从codeigniter中的路径获取文件名

Question

HI guys i am trying to get the image/video filename from path

Here i have paths in my php variable like this from database

E:/xampp/htdocs/pes/new/movie.mp4
E:/xampp/htdocs/pes/new/flowers.jpg

And i am trying to get new/movie.mp4 or new/flowers.jpg from above paths and display it in img tag or video tag

 <img src="new/flowers.jpg" alt="Trulli" width="500" height="333">

or

<video width="400" controls>
      <source src="new/movie.mp4" type="video/mp4">
      Your browser does not support HTML5 video.
 </video>

Can anyone help me how to do that

Thanks in advance

Answer 1

Try with base_url() . Set your base_url in config.php for example www.xyz.com/ Update your code like this:

<img src="<?php echo base_url() ?>new/flowers.jpg" alt="Trulli" width="500" height="333">

And video frame:

<video width="400" controls>
   <source src="<?php echo base_url() ?>new/movie.mp4" type="video/mp4">
                        Your browser does not support HTML5 video.
</video>

Hope this will help you!

Answer 2

If all paths are relative to FCPATH , FCPATH = 'E:/xampp/htdocs/pes' , you can use:

<img src="<?php echo base_url(str_replace(FCPATH, '', 'E:/xampp/htdocs/pes/new/flowers.jpg')) ?>" alt="Trulli" width="500" height="333">

or

<video width="400" controls>
    <source src="<?php echo base_url(str_replace(FCPATH, '', 'E:/xampp/htdocs/pes/new/movie.mp4')) ?>" type="video/mp4">
    Your browser does not support HTML5 video.
</video>

Answer 3

Hope this will help you :

Set your base_url in your config.php

$config['base_url']='http://localhost/pes/'

and ur image src should be like this :

<img src="<?=site_url('new/'.$image); ?>" alt="Trulli" width="500" height="333">