unable to fetch image from the database

Question

I am trying to display images on my web page, the content is getting fetched from my database, but the issue I'm facing is in displaying the image. please, can anyone guide me how should I display the image?

I mean to say the path what I should give

here 'image' is my column name and this is my view

        <?php 
            if( !empty($results) ) {

                foreach($results as $row) {?>
                <div class="col-sm-4">
                  <img src="<?php echo base_url('uploads/');$image?>" alt="">
                  <h3><?php echo $row->title; ?></h3>
                  <p><?php echo $row->content; ?></p>
                </div> 
               <?php 

               } ?>

            <?php }
        ?>

what is name of column for the image in your table? i mean from where `$image` is coming it is from table or it is static — Pradeep, Jul 05 '18 at 11:06
from table im fetcing the image i mean to say the uploaded image is saved in the upload folder and the name is saved in database as 1.jpg,2.jpg like this — keerthi patil, Jul 05 '18 at 11:12
what is $image? is it a name of the image file on drive or encoded format like base64? — Mangesh Sathe, Jul 05 '18 at 11:13

Pradeep · Accepted Answer · 2018-07-05T11:19:19.447

2

Hope this will help you :

Use this : <img src="<?php echo base_url('uploads/'.$row->image);?>" alt="">

The whole code should be like this :

<?php 
            if( !empty($results) ) {

                foreach($results as $row) {?>
                <div class="col-sm-4">
                  <img src="<?php echo base_url('uploads/'.$row->image);?>" alt="">
                  <h3><?php echo $row->title; ?></h3>
                  <p><?php echo $row->content; ?></p>
                </div> 
               <?php 

               } ?>

            <?php }
        ?>

edited Jul 05 '18 at 11:19

answered Jul 05 '18 at 11:18

Pradeep

9,667
13
27
34

1

pleasure is mine happy coding , pls check it as green also – Pradeep Jul 05 '18 at 11:20
https://stackoverflow.com/questions/51220674/unable-to-store-the-multiselect-data-to-database – keerthi patil Jul 07 '18 at 06:50
can u check this question @pradeep – keerthi patil Jul 07 '18 at 06:50

Archana · Answer 2 · 2018-07-05T11:20:30.627

1

If $image is the column name(which contains image name) then Replace

<img src="<?php echo base_url('uploads/');$image?>" alt="">

with

 <img src="<?php echo base_url();?>uploads/<?php echo $row->image;?>" alt="">

edited Jul 05 '18 at 11:20

answered Jul 05 '18 at 11:11

Archana

233
1
2
7

Ahsan Zameer · Answer 3 · 2018-07-05T11:11:19.830

0

If $image has image path then remove ';' and add '.' on echo base_url('uploads/');$image?> line

edited Jul 05 '18 at 11:11

answered Jul 05 '18 at 11:10

Ahsan Zameer

29
2

Not the only issue and already has an answer to this affect, please see my answer – Can O' Spam Jul 05 '18 at 11:10

score -1 · Answer 4 · answered Jul 05 '18 at 11:09

You issue here is the following code;

<?php echo base_url('uploads/');$image?>

This is because you are not concatenating the string, rather, just saying it's variable name, so, use of the following will provide the output;

<?php echo base_url('uploads/') . $image' ?>

This replaces the semi-colon part way through for a period (.) which is the PHP concatenation operator

Obviously, there is the issue you are not setting $image, which would require (before usage) of;

$image = $row['image_column_name'];
// Or
$image = $row->img_column_name

unable to fetch image from the database

4 Answers4