[Resolved] Unable to display pictures in a table

rachelkoh

Hi I'm new to PHP and MySQL. Please help me with this... Thanks alot! I'm having problems with displaying images whose binary data are stored in database. I'm using <img src="simple_picture.php"> to produce the image. If you have seen my code, I'm trying to pass $pictureId to simple_picture.php so that the query can be carried out. There are no forms so I can't use HTTP_POST_VARS.

I would like to display a table like this:

Picture Filename

imageA ABC.gif
imageB XYZ.gif
imageC qwert.jpeg

My limitations:
1) register_global=off.
2) actual image binary data must be stored in database.

So far, I've written this:

//get the pictures this user has saved
if (!($conn = db_connect()))
return false;
$result = mysql_query( "select pictureId, filename
from pictures
where username = '$valid_user'");
if (!$result)
echo "Error retrieving data.";

//create an array of the result
$picture_array = array();
for ($count = 0; $row = mysql_fetch_row ($result); ++$count)
{
$picture_array[$count] = $row;
}

//display the table of pictures
?>
<br><br>
<table border="10" cellpadding=2 cellspacing=0>
<tr>
<td>Picture</td>
<td>File name</td>
</tr>
<?php
if (is_array($picture_array) && count($picture_array)>0)
{
foreach ($picture_array as $picture)
{

echo "<tr><td>";
session_register("pictureId");
$pictureId = $picture[0];
?><img src = "simple_picture.php"><?php
echo "</td>";
echo "<td>".$picture[1]."</td>";
echo "</tr>";
}
}
else
{
echo '</table>';
echo "<tr><td>No pictures on record</td></tr>";
}
?>
</table>

AND IN SIMPLE_PICTURE.PHP:

<?php
require_once("all_functions.php");
session_start();
if (!($conn = db_connect()))
return false;
$valid_user = $HTTP_SESSION_VARS["valid_user"];
$pictureId = $HTTP_SESSION_VARS["pictureId"];

if(!$pictureId)
echo $pictureId;
$result = mysql_query( "select photoBinary, filetype
from pictures
where username = '".$valid_user."' and pictureId = '".$pi."'")
or die("Couldn't get image");

$photoBinary = @mysql_result($result,0,"photoBinary");
$filetype = @mysql_result($result,0,"filetype");

mysql_free_result($result);

Header("Content-type: $filetype");
echo "$photoBinary";
?>

THE ERROR IS:
Always a red X appear in the Picture field.

steadyguy

Have you tried the GET vars? (ie $GET superglobal array?) That is, build it into the URL: <img src="source.php?id=1" /> Then just use $GET['id'] to find which graphic is being requested with a query something like:

$query = "SELECT binaryData FROM imagesTable WHERE id=" . $_GET['id'];

Be sure to check that the value is strictly within the range of allowed values - in this case, probably numbers only, otherwise you'll be opening yourself up to SQL injections.

A bit off the topic of your question, but IMHO, there are much better methods than storing the image binary data in a database, not to mention a few good reasons not to.

rachelkoh

So for my case, I should put

<img src = "simple_picture.php?id=".$pictureId>

And in simple_picture.php

$result = mysql_query( "select photoBinary, filetype

from pictures

where username = '".$valid_user."' and pictureId =" . $_GET['pid'])

  or die("Couldn't get image");

I've tried and I still get the RED X in the picture field 😕

I couldn't write to the local file directory so got to store images in a database. 🙁

rachelkoh

sorry, again:

In simple_picture.php:

<?php
require_once("all_functions.php");
session_start();
if (!($conn = db_connect()))
    return false;
$valid_user = $HTTP_SESSION_VARS["valid_user"]; 

$result = mysql_query( "select photoBinary, filetype 
                          from pictures
                          where username = '".$valid_user."' and pictureId =" . $_GET['id'])
  or die("Couldn't get image");

$photoBinary = @mysql_result($result,0,"photoBinary");
$filetype = @mysql_result($result,0,"filetype");

mysql_free_result($result);

Header("Content-type: $filetype");
echo "$photoBinary";
?>

steadyguy

Try using the image functions:

<?php
$user = 'username';
$pass = 'password';
if ($conn = mysql_connect('localhost', $user, $pass)) {
	$query = "SELECT photoBinary, filetype FROM pictures WHERE username='$valid_user' AND pictureId=" . $_GET['id'];
	if ($result = mysql_query($query)){
		$arr = mysql_fetch_array($result);
		extract ($arr); // now you have $photoBinary, and $filetype
		$src = imagecreatefromjpeg($photoBinary); // create an image from the data
		header("Content-type: $filetype"); //send the filetype - but if it's not 'image/jpeg' it won't display properly
		imagejpeg($src); //send the image
		imagedestroy($src); //clean up
	}
	else
		print "Couldn't create image!";
}
else
	print "Couldn't connect to database!";
?>

rachelkoh

I've tried it out, unfortunately the problem still persist.

I'm sure the problem lies with ".$pictureId in

<img src = "simple_picture.php?id=".$pictureId>

because it works when I replace ".$pictureId with 15 or a number that my query can find.

However, I can't do that in real application cos the number is changing in every row.

(I've tried echo $pictureId to make sure that it has a value, it's fine.)

Does it matter that pictureId is an auto_incremented field in my table?

steadyguy

<img src="simple_picture.php?id=".$pictureId>

Would make more sense as:

print "<img src=\" simple_picture.php?id=$pictureId\" />";

rachelkoh

print "<img src=\" simple_picture.php?id=$pictureId\" />";

resulted in a RED X in a box in the picture field too.

Any idea what could be wrong? What can I do?

Sorry to trouble you... 🙁