So, for this video hosting script I'm working on, ive ran into this problem.

I'm a very visual learner so if i can't "see" the MySQL Data i don't like to use them, and this script can be used by people who don't have access to a MySQL database.

Problem: Ive got the videos set up in an array so they are "sorted", and ive got the .flv link, discription, and tag in an array inside of that.

Ex:

PHP Code:

// $video[$n] = array("file.flv","Description","tag");
$video[0] = array("intro.flv","Intro To Decaf Video","Help");
$video[1] = array("Michael_Angelo_Batio_Double_Guitar_Solo.flv","Dual Hand Gutiar Solo","Music");
$video[3] = array("amazing_guitar_player.flv","Guy plays with his feet","Music");
$video[4] = array("BBC_Horizon_The_Six_Billion_Dollar_Experiment.flv","Particle Acclerator","Educational");
$video[5] = array("Bush_Shoe_Thrown.flv","Bush Shoe Throw","News");
$video[6] = array("Charlie_the_Unicorn.flv","Charlie The Unicorn","Funny");
$video[7] = array("Charlie_the_Unicorn_2.flv","Charlie The Unicorn 2","Funny"); 


My question is how can i get that description to print?

I have tried:

PHP Code:

print "$video[$n][2]";
// And
print "$video[$n]array[2]"; 


Fyi, the $n is just the video number called from the URL/Random Generator.

Any help would be appericated, thanks.

Ditch the quotes.

PHP Code:

$n=7;
print $video[$n][2]; // funny 


Thanks max, that helped.

If you need to print the value inside a string you should use curly brackets.

PHP Code:

echo "Our most recent video {$video[$n][2]} has already got 1 million views"; 


Yep, thanks guys. I think i understand it enough now to make this guide. And my video sharing script is using it for now (untill i get around to making it mySQL).

Thanks guys!