
?>
<br><a href="download.php?id=<?php echo $id;?>">download your file</a></br>
<?php
}
}
mysqli_close($con);
?>
但它仍将我带到上一页。为什么会这样？
更新 2：所以当我下载文件时，我下载了损坏的文件。这是我的 uploadfile.php:
<html>
<body>
<?php
session_start();
$username =$_SESSION["uname"];
?>
<title>Uploading Page</title>
<b>Hello again,<?php echo $username ?>. From here, you can select your image or file and upload our database.</b>
<form method="post" action ="uploaded.php" enctype="multipart/form-data">
<table width="350" border="0" cellpadding="1" cellspacing="1" class="box">
<tr>
<td width="246">
<input type="hidden" name="MAX_FILE_SIZE" value="2000000">
<input name="userfile" type="file" id="userfile">
</td>
<td width="80"><input name="upload" type="submit" class="box" id="upload" value=" Upload "></td>
</tr>
</table>
</form>
</body>
</html>
这是我的 uploaded.php 来确认更新操作：
<html>
<body>
<title>Uploading Page</title>
<?php
session_start();
$username =$_SESSION["uname"];
?>
<?php
$con=mysqli_connect("localhost","root","","webpage");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0)
{
$fileName = $_FILES['userfile']['name'];
$tmpName = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
$fp = fopen($tmpName, 'r');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);
if(!get_magic_quotes_gpc())
{
$fileName = addslashes($fileName);
}
$query = "INSERT INTO datas (username,imagesnotes,size,type,content) ".
"VALUES ('$username','$fileName', '$fileSize', '$fileType', '$content')";
$res=mysqli_query($con,$query);
$res or die('Error, query failed');
echo "<br>File $fileName uploaded<br> <a href = 'default.php'>Return</a> ";
}
mysqli_close($con);
?>
</body>
</html>
我做错了什么？谢谢。
更新 3：现在我将文件上传到服务器而不是数据库。但现在我遇到了一个问题，我无法下载该文件 link。我怎样才能做到这一点？
这是我的 downloadyours.php:
<html>
<body>
<title>Download your uploads</title>
<?php
session_start();
$username =$_SESSION["uname"];
?>
<form action="downloadyours2.php" method="post">
<?php
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql= mysqli_query($con, "SELECT imagesnotes FROM datas where username='$username'");
echo "File or Image";
echo'<select name="imagesnotes">';
echo'<option value="" selected="selected">Select a File</option>';
while($row = mysqli_fetch_array($sql))
{
echo'<option value="' . $row['imagesnotes'] . '">'. $row['imagesnotes'] .'</option>';
}
echo'</select></p><p>';
mysqli_close($con);
?>
<td width="80"><input name="download" type="submit" class="box" id="download" value=" download "></td>
</form>
</body>
</html>
这是我的 downloadyours2.php:
<!DOCTYPE html>
<html>
<head>
<title>Your download link is ready</title>
</head>
<body>
<?php
// No need to query the database again, you get the filename from POST data
echo '<br /><a href="downloadedyours.php?imagesnotes=' . $_POST['imagesnotes'] . '">Download your file</a><br />';
?>
</body>
</html>
这是我的 downloadedyours.php:
<?php
session_start();
$username =$_SESSION["uname"];
?>
<?php
if(isset($_GET['imagesnotes']))
{
$con = mysqli_connect("localhost", "root", "", "webpage");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$imagesnotes = $_GET['imagesnotes'];
// Set database name
$db = "webpage";
// If $_GET['imagesnotes'] is set then get the file with that filename from database
