How to display image path dynamically?

Question

I have the following code

<figure>
<?php if ($data[$i]->thumbnail) { ?>
  <img src="<?php echo $img ?>" width="465" height="300" alt="news-report-juli19" title="" />
<?php } ?>
</figure>

I want to display images from mysql database dynamically, the image_field_name=thumbnail. I have 10 records in db and all have different images.

please read this, it explains it very well: https://codeigniter.com/user_guide/tutorial/news_section.html#display-the-news, where news stands for image — Vickel, Nov 07 '19 at 12:21
hey @Vickel,i am display content and text very well but problem is with image,its img-src doesn't exist and unable to locate image form media folder — mukesh, Nov 07 '19 at 12:27

score 0 · Answer 1 · 2020-06-15T19:23:14.320

0

use query variale with db colmn where images are present

use this in source img src

echo $query_variable['db_coln_name']

edited Jun 15 '20 at 19:23

answered Nov 07 '19 at 15:38

score 0 · Answer 2 · answered Nov 07 '19 at 17:25

0

You should have to given the image path as well when you want to display.

<?php if ($data[$i]->thumbnail) { ?>
  <img src="<?php echo base_url('path_of_img/$img'); ?>" width="465" height="300" alt="news-report-juli19" title="" />
<?php } ?>

answered Nov 07 '19 at 17:25

Anand Pandey

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Happy Son · Answer 3 · 2019-11-22T01:22:27.063

If you use the Codeigniter framework, please try this. In controller

$data['images'] = $this->model->get_images();
$this->load->view('view_name', $data);

In model you should make get_images() method that get image urls.

In view

<?php
    foreach ($images as $value) {
       echo '<img src="'.$value['thumbnail'].'" width="465" 
               height="300" alt="news-report-juli19" title="" />';
    }
?>

Good luck.

How to display image path dynamically?

3 Answers3