I am new to PHP and have been trying to do some coding,but something is just not right with my first PHP code.Please help me point out my error(s) O!
<?php
$user_name="root";
$password="nbuser";
$database="Meet_A_Geek";
$server="localhost";
mysql_connect("$server","$user_name","$password" or die("cannot connect";
mysql_select_db("$database" or die("cannot select DB";
$mypin=$_POST['pin'];
$myserialnum=$_POST['serial'];
//check to see if PIN and SERIAL number indeed exist
$sql="SELECT*FROM cardserials WHERE cardpins='$mypin' AND serialnumber='$myserialnum'";
$result=mysql_query($sql);
$count=mysql_num_rows($result);
//if ppin and SERIAL exist,check to see if it has been used by another person.
if(count==1){
$myresult=mysql_query("SELECT*FROM usepins WHERE card_pin='$mypin' AND serial_no='myserialnum'";
$finite=mysql_num_rows($myresult);
//if it has been used,notify user
if($finite==1){
echo"Card has been used";
}else{
mysql_query("INSERT INTO usepins (card_pin,serial_no) VALUES('$mypin','$myserialnum')";
}
}else{
echo"check your details";
}
?>

on line 16 close the bracket u opened for the mysql_query() function.

i have highlighted the bracket tht was missing below

$myresult=mysql_query("SELECT*FROM usepins WHERE card_pin='$mypin' AND serial_no='myserialnum'");

sorry sir ,  at first your db  connection wont work , 
in PHP ,  if you write
$name = "segun"; // a string ,  right?
and later say echo "$name";
coz you think $name holds a string value ,  pls dont make that mistake anymore in php,  you dont put variables in double quote ,  php sees it as a string ,  hope you got that point sir,
why not write your dbconnect simply as
mysql_connect("localhost", "user_name" , "password;
mysql_db("my db"
simple!
well, good other line of code for a php starter,

you can go get some other php ebook/material from my website www.segsworld.com
it will help ! just register to access my software downloads page!

segsalerty:

sorry sir , at first your db connection wont work ,
in PHP , if you write
$name = "segun"; // a string , right?
and later say echo "$name";
coz you think $name holds a string value , pls dont make that mistake anymore in php, you dont put variables in double quote , php sees it as a string , hope you got that point sir,
why not write your dbconnect simply as
mysql_connect("localhost", "user_name" , "password;
mysql_db("my db"
simple!
well, good other line of code for a php starter,

you can go get some other php ebook/material from my website www.segsworld.com
it will help ! just register to access my software downloads page!

Thanks sir.
Being trying to check out your site but being having little trouble opening it,but will sure try it later.
Now time to try out your advice.

you welcome buddie ,,
my site might not work , i was uploading some stuffs ,
not sure thats why my site isnt opening coz i can access it here
anytime , just contact me

Something is wrong somewhere,
I can't connect your site from my system.
uhm,bro.how do I connect?

uhmmm, i used to face that kind of problem too, na my host dey do something i dont uunderstand ,
try this , it helps sometimes when my website appears to me unexisting
try clearing all caches and cookies of your web browser then reboot system after cleaning everything
then , revisit my site ,
it sucks
How was your code? did it work fine ?
let me know

No it didn't work.But i am still trying to get it work.
This is what I actually have in mind to design.A Fictitious
School's website for registration.
1 )Get your scratch card
2)goto schools site
3)Enter your PIN and serial number and submit
4)Goes to database and check out from a table if PIN indeed exist.
5)check to make sure that PIN has not been used by another person from A TABLE(usepins) that contains students details
6)if PIN has been used echo(pin has been used)
7)Else redirect to registration form.
I don' know if I am in line from my code.
Please help me out I really need this first code to work to serve as a motivation.
Thanks for help so far.I will do exactly as you instruct.

cool job indeed
you are kinda on a right part , but you have to take note of few things , how many times do you want student to use the card to access registration panel ?
looking at your code,
its not on the shot at all
well, i can help you !
but , you said you are new to PHP , what have you really covered? how good are you with SQL , designing / setting up your db / tables matters now
Uhmm
if you can chat with me on segsalerty@yahoo.com
am online now
i can put you thru what to do, coz i dnt really like give some1 fish but i can teach you how you catch ! that was how i learned too , i am still learning self , nobody know it all
i wanna stck my nose into Ajax now into Ajax now ,
thoum not going to be for my website use , i better be get expensive for client to fetch ,
you see how my site no fine ? lol

Your site is simple and cool,I like it.
I registered my membership but couldn,t activate it
via the link sent to my mail.Keep getting wrong parameter.
Bro's weyting I go do,although I send mail to your support
sha.

you activate your membership already , trying to do so again will give you wrong parameter ( i should have coded it to tell you already cofirmed membership) sorry , havent got much time for the damn site for a while self, so, go logon and surf my software page to download and enjoy !

segsalerty:

sorry sir ,  at first your db  connection wont work , 
in PHP ,  if you write
$name = "segun"; // a string ,  right?
and later say echo "$name";
coz you think $name holds a string value ,  pls dont make that mistake anymore in php,  you dont put variables in double quote ,  php sees it as a string ,  hope you got that point sir,
why not write your dbconnect simply as
mysql_connect("localhost", "user_name" , "password;
mysql_db("my db"
simple!
well, good other line of code for a php starter,

you can go get some other php ebook/material from my website www.segsworld.com
it will help ! just register to access my software downloads page!

on the contrary, you can use a variable in a double quote, But cant use it in single quotes
$var="hello";
echo "I feel like saying $var";

is valid and will return "I feel like saying hello"



secondly, i guess u meant mysql_select_db()

Dude,

Did you check to make sure that you can see mysql in "phpinfo()"?
I think you might need to check that first, then make a connection to the db.

Thanks

Looks like i missed the action here. Anyway, is it safe to assume that the problem has been solved?
'cos i can see the gurus got here before me, i am out. . .

I guess i have to follow so i can learn from the gurus.

But that first tymer coder e long small o!