Converting MySQL rows to XML with PHP - php

Question

I'm working on my php script to output the information from mysql database. I want to output these echo results in my php page as a xml file as i want to make it looks like this:
<?xml version="1.0" encoding="UTF-8" ?>
<tv generator-info-name="www.mysite.com/xmltv">
<channel id="">
<display-name></display-name>
<programme channel="" start="" stop="">
<title lang="en"></title>
<sub-title lang="en"></sub-title>
<desc lang="en"></desc>
<category lang="en"></category>
</programme>
</channel>
Here's my PHP:
<?php
function db_connect()
{
define('DB_HOST', 'localhost');
define('DB_USER', 'myusername');
define('DB_PASSWORD', 'mypasword');
define('DB_DATABASE', 'mydbname');
$errmsg_arr = array();
$errflag = false;
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$link)
{
die('Failed to connect to server: ' . mysql_error());
}
$db = mysql_select_db(DB_DATABASE);
if(!$db)
{
die("Unable to select database");
}
}
db_connect();
function clean($var)
{
return mysql_real_escape_string(strip_tags($var));
}
$channels = clean($_GET['channels']);
$id = clean($_GET['id']);
if($errflag)
{
$_SESSION['ERRMSG_ARR'] = $errmsg_arr;
echo implode('<br />',$errmsg_arr);
}
else
{
$insert = array();
if(isset($_GET['channels']))
{
$insert[] = 'channels = \'' . clean($_GET['channels']) .'\'';
}
if(isset($_GET['id']))
{
$insert[] = 'id = \'' . clean($_GET['id']) . '\'';
}
if($channels && $id)
{
$qrytable1="SELECT id, channels, links FROM tvguide WHERE channels='$channels' && id='$id'";
$result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error());
while ($row = mysql_fetch_array($result1))
{
zap2it($row);
}
mysql_close();
}
else if(!$channels && ! $id)
{
$qrytable1="SELECT id, channels, links, streams FROM tvguide";
$result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error());
echo '<?xml version=""1.0"" encoding="UTF-8" ?>';
echo '<tv generator-info-name="www.mysite.com/xmltv">';
echo '<channel id="">';
echo '<display-name></display-name>';
echo '<programme channel="" start="" stop="">';
echo '<title lang="en"></title>';
echo '<sub-title lang="en"></sub-title>';
echo '<desc lang="en"></desc>';
echo '<category lang="en"></category>';
echo '</programme>';
echo '</channel>';
while ($row = mysql_fetch_array($result1))
{
echo "<p id='channels'>".$row["id"]. " " . $row["channels"]. "</p>";
echo "<p id='links'>";
echo "http://www.mysite.com/get-listing.php?channels=" . $row["channels"] . "&id=" . $row["id"] .'</p>';
}
}
}
?>
Here's what my php output looks like:
<?xml version="1.0" encoding="UTF-8" ?><tv generator-info-name="www.mysite.com/xmltv"><channel id=""><display-name></display-name><programme channel="" start="" stop=""><title lang="en"></title><sub-title lang="en"></sub-title><desc lang="en"></desc><category lang="en"></category></programme></channel>
Edit: I have got a problem with echo. I can't be able to echo for the channels in the database.
if(!$channels && ! $id)
{
$qrytable1="SELECT id, channels, links, streams FROM tvguide";
$result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error());
while ($row = mysql_fetch_array($result1))
{
}
mysql_close();
}
Here's the error I have got: error on line 12 at column 11: XML declaration allowed only at the start of the document
I will get the error when I'm using this under the while statement:
while ($row = mysql_fetch_array($result1))
{
echo '<?xml version="1.0" encoding="UTF-8" ?>
<tv generator-info-name="www.mysite.com/xmltv">
<channel id="">
<display-name></display-name>
<programme channel="" start="" stop="">
<title lang="en"></title>
<sub-title lang="en"></sub-title>
<desc lang="en"></desc>
<category lang="en"></category>
</programme>
</channel>
</tv>';
}
The output for my php is show as blank page. How do you use the code to allow me to print these xml output in my php?

Answer 1

if channel = 0 and id = 0 your output is correct. It may not be tabbed, but it is correct to your logic.
Depending on the browser XML might show a blank page ... make sure you do view code!
You need to rethink your logic though $channel && $id and then do !$channel && !id is suspiciously odd.

Answer 2

It looks like you are already successfully generating XML. Your problem is that it shows up as a blank page. This is expected, because your web browser is going to interpret the XML as a bunch of HTML tags, and tags don't get displayed.
What you probably want to do is set the content type to XML so that the web browser knows it's XML and not HTML. You can do this by adding the header before you send any other output (i.e. before any echo statements):
header('Content-type: application/xml');
This is necessary because if you don't specify the content type explicitly, the server will automatically send a content type header that tells the browser that the content is HTML.
You can use your browser's debugging console (press F12 to open it in most modern browsers) to inspect the HTTP headers that your script is sending and verify that it's declaring the correct content type.
The other alternative is to format the XML as HTML. This probably isn't really desirable, because the purpose of XML is the be processed by some other client and turning it into HTML makes it useless. However, if you really want to, it can be done by putting all the XML in a string and then calling the htmlspecialchars function to format it as HTML. (This turns the < and > into HTML entities, causing them to be displayed in your browser.)