Accessing Images in a Database

ajg3628

I have a parts database where the user selects a category from a menu then a list of parts in that category are output to the page. If the part includes an accompanying photo, then a "view photo" button appears along with the part description (if the part includes a photo. My problem is that I can't get the image to appear when the button is pushed. I just get a blank screen. Here is a sample of the code that I'm using to try to show the image:

<?php

if ($id) {

$dbcnx = mysql_connect('localhost', 'root','thegman');
mysql_select_db('skymusik_com');

// The basic SELECT statement
$get_image = "select filedata, mimetype from parts where ID='$id'";
$get_image_result = mysql_query($get_image) or die("Couldn't get image.");

$filedata = @mysql_result($get_image_result,0,"filedata");
$mimetype = @mysql_result($get_image_result,0,"mimetype");

echo $filedata;

};

I'm using the <A href tag refering to the above script (showimage.php) with the photo button.

Any thoughts or advice?

Thanks,
Anthony

barand

Write a php script, getimage.php, which gets the data from the table and creates an image.

Display it on the page with HTML <img> tag

hth

ajg3628

I've used the php script that I had previously posted which is called "showimage.php". It is supposed to get an image from the database. Am I on the right track with that? I also have a script that adds images to the database which seems to be working. (Images seem to be getting placed there).

Also, can you explain what you mean by creating an image?

Where does the html go? I assume on the pages where my get photo buttons are.

Thanks a lot!
Anthony

barand

Your script needs content-type header.

Nearly there, see this

hth

treds1

The way i retrieve images from a database is as follows (however you are going about it the right way)

include 'dbcon.php';


$image_brand = $_GET['image_brand'];
$image_name= $_GET['image_name'];

$result = mysql_query("SELECT * FROM image WHERE image_brand='$image_brand' AND image_name='$image_name'",$db);


while($numRow = mysql_fetch_array($result))
{
$itype = $numRow['image_type'];
Header("Content-type: $itype");
$image =$numRow['image_data'];
$height=$numRow['image_height'];
$width=$numRow['image_width'];
echo $image;
flush();
}

then you would view the image by using

inserted into your htmlof php as you woudl to view a normal image.

hope this helps

ajg3628

OK. Thanks a lot. I'll keep working on it.

Only one other note, concerning the Content Header. I keep getting error messages when I place them in the code, something like -
"Header alreay exists or was sent (something like that).

Thanks again!

BeastRider

Originally posted by ajg3628
OK. Thanks a lot. I'll keep working on it.

Only one other note, concerning the Content Header. I keep getting error messages when I place them in the code, something like -
"Header alreay exists or was sent (something like that).

Thanks again!

Your script must allow absolutely nothing, not even a space, before issuing the header, otherwise, a text header is automatically issued, and you get your error message.

This is my image retrieval script which is in use for the thumbnails at this website.

http://www.herbhost.com/seasia/index.htm

<?php
ob_start(); // turn output buffering on
// open the database and retrieve the record
$dbh=mysql_connect ("localhost",
                       "YOUR_USERNAME_HERE",
                       "YOUR_PASS_HERE");
mysql_select_db("YOUR_DBNAME_HERE",$dbh);
$query = "SELECT thumbnail FROM thumbs
          WHERE thumbid = $thumb";
$result = @ mysql_query ($query, $dbh);
$data = @mysql_fetch_array($result);
$image = $data["thumbnail"];
header("Content-Type: image/jpeg");
echo $image;
ob_end_flush();
?>

ajg3628

Hi:
I tried playing around with the tutorial that you were referring to. I created the database, used the two files, stored a couple of images, and I'm still getting a blank page upon retrieval. (The same problem that I've been having with my scripts). Do you know what this could be? I checked the database and the binary data is thre. Do you think that I need to adjust a setting in php.ini?