// Expecting a single result row: let's use mysqli_fetch_array(query(...), MYSQLI_ASSOC) function
$sql = mysqli_fetch_array(mysqli_query($con, "SELECT type, size FROM datas WHERE imagesnotes='$imagesnotes'"), MYSQLI_ASSOC);
header("Content-length:" . $sql['size']);
header("Content-Disposition: attachment; filename=" . $imagesnotes);
header("Content-Transfer-Encoding: Binary");
header("$target_file");
echo $sql['content'];
}
mysqli_close($con);
?>
</body>
</html>
<?php=$id;?>
应该是
<?php echo $id; ?>
就这样：
header('Location: uploaddir/'.$_GET['id']);
例如，如果文件是：document.docx
header('Location: uploaddir/document.docx');
会起作用 如果 .htcaccess 配置正确，该文件将自动下载。 因此，如果这对您还不起作用：
How to configure .htcaccess
我在你的 download.php（第一个文件）中没有看到 $_GET['id'] 的管理并且你在 [=118= 中创建了一个 link 到 download.php（第一个文件） ]（第二个文件）。所以它只打印 download.php 页面内容。
您应该重定向到管理 $_GET['id'] 的另一个页面（download3.php？）以重定向到真实文件 link，或者首先在 download.php 中执行并打印 download.php 如果没有提供 $_GET['id']。
如果您还没有这样做，请像这样在您的 MySQL table 中分配和 ID：
id | filename
-----------------
1 | myfile.zip
2 | myfile2.rar
然后将其放在 download.php 之上或新的 download3.php 中（不要忘记在 download2.php 中上传 link）：
<?php
if ($_GET['id'] != "") {
$id = $_GET['id'];
// Query your database to get the filename corresponding to $id and save it in a $filename variable
header('Location: uploads/'.$filename);
// Dont forget to change "uploads" to your real uploads folder
}
?>
还有linkid打印在download2.php是错误的，使用：<?php echo $id; ?>
编辑：根据您更新的问题。请阅读我写的代码里面的注释。
文件名: "mysql.php"
目的：压缩 mysqli 函数，避免在更改数据库时编辑每个文件user/password。
注释: 编辑 define() 变量来设置你的数据库 user/password.
<?php
define("MYSQL_SERVER", "localhost");
define("MYSQL_USER", "root");
define("MYSQL_PASS", "YourPassword");
// Query function to use when expecting single result row or no result
// USAGE when needing query result:
// $result = mysqli_fetch_array(query("db_name", "SELECT column1, column2 FROM table_name WHERE column3='$something'"), MYSQLI_ASSOC);
// or
// USAGE when NOT needing query result:
// query("db_name", "INSERT INTO table_name (column1, column2) VALUES ('$column1_value','$column2_value') ");
function query($db, $arg) {
$con = mysqli_connect(MYSQL_SERVER, MYSQL_USER, MYSQL_PASS, $db)
or die("Cannot connect: " . mysqli_error());
$query = sprintf($arg);
$result = mysqli_query($con, $query);
mysqli_close($con);
}
// Creates a multidimensional array from query result, to use when expecting multiple result rows
// $type is the array type, accepted values are MYSQLI_ASSOC or MYSQLI_NUM or MYSQLI_BOTH
// usually MYSQLI_ASSOC is good
// USAGE:
// mysqli_fetch_alls("db_name", "SELECT column1, column2 FROM table_name WHERE column3='$something'", MYSQLI_ASSOC);
function mysqli_fetch_alls($db, $arg, $type) {
$data = query($db, $arg);
$result = array();
$i = 0;
while($row = mysqli_fetch_array($data, $type))
{
$result[$i] = $row;
$i++;
}
return $result;
}
?>
文件名: "download.php"
目的：创建一个包含所有可下载文件列表的表单。
注意事项: 检查mysqli_fetch_alls()并确保数据库table名称正确。
<!DOCTYPE html>
<html>
<head>
<title>Download your friend's uploads</title>
</head>
<body>
<form action="download2.php" method="post">
<?php
// Let's load mysql.php where are defined MySQL database parameters and query functions
if (!defined('ABSPATH'))
define('ABSPATH', dirname(__FILE__) . '/');
