i am using a script, to get the first image, inside a post.
Here is the script.
$first_img = '';
$my1content = $row['post_content'];
$output = preg_match_all('/<img.+src=[\'"]([^\'"]+)[\'"].*>/i', $my1content, $matches);
$first_img = $matches [1] [0];
if(empty($first_img)){ //Defines a default image
$first_img = "/img/default.png";
}
This script echo the full image link, for example: http://mywebsite.com/images/thisistheimage.jpg
is possible to divide the image link , image name , and image extenction so i need to get 3 results
the link, example:http://mywebsite.com/images/ the image name, example: thisistheimage the image extenction, example: .jpg
Please let me know if its all clear, thanks for reading.
<?php
$image_name = pathinfo( $first_img, PATHINFO_FILENAME );
$image_extension = pathinfo( $first_img, PATHINFO_EXTENSION );
$image_with_extension = basename( $first_img );
$image_directory = dirname( $first_img );
?>
Check out the built-in PHP function pathinfo. Looks like just what you need.
You can use the built-in function pathinfo()
to parse the src
for what you want.
$path_parts = pathinfo('/img/default.png');
echo $path_parts['dirname'], "
"; // /img
echo $path_parts['basename'], "
"; // default.png
echo $path_parts['extension'], "
"; // .png
echo $path_parts['filename'], "
"; // default
The PHP reference is here