Any one outthere done this? Hope so!

Med

I am trying to perform DB operation with images and have used some of the scripts available on-line.

Mysteriously, I get no error messages whatsoever, neither do I get any results. Even when I insert a row manually consisting of an image, and when I attempt to display the query nothing comes up. No image, no error; just a blank page.

Where can I start looking into to find out what's going on?

This is the index page that supposed to view images with a link to upload:

PHP:

<!DOCTYPE HTML PUBLIC
"-//W3C//DTD HTML 4.0 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Browse Upload Files</title>
</head>
<body bgcolor="#CCFF99">

<?php
include 'db.inc';

$query = "SELECT id, shortName, mimeName FROM files";

if (!($connection = @ mysql_pconnect($hostName,

$username,
$password)))
showerror();

if (!mysql_select_db("files", $connection))
showerror();

if (!($result = @ mysql_query ($query, $connection)))
showerror();
?>
<h1>Image database</h1>

<h3>Click <a href="insert.php">here</a> to

upload an image.</h3>
<?php

if ($row = @ mysql_fetch_array($result))
{
?>

<table> 
<col span="1" align="right"> 
<tr> 
   <th>Short description</th> 
   <th>File type</th> 
   <th>Image</th> 

</tr>

<?php
do

{
?>
<tr>
<td><?php echo "{$row["shortName"]}";?></td>

<td><?php echo "{$row["mimeName"]}";?></td>
<td><?php echo "<img src=\"view.php?file={$row["id"]}\">";?></td>
</tr>
<?php
} while ($row = @ mysql_fetch_array($result));
?>
</table>
<?php
} // if mysql_fetch_array()
else
echo "<h3>There are no images to display</h3>\n";
?>
</body>
</html>

And here's the view.php:
PHP:

<?php
include 'db.inc';

$file = clean($file, 4);

if (empty($file))
exit;

if (!($connection = @ mysql_pconnect($hostName,
$username,
$password)))
showerror();

if (!mysql_select_db("files", $connection))
showerror();

$query = "SELECT mimeType, fileContents FROM files

WHERE id = $file";

if (!($result = @ mysql_query ($query,$connection)))
showerror();

$data = @ mysql_fetch_array($result);

if (!empty($data["fileContents"]))
{
// Output the MIME header
header("Content-Type: {$data["mimeType"]}");
// Output the image
echo $data["fileContents"];
}
?>

Here's the table difinition:
PHP:

CREATE TABLE files (
id int(11) NOT NULL auto_increment,
shortName varchar(50) default NULL,
mimeType varchar(30) default NULL,
mimeName varchar(50) default NULL,
fileContents blob,
PRIMARY KEY (id)
) TYPE=MyISAM;

I have a few rows loaded manually, but can't load via the page. Again, no error no result.

Thanks in advance.

dbchip2000

One main thing to do is to remove the @ symbol. It tells PHP not to report errors and that would be why you are not getting any errors. (I am very sure about this, but not actually ever using the @ symbol, I have never have needed to confirm this, so tell me if that was the problem)

Another suggestion to try, to see if the viewer is outputting right is to comment out the header("Content-type: {$data["mimeType"]}"); and then you will be able to view the source of the page. It won't look pretty, but it will tell you if it is outputting things right.

Med

Thanks dbchip2000 for the reply,

I took both your suggestions into account and made the changes. Unfortunately, nothing changed. Still no errors, no results. 😕

daimoore

I have done a similar thing for a content management system i built. I also initially tried to upload file in blob but then changed this and uploaded the img file via a form and stored the path in the database. This may be the way to go for you.

If you wish to do this use the $file variables to create the uploaded image in a dir on the server (make sure you have the permissions set up and safe_mode).

It is also worth adding a delete page where you (or users) can delete the images.

If you want more info on this let me know 🙂

Med

daimoore,

Your suggestion is a more practical choice due to the file type, and I did consider that option. I just didn't know how to establish a link between the DB field (representing a file) and the file itself on the file system.

Could you post the snippet for doing this. Also if you could provide the DB table deifinition.

Appreciate the help.

treds1

to add the image (not the url) you need to apply somehting like this to the image

$image = addslashes(fread(fopen($fileinput,"r"), $filesize));

the $fileinput variable is teh location of your image i use

$fileinput = $_FILES['image']['tmp_name'];

(tmp_name) is the tempory location where php stores teh fiel it is preparing to upload.

daimoore

treds1 has the right idea, do you need me to expand on this med?
I am at work at the mo so am a bit busy but can post later today if you want

Med

Somehow I must've goofed up the mySql table. I just can't pull anything out of this one table. I'll do some digging.

Thanks for the replies guys.

treds1

if you want i can post more detailed code example using the method that i explained before

scuk

How big are your images? Are you sure you don't need a LONGBLOB instead? try

ALTER TABLE `files` CHANGE `fileContents` `fileContents` LONGBLOB NOT NULL

Med

Each image is no more than 100K and I have only 2 or 3 in the table.

Additional sample code would be helpful, and appreciated if you guys don't mind posting it.

Thanks again.