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]

Re: Pulling jpeg name images out of MySQL onto PHP page

Message-ID: <1149628332.042 353.148890@y43g 2000cwc.googleg roups.com> from
MS contained the following:
[color=blue]
>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
>straightforwar d code to show me how to do this? Is there a better way
>that is not too complicated? Remember, I am a newbie.[/color]

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";
$result=mysql_q uery($sql);
while($myrow=my sql_fetch_array ($result)){
echo "<img src='$path".$my row['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.
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[ctclibby]

Re: Pulling jpeg name images out of MySQL onto PHP page


MS wrote:
[snip][color=blue]
> straightforward code to show me how to do this? Is there a better way
> that is not too complicated? Remember, I am a newbie.[/color]

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:

$Image = imagecreatefrom jpeg($ImageName .jpg)

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

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

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

$Query = "select DirFilePath from FileDatabase where
imagename='Imag eName.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!