Thank you

BeastRider

The text file below contains the complete code that runs this website. Perhaps this will solve your problem.

http://www.herbhost.com/seasia/index.htm

ajg3628

Hi:
I'm working with the tutorial that you referred me to (By Florian Dittmer). I think that I'm making some progress. I'm not getting an image yet, however, I am getting the red X where an image is suppose to be. Do you have any clues?

Here is my latest version -

Thanks again!

<html>
<head>
<title>Get The Image</title>
<head>
<body>

<?php

if ($id) {

$dbcnx = mysql_connect('localhost', 'root','thegman');
mysql_select_db('skymusik_com');

// The basic SELECT statement
$get_image = "select filedata, mimetype from parts where ID='$id'";
$get_image_result = mysql_query($get_image) or die("Couldn't get image.");

$filedata = @mysql_result($get_image_result,0,"filedata");
$mimetype = @mysql_result($get_image_result,0,"mimetype");

Header( "Content-type: $mimetype");
echo $filedata;

};

</body>
</html>

BeastRider

You are getting a red X because your script that retrieves the image from the database is outputting more than the image itself.

barand

In showimage.php

<?

if ($id) { 

$dbcnx = mysql_connect('localhost', 'root','thegman'); 
mysql_select_db('skymusik_com'); 


// The basic SELECT statement 
$get_image = "select filedata, mimetype from parts where ID='$id'"; 
$get_image_result = mysql_query($get_image) or die("Couldn't get image."); 

$filedata = @mysql_result($get_image_result,0,"filedata"); 
$mimetype = @mysql_result($get_image_result,0,"mimetype"); 

Header( "Content-type: $mimetype"); 
echo $filedata; 

}
?>

in dispimage.php

<html> 
<head> 
<title>Get The Image</title> 
<head> 
<body> 


<img src="showimage.php?id=$id width=600 height = 400"> 

</body> 
</html>

hth

ajg3628

As always thanks again everyone. I'll keep crackin' at it

BeastRider

Originally posted by ajg3628
As always thanks again everyone. I'll keep crackin' at it

Seriously, in your show image, you cannot have so much as

a) a space line before your <?php statement - this causes a space to be output before the header statement which gives you the header already output message

b) a space or new line after your final ?> statement as this will cause a space or new line to be added to the image. The image will no longer be valid in terms of size as defined by the header and the red X will appear.

It is also pretty worthless to put any error messages into your showimage.php script. After all, they will be interpreted as image data because of the header and all you will ever see is a red X, not errors.

Finally, I found that I had to uses buffers to get the thing to work.

ajg3628

I'm sorry to keep beating this into the gound, but, my dilemma is that I'm trying to use the following script (results.php) with "showimage.php". I'll try to be more specific. The results script outputs the name and description of the part along with a button for more info and a button for a photo if the part has an image in the database. This script is working fine as far as outputting the part description and the buttons. But I still can't get the photo button to work with "showimage.php".

I was told to use this line:
<img src="showimage.php?id=$id width=600 height = 400">

But I've tried this (using A HREF), to output a button if there is an image to accompany the part, because I want to use the button to to display the image on its own page:

<?php if ($image != "") { ?>
<TD><A HREF="showimage.php?ID=$id width = 640 height = 480"><IMG SRC="b-photo.gif"></A></TD>
</TR>
<p>
<?

Here is the whole "results.php:

<html>
<head>
<title> Skymusik Search Results </title>
</head>
<body BGCOLOR="#000000" >
<CENTER><IMG SRC="skygrasm1.jpg"></CENTER><BR>
<h2><Font Color="#C6EFF7">Skymusik Search Results</Font> </h2>
<br><a href="index.php"><Font Color="Red">New Search</Font></a>
<?php

$dbcnx = mysql_connect('localhost', 'root','thegman');
mysql_select_db('skymusik_com');

// The basic SELECT statement
$select = "SELECT DISTINCT ID, name, filedata, mimetype";
$from = " FROM parts";
$where = " WHERE ID > 0";

