Last updated on January 9th, 2020 at 12:31 pm
Today I am going to tell u how to create a simple RSS feed with
Php and MySql. This blog is for those who want to add their website or blog to Google
feed burner.
About Google Feed burner :
Google Feed burner is one of the best place to promote your
blog mainly. It will help you to reach your goal. It help your user to get your
latest news feed you created. You don?t need to add single link one by one it
will automatically access your newly created blog. You simply just create XML file
that contains your all news feed links of your blog.
Simple Xml File Structure :
<?xml version="1.0" encoding="ISO-8859-1"?> <rss version="2.0"> <channel> <title>Site Title</title> <link> Your Blog address</link> <description>Little Description about your blog</description> <language>en-us</language> <copyright>Copyright (C) </copyright> <item> <title>Your Blog Title </title> <description><![CDATA[ Short Description of your blog ]]></description> <link>Complete link of your blog</link> </item> </channel> </rss>
First create a table in your data base
Xmlfeed.sql
-- -- Table structure for table `feed` -- CREATE TABLE IF NOT EXISTS `feed` ( `sr` int(11) NOT NULL AUTO_INCREMENT, `title` text NOT NULL, `desc` text NOT NULL, `link` text NOT NULL, PRIMARY KEY (`sr`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ; - - Dumping data for table `feed` -- INSERT INTO `feed` (`sr`, `title`, `desc`, `link`) VALUES (1, 'welcome to trinity blogs', 'Trinity blog is best programming blog for Beginners', 'your-blog-links.php'), (2, 'About trinity blog', 'Trinity blogs are focused on window application, PHP, MySql ', 'your-page-link2.php');
Now we write code in Php and simply fetch the data from feed
table and insert into XML layout.
I hope you have basic knowledge of Php and Mysql.
Feed.php
<?php
/*
*
Simple and easy
*/
$conn = mysql_connect("localhost",
"root", "");
$select_db = mysql_select_db("xmlfeed");
header("Content-Type: application/xml;
charset=UTF-8"); // To output file in xml Format
$rssfeed = '<?xml version="1.0"
encoding="ISO-8859-1"?>';
$rssfeed .= '<rss version="2.0">';
$rssfeed .= '<channel>';
$rssfeed .= '<title>Trinity Blog</title>';
$rssfeed .=
'<link>http://www.trinityblog.in</link>';
$rssfeed .= '<description>RSS feed of
trinityblog.in</description>';
$rssfeed .= '<language>en-us</language>';
$rssfeed .= '<copyright>Copyright (C)
trinityblog.in</copyright>';
$query = "SELECT * FROM `feed` ORDER BY sr DESC";
$result = mysql_query($query) or die ("Could not
execute query");
while($row = mysql_fetch_array($result)) {
extract($row);
/* Get
category name */
$rssfeed
.= '<item>';
$rssfeed
.= '<title>' . $title . '</title>';
$rssfeed
.= '<description><![CDATA['. $desc. ']]></description>';
$rssfeed
.= '<link>' . $link. '</link>';
$rssfeed
.= '</item>';
}
$rssfeed .= '</channel>';
$rssfeed .= '</rss>';
echo $rssfeed;
?>And Final Output
This XML file does
not appear to have any style information associated with it. The document tree
is shown below.
<rss version="2.0"> <channel> <title>Trinity Blog</title> <link>http://www.trinityblog.in</link> <description>RSS feed of trinityblog.in</description> <language>en-us</language> <copyright>Copyright (C) trinityblog.in</copyright> <item> <title>About trinity blog</title> <description> <![CDATA[ Trinity blogs are focused on window application, PHP, MySql ]]> </description> <link>your-page-link2.php</link> </item> <item> <title>welcome to trinity blogs</title> <description> <![CDATA[ Trinity blog is best programming blog for Beginners ]]> </description> <link>your-blog-links.php</link> </item> </channel> </rss>
Smart and simpleJ