I need my PHP page to show my BLOB image from mysq

Question

So my last question was on how to have my techID shown from a search :

I am trying to have my "Details" page to reference two seperate parts of my server that are linked via techID

My new question is still on this page. I have added in an echo image as well. but am having trouble using Blob and having it display my image and not binary JPEG data. I've been trying to find another instance of this but cannot find any that fix my error.

//Header ('Content-type: image/jpeg')
echo "<dt><strong>Technician Image:</strong></dt><dd>" . '<img src='.$row2['image'].' width="290" height="290">' . "</dd>";

and

$query_Recordset2 = "SELECT * FROM technician WHERE techID=" . $row1["techID"] ;
$Rs2 = mysql_query($query_Recordset2) or die(mysql_error());

Are the only changes I have put in so far from my last question (obviously including the fix I was given that worked).

What I do not understand is where and how to put 'Content-type: image/jpeg' to have my page recognize the image being linked is it's MIME TYPE image/jpeg.

What I am seeing on my page is this

Technician Image: �E��j��i`=7f$D��o"�������b���Ckkc��R��^M�;n~��0&m)J��R��E)JDR��E)JDR��E)JDR��E)JDR��E)JDR��E)JDSjR��)���+��N��.R,u����i��n9,���QX~ ����{(����̮�:���2�12��"��aV7�6���{���LP[�W�����گ� R$+� ��LMc'hM�5�o�PA����|���ګ���.8��E��ģ��Rn ��1�[��{��3>�rY��X�ۜ;�Ǖ����u���z��'�vf�N葟 ��z�Q�����k��3���O��ܨ�ۀ�?S���,N� �����[{+D� �;�'�$�$�&�iJR��)JR��)JR��)JR��)JR��)JR��)JR��)JR��)JR��)JR��)JR��)JR��)�� width="290" height="290">

Obviously I have deleted a middle chunk so it's not massive. there is a little "Broken image" box that appears infront and when I right click and choose "Open image in new window" the URL it puts in is simply Content-type: or I get a forbidden access page with the url http:// localhost/Sim5Server/Pages/%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD%10JFIF%EF%BF%BD%01%02%EF%BF%BD%EF%BF%BDd%EF%BF%BDd%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BD%EF%BF%BDC%EF%BF%BD

I have put a space in that url since it is not a link for the internet.

I have only used normal BLOB type as I just need it as a small less than 64Kb image

Answer 1

The only way you can output an image from the same page as the document is with a data uri.

echo "<dt><strong>Technician Image:</strong></dt><dd>" . 
     '<img src="data:image/jpeg;base64,'.
      base64_encode($row2['image']).
      '" width="290" height="290">' . "</dd>";

Answer 2

I think best solution is to store path of the image in the database instead of storing the whole image as BLOB. And then for showing the image on webpage set the src property of the img tag to the path stored in database.

Answer 3

The right code should be as below. You have to use stream_get_contents() to change resource id into string.

<img src="data:image/jpeg;base64,<?php echo base64_encode(stream_get_contents($row2['image'])); ?>" />

Answer 4

In your current case, you have two upfront options.

The first, and the one I don't recommend if you have numerous images like this, is to use inline base64 encoding. This is done with:

<img src="data:image/jpeg;base64,<?php echo base64_encode($image); ?>" />

A copy/paste version, using your existing code:

echo '<dt><strong>Technician Image:</strong></dt><dd>'
     . '<img src="data:image/jpeg;base64,' . base64_encode($row2['image']) . '" width="290" height="290">'
     . '</dd>';

The second method is to create an "image" PHP file that takes the ID of the image in the database as a query-string parameter and outputs the image. So, your HTML would look something like:

<img src="image.php?id=<?php echo $image_id; ?>" />

And your PHP page would look something similar to:

<?php
$id = (isset($_GET['id']) && is_numeric($_GET['id'])) ? intval($_GET['id']) : 0;
$image = getImageFromDatabase($id); // your code to fetch the image

header('Content-Type: image/jpeg');
echo $image;
?>