我将如何做这样的工作?
<input name="downvid2" type="button" id="downvid2" onclick="
<?php
header('Content-disposition: attachment; filename=file.pdf');
header('Content-type: application/pdf');
readfile('file.pdf');
?>" value="Download Story" />
问题是我会做一个表单并提交,但我需要一些 PHP 变量的值,我不想在移动页面时丢失它们。
完整的代码是:
<title>Legendmaker - Your Legendmaker Adventure Starring You:</title>
<?php
if( $_POST )
{
$username="***";
$password="*****";
$con = mysqli_connect("storycodes.db.10339998.hostedresource.com",$username,$password);
if (!$con)
{
die('Could not connect: ' . mysqli_error());
}
mysqli_select_db($con, "storycodes");
$code = $_POST['codeInput'];
$code = mysqli_escape_string($con, htmlspecialchars($code)); //May not acually need htmlspecialchars
$query = "SELECT story,video FROM `storycodes` WHERE `code` = '$code'";
$result = mysqli_query($con, $query);
if (mysqli_num_rows($result))
{
$row = mysqli_fetch_assoc($result);
mysqli_free_result($result);
extract($row);
echo $story . $video;
}
else
{
echo "No Data Found. Please check your serial code to ensure that you have not incorrectly entered it. If the code is correct please email the website administrator for further assistance";
}
mysqli_close($con);
}
?>
<div align="center">
<p><span class="linkText"><a href="/index.html">Home</a> <a href="/contact-us.php">Contact Us</a> <a href="/payments.html">Products</a><a href="/products.html"></a></span> </p>
<p> </p>
<h2 class="headingText"><img alt="legendmaker - makes legends: banner" width="728" height="90" /></h2>
<h2 class="headingText"> </h2>
<h2 class="headingText">Your story</h2>
</div>
<p> </p>
<label>
<input type="button" name="downvid" id="downvid" value="Download Video" />
</label>
<input name="downvid2" type="button" id="downvid2" onclick="
<?php
header('Content-disposition: attachment; filename=file.pdf');
header('Content-type: application/pdf');
readfile('file.pdf');
?>" value="Download Story" />
我想我必须让它处理涉及文件 PHP 的所有事情,因为我想让这些文件不能通过网络访问,这样只有那些在以前的表单中输入序列码的人才能访问它。