Images from MySQL

[Brian]

Hi All

I'm trying to store images in a MYSQL db, the images at being uploaded
for an external program in a blob field.
I have had a look round and found a few scripts but none of them seem
to work.
All the images are JPG, and what I need to do is at a given point in a HTML
page I want to say "get this image from this table and display it here". The
trouble
I have been having is some of the scripts what to print out the
header("Content-Type: image/jpeg"); but the page will not allow that because
it
already done all its header.
If anybody has some code that does work that would be great ! :)

Brian

[J.O. Aho]

Brian wrote:

> Hi All
>
> I'm trying to store images in a MYSQL db, the images at being uploaded
> for an external program in a blob field.
> I have had a look round and found a few scripts but none of them seem
> to work.
> All the images are JPG, and what I need to do is at a given point in a HTML
> page I want to say "get this image from this table and display it here". The
> trouble
> I have been having is some of the scripts what to print out the
> header("Content-Type: image/jpeg"); but the page will not allow that because
> it
> already done all its header.
> If anybody has some code that does work that would be great ! :)
>
> Brian
>
>

I have done it in the following manner, first in the page where I have all the
HTML, I have the following link:

<IMG SRC=<?PHP echo "\"img/showimg.php?pic=" . $picid ."&\""; ?>>

and the showimg.php looks something like this:

/* include connect_db() so we make a connection to the database with one
command */
include_once("../libs/db.php");
/* check that we have type and company varibales */
if(strlen($pic)) {
/* connect to database */
$link=connect_db();
/* Do query */
$query="SELECT PictureData, PictureType FROM Pictures WHERE PicID='$picid'
LIMIT 1";
$result=mysql_query($query);
/* save picture to variable */
$picture=mysql_fetch_array($result);
/* close database */
mysql_close($link);
/* send content type and then the data */
header( "Content-type: " . $picture["PictureType"]);
echo $picture["PictureData"];
}
?>


As you see, the picture itself isn't included in the main php page, but has
it's own file, which is loaded each time a picture from the database is
displayed and of course, there aren't any limitation how many pictures at the
time you show (of course long query times and a lagish database can affect this).



//Aho

[ebobnar]

I also had this problem, and I solved it similarly to the way J.O. Aho
did. The problem I am having now is a simple one. If there is no image
found in the database I want to display a "image not found" graphic,
like so:

if (mysql_num_rows($result) < 1)
{
header("Content-type: image/gif");
print("photo_add.gif");
}

else
{
$row = mysql_fetch_array($result);
header("Content-type: " . $row['photo_format']);
print($row['photo_image']);
}

The line "print("photo_add.gif");" is where I am having the problem.
Which is the correct way to phrase this so that photo_add.gif is
returned to the calling file.

The code above is called from another html file like so:

<img src="display_photo.php"></img>

This displays the image fetched from mysql, but not the photo_add.gif
image.

Thanks,

Esoos

[J.O. Aho]

ebobnar wrote:

> I also had this problem, and I solved it similarly to the way J.O. Aho
> did. The problem I am having now is a simple one. If there is no image
> found in the database I want to display a "image not found" graphic,
> like so:
>
> if (mysql_num_rows($result) < 1)
> {

$fullPath="photo_add.gif";
if ($fd = fopen ($fullPath, "rb")) {

> header("Content-type: image/gif");

fpassthru($fd);
fclose($fd);
}

> }
>
> else
> {
> $row = mysql_fetch_array($result);
> header("Content-type: " . $row['photo_format']);
> print($row['photo_image']);
> }
>
> The line "print("photo_add.gif");" is where I am having the problem.
> Which is the correct way to phrase this so that photo_add.gif is
> returned to the calling file.

print is the same as echo and does only handle strings, so to get this to
work, you would need to do the modifications to your code that I have done, I
went a step futher and do only send data if the photo_add.gif is found, in the
case there aren't any picture in the database.
The $fullPath, is from the location from the php file (../img/photo_add.gif)
or an absolute path (/home/user12/pictures/website/photo_add.gif).


//Aho