[SOLVED] Retrieving image location

[Scotty2024]

I have the location of images saved in a database. The images themselves are not stored in the database, only their location. The picture ID and location is stored in the database. The idea is that you search for a picture based on its ID and it's location is returned. So I built a retrieve.php page that finds the location and a display.php page which shows the image. For example

 

display.php

<img src="retrieve.php?id=<?PHP echo picID; ?>">

 

retrieve.php

$id = htmlentities($_GET['id']);
if($id != '')
{
   $query = "SELECT location FROM pictures WHERE id='$id'";
   $result = mysql_query($query) or die(mysql_error());
   if(mysql_num_rows($result) > 0)
   {
      $imageData = mysql_fetch_array($result);
      echo $imageData['location'];
   }
}

This echos the correct path of the image (which is just text), but it fails to display in display.php. How can I get the location from retrieve.php to display.php?

Thanks!

[Alex]

What you're trying to do is the set the source of an image to a page containing only html. What you're going to want to do is look into the PHP GD Library. What you can do is using imagecreatefrom*() (imagecreatefrompng, etc..) create the imageresource from the location then output it to the browser using image*() (imagepng(), etc..).

 

Assuming it's a .png image it would be something like this:

 

$id = htmlentities($_GET['id']);
if($id != '')
{
   $query = "SELECT location FROM pictures WHERE id='$id'";
   $result = mysql_query($query) or die(mysql_error());
   if(mysql_num_rows($result) > 0)
   {
      $imageData = mysql_fetch_array($result);
      $im = imagecreatefrompng($imageData['location']);
      header('Content-type: image/png'); // Set Content-type header to image/png to output an image
      imagepng($im);
   }
}

[PFMaBiSmAd]

To just output an image as is, use readfile after the Content-type header. You will also either need to store the correct content-type with each image (assuming the can be different types) or you will need to test the file extension to determine the correct content-type header to output.

 

Using the GD functions takes a huge amount of server resources because it creates an uncompressed bitmap image of the file. If you are not manipulating the image, there is no need to use GD.

[Scotty2024]

Hello AlexWD and PFMaBiSmAd.

 

Alex, I was able to get your method to work! Thanks!

 

PFM, I couldn't get readfile() to work. Do I switch readfile() with imagepng() and add ob_clean() flush() in Alex's example? And can I still use <img src="retrieve.php?id=<?PHP echo picID; ?>"> in display.php to retrieve the image?

 

Thanks!

[Alex]

No, my example was to get it from the url, not a blob field.

 

For PFMaBiSmAd's method you'd do this:

$id = htmlentities($_GET['id']);
if($id != '')
{
   $query = "SELECT location FROM pictures WHERE id='$id'";
   $result = mysql_query($query) or die(mysql_error());
   if(mysql_num_rows($result) > 0)
   {
      $imageData = mysql_fetch_array($result);
      header('Content-type: image/png'); // Set Content-type header to image/png to output an image
      readfile($imageData['location']);
   }
}

[Scotty2024]

Thanks again Alex. I got both methods to work!