I have a site that currently uses images on a file server. The images appear on a page where the user can drag and drop each as is needed. This is done with jQuery and the images are enclosed in a list. Each image is pretty standard:
<img src='//network_path/image.png' height='80px'>
Now however I need to reference images stored as a BLOB in an Oracle database (no choice on this, so not a merit discussion). I have no problem retrieving the BLOB and displaying on it's own using:
$sql = "SELECT image FROM images WHERE image_id = 123";
$stid = oci_parse($conn, $sql);
oci_execute($stid);
$row = oci_fetch_array($stid, OCI_ASSOC+OCI_RETURN_NULLS);
$img = $row['IMAGE']->load();
header("Content-type: image/jpeg");
print $img;
But I need to [efficiently] get that image as the src attribute of the img tag. I tried imagecreatefromstring() but that just returns the image in the browser, ignoring the other html. I looked at data uri, but the IE8 size limit rules that out.
So now I am kind of stuck. My searches keep coming up with using a src attribute that loads another page that contains the image. But I need the image itself to actually show on the page. (Note: I say image, meaning at least one image but as many as eight on a page).
Any help would be greatly appreciated.
Well, you can do a few things. You can either make a page that will render the image
<img src="image.php?id=123" />
That image.php page would have this:
$sql = "SELECT image FROM images WHERE image_id = " . (int) $_GET['id'];
$stid = oci_parse($conn, $sql);
oci_execute($stid);
$row = oci_fetch_array($stid, OCI_ASSOC+OCI_RETURN_NULLS);
if (!$row) {
header('Status: 404 Not Found');
} else {
$img = $row['IMAGE']->load();
header("Content-type: image/jpeg");
print $img;
}
Or, you could base64 encode it into the src (note, not all browsers handle this well):
<img src="data:image/jpeg;base64,<?php echo base64_encode($img); ?>" />
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