Pulling jpeg name images out of MySQL onto PHP page

[MS]

I am new to PHP and MySql. What I would like to do is have my PHP page
pull the name of an image (jpeg, bmp, etc.) out of MySQL and display
image from another directory. I don't want the actual image saved in
the DB, just the name.

I have the connection established from the Web Page to MySql, all I
want is how to display the image part. Does anyone have any
straightforward code to show me how to do this? Is there a better way
that is not too complicated? Remember, I am a newbie.

Thanks.

[Geoff Berrow]

Message-ID: <11**********************@y43g2000cwc.googlegroups .com> from
MS contained the following:

I have the connection established from the Web Page to MySql, all I
want is how to display the image part. Does anyone have any
straightforward code to show me how to do this? Is there a better way
that is not too complicated? Remember, I am a newbie.

Assuming you are storing image names and alt text, this should work

$path="/images/"; // absolute path to images directory
$sql="SELECT imagefield, alt_text FROM imagetable";
$result=mysql_query($sql);
while($myrow=mysql_fetch_array($result)){
echo "<img src='$path".$myrow['imagefield']."'
alt='".$myrow['alt_text']."'><br>";
}

--
Geoff Berrow (put thecat out to email)
It's only Usenet, no one dies.
My opinions, not the committee's, mine.
Simple RFDs http://www.ckdog.co.uk/rfdmaker/

[ctclibby]

MS wrote:
[snip]

straightforward code to show me how to do this? Is there a better way
that is not too complicated? Remember, I am a newbie.

Straightforward? Boy, am not sure if you want to display this image
inline or as an image src...

For inline, you need to open the image file:

$Image = imagecreatefromjpeg($ImageName.jpg)

..... do stuff with the image here if you wish...

header("content-type: image/jpeg")
imagejpeg($Image)
?>

Image already exists as a .jpg on disk NOT in a database, create a link
to it.

$Query = "select DirFilePath from FileDatabase where
imagename='ImageName.jpg'";
$db = $chk->query($Query);
.....db error glue stuff here.
$info = $chk->fetchrow();

for ease of use, combine your DirFilePath with ImageName.jpg
( /usr/local/images/imagename.jpg ) or something like that.
<img src="/usr/local/images/imagename.jpg">
..... more html/php code here.

your results may vary!