help with displaying images

[vmicchia]

So I am trying to display an image from a database (I know it's not the best idea but it's what I was told to do). Anyway I'm having problems displaying it. Here is the code for displaying the image.

$cat = $_GET['cat'];
include 'includes/openDbConn.php';
$sql = 'SELECT * FROM CushionsCategories WHERE CushionCategory = "'.$cat.'"';
echo $sql;
$result = mysql_query($sql); 
echo '<table>';
while($val = mysql_fetch_assoc($result)){
	$cush = 'SELECT * FROM Cushion WHERE SKU = "'.$val['CushionSKU'].'"';
	echo $cush;
	$cushres = mysql_query($cush);
	$cushval = mysql_fetch_assoc;
	$img = 'SELECT * FROM Images WHERE SKU = "'.$val['CushionSKU'].'"';
	echo $img;
	$imgres = mysql_query($img);
	$imgval = mysql_fetch_assoc($imgres);
	if( $i % 3 == 0 ) {
	      echo '</tr><tr>';
  			}
	echo '<td><img src="image.php?sku='.$val['CushionSKU'].'" name="'.$imgval['FileName'].'" description="'.$imgval['Description'].'" /></td>';
}
echo '</tr></table>';

 

and the page that gets the image:

$sql = 'SELECT Image FROM Images WHERE SKU = "'.$sku.'"';
//echo $sql;
$result = mysql_query($sql) or die("Invalid query: " . mysql_error());

        // set the header for the image
        header("Content-type: image/jpeg");
        echo mysql_result($result, 0);

 

Thanks for any help in advance.

 

 

[chrispos]

Just put the image name in the database ie beach.jpg under a table called image run the query and do the image source as

		  echo "<img src='$image'>";

 

[BlueSkyIS]

Just put the image name in the database ie beach.jpg under a table called image run the query and do the image source as

echo "<img src='$image'>";			

 

already stated that image is stored in the database and needs to pull image from database and send directly to browser, thus no possibility for a simple image link to a file.

[vmicchia]

Yes unfortunately it's in the database I for some reason can't get it to display with that code( I was following a tutorial)  and was hoping for some help with that.

[denno020]

Give this link a go mate:

http://www.codewalkers.com/c/a/Database-Articles/Storing-Images-in-Database/3/

 

It was easy enough to find using Google...

 

Denno

[vmicchia]

I had looked at this link before, it seemed a lot like the tutorial I was using. Anyway I tried it the way this shows. I changed the code to how this tutorial shows and it is still not working for me. When I go to the image.php it echos out Array like I assume it should but on the page the image is displayed I still get a broken link. here is what the code looks like now:

$sql = 'SELECT Image FROM Images WHERE SKU = "'.$sku.'"';
//echo $sql;

$result = mysql_query($sql); 
if (mysql_num_rows($result) != 0) { 
  $row = @mysql_fetch_array($result); 
  $image_type = $row["Type"]; 
  $image = $row["Image"]; 
  header ("Content-type: $image_type"); 
  print $image; 
}

 

Edit: Also when I do a print_r on the results it still only says array rather than giving me the contents of the array.

 

[PFMaBiSmAd]

What's your actual code in image.php, because so far you are not setting the $sku variable to anything and your query won't match anything.

[vmicchia]

Here's the actual code. I got it figured out though. I was just sending a variable wrong.

 

$sku = $_GET['sku'];
//echo $sku;
include 'includes/openDbConn.php';
$sql = 'SELECT * FROM Images WHERE SKU = "'.$sku.'"';
//echo $sql;

$result = mysql_query($sql); 
if (mysql_num_rows($result) != 0) { 
  $row = mysql_fetch_assoc($result); 
  $image_type = $row["Type"]; 
  //echo $image_type;
  $image = $row["Image"]; 
// echo $image;
  header ("Content-type: $image_type"); 
  echo $image; 
}