How do I share the images and the title of the content to social media using PHP?

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PHP

HTML5

Hi all,

I've created one web page in that I need to share some contents to the users via social media for that I've tried,

<ul class='share-story'>
<?php $site_url = "http://" . $_SERVER['HTTP_HOST'] . $_SERVER['REQUEST_URI']; 

echo "<li><a class='tips' data-title='Facebook' href='http://www.facebook.com/sharer.php?u=  ".$site_url."&title=Welcome to our Blog' rel='nofollow' target='_blank'><img src='https://lh4.googleusercontent.com/-ZLBXhszmKcM/UhRopVveCXI/AAAAAAAACgk/A4UMB2mIVpc/s33/fb.jpg'/></a></li>";


echo "<li><a class='tips' data-title='Linkedin' href='https://www.linkedin.com/cws/share?url=".$site_url." '><img src='https://lh6.googleusercontent.com/-ffL-1GnpCbQ/UhRoqObh1DI/AAAAAAAACg4/DPcWbZiXIPM/s33/linkedin.jpg'/></a></li>";


echo "<li><a class='tips' data-title='Twitter'                 href='http://twitter.com/share?url=".$site_url."' rel='nofollow' target='_blank'><img src='https://lh6.googleusercontent.com/-LWZCI166jr4/UhRorVB4MdI/AAAAAAAAChI/V3ir2xRth7U/s33/twitter.jpg'/></a></li>";


echo "<li><a class='tips' data-title='Google+'        href='https://plus.google.com/u/0/share?url=".$site_url."' rel='nofollow' target='_blank'><img src='https://lh6.googleusercontent.com/-GnP5yhTBMQU/UhRop_RqX7I/AAAAAAAACg8/z1oV5uQeGdk/s33/gplus.jpg'/></a></li>";
?>

</ul>

This code run correctly but what is my problem is, it will share only the link of the specified page to the user and does not share any images and the content.


Help me to do this one.


Thanks in advance,
Sanjita