require_once(ABSPATH . 'mysql.php');
session_start();
$username = $_SESSION["uname"];
// Set database name
$db = "webpage";
// Query the database. Expecting multiple result rows: let's use mysqli_fetch_alls() function
$sql = mysqli_fetch_alls($db, "SELECT imagesnotes FROM userdata WHERE sharedpeople='$username'", MYSQLI_ASSOC);
echo "File or Image";
echo'<select name="imagesnotes">';
echo'<option value="" selected>Select a File</option>';
// Print results
for ($i = 0; $i < count($sql); $i++)
echo'<option value="' . $sql[$i]['imagesnotes'] . '">'. $sql[$i]['imagesnotes'] .'</option>';
echo'</select>';
?>
<input name="download" type="submit" class="box" id="download" value="download">
</form>
</body>
</html>
文件名: "download2.php"
目的：打印link下载文件。
备注：没什么好说的。
<!DOCTYPE html>
<html>
<head>
<title>Your download link is ready</title>
</head>
<body>
<?php
// No need to query the database again, you get the filename from POST data
echo '<br /><a href="download3.php?imagesnotes=' . $_POST['imagesnotes'] . '">Download your file</a><br />';
?>
</body>
</html>
文件名: "download3.php"
用途：打印文件内容。
注意事项: 检查query()并确保数据库table名称正确。
<!DOCTYPE html>
<html>
<head>
<title>Download your friend's uploads</title>
</head>
<body>
<?php
// Let's load mysql.php where are defined MySQL database parameters and query functions
if (!defined('ABSPATH'))
define('ABSPATH', dirname(__FILE__) . '/');
require_once(ABSPATH . 'mysql.php');
session_start();
$username = $_SESSION["uname"];
// $_GET['imagesnotes'] is the GET parameter name printed in download2.php
// as "download3.php?imagesnotes=filename"
// If it was "download3.php?somethingelse=filename"
// you should have looked for $_GET['somethingelse']
if(isset($_GET['imagesnotes']))
{
$imagesnotes = $_GET['imagesnotes'];
// Set database name
$db = "webpage";
// If $_GET['imagesnotes'] is set then get the file with that filename from database
// Expecting a single result row: let's use mysqli_fetch_array(query(...), MYSQLI_ASSOC) function
$sql = mysqli_fetch_array(query($db, "SELECT type, size, content FROM datas WHERE imagesnotes='$imagesnotes'"), MYSQLI_ASSOC);
header("Content-length:" . $sql['size']);
header("Content-type:" . $sql['type']);
header("Content-Disposition: attachment; filename=" . $imagesnotes);
echo $sql['content'];
}
?>
</body>
</html>
回复您的更新 3
首先确保你的上传功能正常。这意味着您应该通过上传 php 脚本上传文件并通过 FTP 客户端下载文件，以查看文件是否已正确上传。如果是这样，那么请着手修复您的下载功能。
您的下载功能有几个错误。
$_FILES["fileToUpload"]["name"]
你为什么期望这会重新运行某些内容？ - 参见 PHP $_FILES
header("$target_file");
没有任何意义 - 请参阅 PHP header() function
echo $sql['content'];
没有打印任何内容，因为您没有存储任何内容。
下载功能应该如下所示。无需调用数据库，您将文件名作为 $_GET 参数传递。
<?php
session_start();
$username = $_SESSION["uname"];
if(isset($_GET['imagesnotes'])) {
$filename = $_GET['imagesnotes'];
$dir = "uploads/";
$file = $dir . $filename;
if (file_exists($filename)) {
header('Content-Description: File Transfer');
header('Content-Disposition: attachment; filename=' . $filename);
header('Content-Transfer-Encoding: Binary');
header('Expires: 0');
header('Cache-Control: must-revalidate');
header('Pragma: public');
header('Content-Length: ' . filesize($file));
ob_clean();
flush();
readfile($file);
}
else
echo "File not found!";
exit;
}
?>