Help with PHP?

This problem is solved by using a SQL query rather than using PHP statements. That doesn't mean you need not use PHP statements . It is necessary but it is the mysql (i hope you are using mySql) connection part of PHP. Ok so you need to create a database lets name it 'movies'. Then we need to create a table in it called 'movie_list'. The fields in the tables are 1) Id int(11) 2) Name varchar(250) 3) Genre varchar(100) So our database is setup. I hope you understood all these things. Next is the PHP and sql part. use the following code to get a list of all movies whose genre is "Action". $connection= mysql_connect('localhost','root','password'); the above code would create a connection to your mysql server. The first argument is your hostname, second is the mysql user name and the last one is the password for the mysql server. We save this connection settings to a variable called $connection mysql_select_db('movies',$connection); this line creates a connection to the database movies on server ($connection) that we had made the connection in the first step. $query= "SELECT* FROM movie_list WHERE Genre= 'Action'";$result= mysql_query($query); The first step above is a standard SQL query that gets all the list of movies from a table called movie_list where the Genre is Action. This SQL is as such not exicuted it is only saved as a string to a variable called $query. The second step is where we execute the SQL query and the result is stored to the variable $result. while ($row=mysql_fetch_object($result)) { echo "$row-Id, $row-Name";} This code fetches the results from the $results as an object. This is put in a while loop so that the loop is executed till the last object and then exits the loop. Inside the loop we print the Id and the Name of the movies that we got after th query. So all the above said code is given below $connection= mysql_connect('localhost','root','password'); mysql_select_db('movies',$connection); $query= "SELECT* FROM movie_list WHERE Genre= 'Action'"; $result= mysql_query($query);while ($row=mysql_fetch_object($result)) { echo "$row-Id, $row-Name"; } I think that was a good introduction for you to PHP and mysql.I haven't tested this code, but it should work. If you are getting any error, post it here and I would help you. Let me tell that PHP is very easy to learn available and not that difficult as you think. By the the way dont for get to add the opening and closing PHP tags for the above code. I was not able to do that since it wont be printed here in mylot.

Why dont you try oscommerce, you can stop learning php, and star reading tutorials about the configuration, is easy and you can have runing your store in a few days.

this is called a catalog online. you need a category table and a video table. category to save the video genre. and video table is to save the video detail. this two table must be link using a key. then u need a PHP algorithm to do this. i already make this site. but for furniture site. if you interest you can contact me. thank you

INfact this is quite a customized thing u require thats why u are getting harder with it, u must comtact some PHP developer or programmer, no book can do it 100% for u

So You need to have categories and under each categoies titles with description... You can use for it files or mysql. So You must know how to read from input, for example.. index.php?cat=action , so when you recognize that cat variable is action You 'll display all titles with action. $cat = $_GET["cat"]; # so $cat will contain 'action', so then You neeed to connect to mysql or to open file.... and read entries and display it...

Visit www.thenetguruz.com and ask a guy called Hashim