// As the categories associated with a particular part are stored in the partLookup
// table, we need to add this table to the query to create a join. To do this, we simply
// tack the name of the table onto the end of the $from variable
if ($cid != "") { // A category is selected
// to complete the join we must also specify that the ID column (in the jokes table)
$from .= ", partlookup";
// must match the JID column (in JokeLookup), so we add this condition to the $where variable.
// finally we require the CID column (in JokeLookup) to match the category ID selected in the form $cid)
$where .= " AND ID=PID AND CID=$cid";
}

$parts = mysql_query($select . $from . $where);
if (!$parts) {

echo("<p>Error retrieving parts from database!<br />".
"Error: " . mysql_error() . "</p>");

exit();
}

$num_results = mysql_num_rows($parts);
?>
<p>
<Font Color="#C6EFF7"> <?php echo "Number of parts found: ".$num_results."</p>";
?></Font>

<?php
while ($row = mysql_fetch_array($parts)) {

$name = htmlspecialchars($row["name"]);
$image = ($row["filedata"]);

echo "<strong>";
?>
<HR Size="2" WIDTH="700" ALIGN="center">

<?php if ($image != "") { ?>
<TD><A HREF="showimage.php?ID=$id width = 640 height = 480"><IMG SRC="b-photo.gif"></A></TD>
</TR>
<p>
<?

}
?>
</TABLE>
<HR Size="2" WIDTH="700" ALIGN="center">
<?php

echo "</p>";

}
?>

</table>

</body>
</html>

Does this give you any more detail in what I'm trying to do?
Thanks!

BeastRider

Start simple. Does showimage.php display an image on an otherwise blank page?

ajg3628

No I've never had it working. The closest that I got to was the Red X.

BeastRider

Originally posted by ajg3628
No I've never had it working. The closest that I got to was the Red X.

Then again, the first thing to do is reread my advice above about making sure that your showimage.php script is not putting out extraneous characters. You might try using my script in place of your showimage script... just change the lookup to ID.

I would note that I have never been able to get an image script to work if it has an include in it.

<?php 
ob_start(); // turn output buffering on 
// open the database and retrieve the record 
$dbh=mysql_connect ("localhost", 
                       "YOUR_USERNAME_HERE", 
                       "YOUR_PASS_HERE"); 
mysql_select_db("YOUR_DBNAME_HERE",$dbh); 
$query = "SELECT thumbnail FROM thumbs 
          WHERE thumbid = $thumb"; 
$result = @ mysql_query ($query, $dbh); 
$data = @mysql_fetch_array($result); 
$image = $data["thumbnail"]; 
header("Content-Type: image/jpeg"); 
echo $image; 
ob_end_flush(); 
?>

ajg3628

Hi:

I've been playing around with your script:

<?php

ob_start(); // turn output buffering on
// open the database and retrieve the record
$dbh=mysql_connect ("localhost",

"root",

"thegman");

mysql_select_db("skymusik_com",$dbh);

$query = "SELECT filedata FROM parts

WHERE ID = $id";

$result = @ mysql_query ($query, $dbh);

$data = @mysql_fetch_array($result);

$image = $data["filedata"];

header("Content-Type: image/jpeg");

echo $image;

ob_end_flush();

?>

No spaces, but I'm still getting the red X. The following line of code, from a file called, "results.php" refers to the above file. It provides the functionality of the user pressing a button to view the image:

Is something wrong with this line?

Thanks,
Anthony

cassius

No spaces, but I'm still getting the red X. The following line of code, from a file called, "results.php" refers to the above file. It provides the functionality of the user pressing a button to view the image: <TD><A HREF="showimage2.php?ID=$id width = 640 height = 480"><IMG SRC="b-photo.gif"></A></TD>

Try changing your line to this:
<td><a href="showimage2.php?ID=$id" width="640" height="480"><img src="b-photo.gif"></a></td>

If this is written inside <?php .... ?> you have to write it with single quotes.