"Tamara Temple" <tamouse.lists@xxxxxxxxx> wrote in message news:7F666311-4BC8-4064-8C70-4F2597E7BF7F@xxxxxxxxxxxx > On Oct 29, 2010, at 2:44 PM, Gary wrote: >> "Adam Richardson" <simpleshot@xxxxxxxxx> wrote in message >> news:AANLkTi=keNXt7yEwrZtcm4+hyifRLQhOZxSe7Ufmqjp=@xxxxxxxxxxxxxxxxx >> On Fri, Oct 29, 2010 at 3:05 PM, Gary <gpaul@xxxxxxxxxxxxxxxx> wrote: >>> I am trying to get the watermark to work, however I am having a problem >>> in >>> that the image is being called from a database (image sits in images >>> file). >>> >>> The script in question is this >>> >>> $image = imagecreatefromjpeg($_GET['src']); >>> >>> However it produces an error message of >>> >>> Warning: imagecreatefromjpeg() [function.imagecreatefromjpeg]: Filename >>> cannot be empty in /home/content/a/l/i/alinde52/html/ imagesDetail.php >>> on >>> line 233 >>> >> >> First things first. It looks like there's nothing in $_GET['src']. >> From where are you getting the value 'src'? >> >> Adam >> >> ** >> Adam >> >> Thanks for your reply, that is my question, what is to replace ['src'] >> if I >> am calling the image from a database. >> >> Gary > > I'd really need to know more about this application to help. If you are > calling the image from a database (does this mean you have the image's > file spec saved in the database, or you actually storing the image data > in the database?), you need to use a query to do that. Is the > imagesDetail.php script being called by something else that already > queried the database and put the file spec in the src query string > argument? Before you dump the query string argument directly into the > imagecreatefromjpeg() funciton, you should verify that it exists: > > if (isset($_GET['src'[) && (!empty($_GET['src'[) { > $src = $_GET['src']; > if (fileexists($src)) { > $image = imageceatefromjpeg($src); > if ($image === FALSE) { > # process error from imagecreatefromjpeg > } > } else { > # process missing image > } > } else { > # process missing query string parameter > } > > This is all prediated on the idea that something is calling your > imageDetail.php script with the source path for the image in question. If > that is not the case, and you need to in fact query the database for the > source path of the image, then you need to do a database query. > > Not knowing anything about how your database is set up, I'll take a stab > at a generic method of doing it. > > Somehow, imageDetail.php needs to know what image to get. Let's assume > you are calling it from a gallery that has several images displayed. Part > of the information needed is some what to identify the image's record in > the database. Let's assume you have images stored with an id, that is not > null and autoincrements when you store a new image. Here's a sample > schema: > > CREATE TABLE `photos` ( > `id` INT AUTO_INCREMENT NOT NULL, > `src` VARCHAR(255) NOT NULL, > `created` TIMESTAMP NOT NULL DEFAULT 0, > `updated` TIMESTAMP NOT NULL DEFAULT 0 ON UPDATE CURRENT_TIMESTAMP > PRIMARY KEY (`id`) > ); > > Of course, you would probably want to store a lot more info about an > image, but this will do for explanation. > > Let's say you have a gallery shown at your application's index.php > program. In the html it generates, you may have something like this for > any particular thumbnail: > > > <a href="imageDetail.php?id=25"><img src="thumbs/imageABC.jpg"></a> > > This is making some assumptions: > * you have image thumbnails for each image stored in the subdirectory > thumbs/ > * you've figured out somehow that the id for thumbs/imageABC.jpg is 25 > > When the user clicks on the image, they get taken to your imageDetail.php > script, with the query string paramter id set to 25. > > In PHP you would then do: > > if (isset($_GET['id'] && (!empty($_GET['id'] && is_numeric($_GET['id']) { > $id = $_GET['id']; > $sql = "SELECT * FROM `photos` WHERE id=".$id." LIMIT 1"; > $result = mysql_query($sql,$db); > if ($result) { > $imagedata = mysql_fetch_array($result,MYSQL_ASSOC); > $src = $imagedata['src']; > if (isset($src) && !empty($src) && fileexists($src)) { > $image = imagecreatefromjpeg($src); > > # do stuff with image > > } else { > > # handle invalid src data > > } > } else { > > # handle error from query > > } > } else { > > # handle invalid id paramter on script > > } > > > Note: some people like to handle error conditions before moving on to > working with the successful state. It's a matter of style. Either works. > > Hope this helps. > > Tamara > Tamara Thanks for your reply. The image is already called, (and if you read the response I posted to tedd, you'll see I am now not sure how it is). I also posted the code that calls the image if that sheds any light. I am digesting all the information to get it to work, if any of the code changes your answer, I would love to hear it! Thank you again. 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