php mysql image not display properly

Question

I have following code for get image from my database table field.

public function getStudentses() {
    $students_data = array();       

    $query = $this->db->query(
        "SELECT * FROM " . DB_PREFIX . "students 
        WHERE customer_id = '" . (int)$this->customer->getId() . "'");   

    foreach ($query->rows as $result) {             
        $students_data[$result['students_id']] = array(
            'students_id'        => $result['students_id'],
            'firstname'          => $result['firstname'],
            'filename'           => $result['filename'],
            'image'              => $result['image'],
        );
    }       
    return $students_data;
}

This is my HTML code for display image:

<img src="<?php echo $result['image']; ?>" />

but when I render my page the image is not display properly its like some symbols like

s4�.���N����萗�p�A@4pbr��]�����F�G�>�v��W

Why its like this? How can I fix my error?

You may need to increase the mine head of the appropriate information — Xingjia Luo, Nov 20 '12 at 09:07
When you post code online, please make sure it is formatted nicely. I've formatted your code, please accept my edit. — Botond Balázs, Nov 20 '12 at 09:12

Prasanth · Accepted Answer · 2012-11-20T09:15:47.693

4

Modern browsers have this support for using raw image data as source.

You can try to use something like:

<img src="data:image/png;<?php echo $result['image']; ?>" />

You should encode the image using base64 and then output something like:

<img src="data:image/png;base64,<?php echo base64_encode($result['image']); ?>" />

I have use image/png for example. If your image is something else, you should use corresponding encType.

Just make sure to read this and this.

edited Nov 20 '12 at 09:15

answered Nov 20 '12 at 09:09

Prasanth

5,230
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61

Shouldn't `$result['image']` be html-escaped? – Vyktor Nov 20 '12 at 09:11
This is the best and the easiest solution I've found ever. Thank you:) – Abdulfattah Alhazmi Jun 09 '14 at 07:30

score 2 · Answer 2 · answered Nov 20 '12 at 09:09

2

You will most likely need to change the header to an image

http://php.net/manual/en/function.header.php

header('Content-Type: image/jpeg');

answered Nov 20 '12 at 09:09

Sandeep Bansal

6,280
17
84
126

1

I suppose it's worth mentioning that in this case the output should also NOT contain any HTML code. – Oleg Nov 20 '12 at 09:17

score 2 · Answer 3 · answered Nov 20 '12 at 09:17

Most common practice for this (what I've seen) is to build html:

<img src="image.png.php?student_id=<?php echo $result['students_id']; ?>" />

And then image.png.php:

$query = $this->db->query("SELECT * FROM " . DB_PREFIX . "students "
            "WHERE students_id = '" .(int)$request->getRequestParam('sudent_id') . "'"); 

// And now display raw content as image
$student = reset( $query->rows);
if( !$student){
   header('HTTP/1.0 404 Not Found');
   die( 'Blah blah... Blah blah blah');
}

header( 'Content-Type: image/jpeg');
echo $student['image'];

This way you'll be able to create links to images usable from whatever place you'll want (without any database access required on remote site).

score 1 · Answer 4 · answered Nov 20 '12 at 09:10

Assuming you are storing the raw binary of the image, you can do this:

'image'              => base64_encode($result['image']),

Then in the html:

<img src="data:image/png;base64,<?php echo $result['image']; ?>" />

This assumes the image is a png. If not, change the mime type accordingly.

score 0 · Answer 5 · edited May 23 '17 at 11:56

if the image is stored by this code

  s4�.���N����萗�p�A@4pbr��]�����F�G�>�v��W

its because you are just copying and pasting this code from database. be sure where it stored the the name/id of the image and the path of the image and get the image by its path and id/name or follow what Prasanth said to you. this my help you How to retrieve images from MySQL database and display in an html tag

php mysql image not display properly

5 Answers5