Images from database wont show, headers already sent [duplicate] - php

Question

This question already has answers here:
How to fix "Headers already sent" error in PHP
(11 answers)
Closed 9 years ago.
$sql="SELECT * FROM `blob` ";
$query=mysql_query($sql);
while($row=mysql_fetch_array($query)) {
$image=$row ['image'];
header("content-type: image/jpg");
echo '<img src="path/'.$image.'" width="360" height="150">';
}
This is the code i'm using to display the image uploaded to my database... however, i get the error message that the headers have already been sent... i have no idea what i'm doing wrong!
i can see on my website where my images are meant to be but no actual image!

Answer 1

In general:
When you get this message it seems that there is an output done before. It can even be a simple "whitespace" character like [:space:].
But:
In addition to that you seem to get some things wrong. You do create HTML (!) but you send a header-message that tells the browser that it get's binary data of an image.
So when you really have binary data in your database you have to output binary data and not any html "thingy". I really really recommend you to read a good book about how a server works and what header(..) actually does.
Please also have a look here:
php:Store image into Mysql blob, Good or bad?
http://pujanpiya.com.np/?q=node/25
PS: I did not get into any more detail about your question neither provide a solution because I do not think this would help you rather than solving this issue of misunderstanding temporarily.

Answer 2

Your code is totally wrong; storing images inside a MySQL database is a bad idea and you're using your image data as a path to a nonexistent image; if you still want to stick with the idea of having images in your database, here's example code that would work :
while ($row = mysql_fetch_array($query)) {
echo '<img src="data:image/jpeg;base64,'.base64_encode($row["image"]).'" />';
}
This encodes the image into base64 and then uses that as a data URI inside an img; no headers needed.
Also, mysql is deprecated; use mysqli or PDO instead.

Answer 3

The problem is that you're doing it in a while loop, which will be executed multiple times:
while($row=mysql_fetch_array($query)) {
$image=$row ['image'];
header("content-type: image/jpg"); // Headers are sent here
echo '<img src="path/'.$image.'" width="360" height="150">'; // Content here
}
When mixing headers like this, what you would normally want to do is put the header before any content. However in your case, this wouldn't make sense either because you are sending "content-type: image/jpg". If you are echoing an image to the browser using HTML, the content-type should be text/html (which is sent by PHP by default). So just leave that line out completely. If you are trying to send the actual binary contents of an image, then you wouldn't use an HTML tag, just send the Content-Type header and then echo the contents for one single image.