Re: How upload files to protected directories?

Kathleen Ballard wrote: > Mig, > This is some code I was sent when I needed to display images that were > located outside of the web dir. > > ~~~~~~~~~~ > If you want the images outside the web-tree, your image tag would be more > like this: > > <IMG SRC=display.php3?imagefile=happy.gif> > > Then display.php3 would look not unlike: > > <?php > header("Content-type: image/gif"); > header("Content-length: " . filesize("C:/hide/images/$imagefile")); > //Caching headers here. > //And Content-disposition > //And Content-filename > //and so on. > readfile("C:/hide/images/$imagefile"); > ?> Hello Kathleen Thanks very much, I checked your idea and must say it works nice .... with files, but when I need pass strings like parameters for PHP program, things become difficult. I will follow your clever idea, because I know it will work, but first need solve two problems (sorry I am newer to PHP and sometimes confused the way PHP treat variables ). First problem is: how store content of image file (jpg or gif) into MySql BLOB fields? Mysql update command uses variable='value' to update (or insert) values. Image data has lot of ' " ' and " ' " in between, so I receive 1054 ERROR when trying to pass update image='$image', because $image variable expands before sending data to database. Second problem is: how to pass later same string $image recovered from MySql to display.php3 program? In your example, your are passing file name, a very single string that wouldn't cause any confusion, but when passing display.php3?stream-image=$image things will become hard. Probably I will need to reopen Mysql database into display.php3 program and so I would pass only record number to display.php3, letting this program to search again record. Anyway, I modified display.php3 program to execute $image = fread(file...) and then echo $image, .... and it works fine.., that's to say, it is just question of string treatment, and really GD library is not necessary. Thanks Miguel