Insert into database error

[duncan_cowan]

Hi I have made a script to insert a row into a database with the computers ip and a random number, but noting will be added to the database. Please could someone help...script:

<?php$ip = $_SERVER["REMOTE_ADDR"];$random = rand(600000000, 900000000); // between * and * inclusive$user_number = $ip.$random;if (isset($_COOKIE["user_online"])){echo $_COOKIE["user_online"];$sql_script = 'SELECT COUNT (*) FROM users_online';$users_online = mysql_query($sql_script);} else {setcookie("user_online", "$user_number");echo "Cookie set";mysql_query("INSERT INTO users_online (rand-num) VALUES ('".$user_number."')");$sql_script = mysql_query("SELECT * FROM users_online");$num_rows = mysql_num_rows($sql_script);echo $num_rows;}?>

Thankyou in advance...

[Mr_CHISOL]

Have you tried

echo mysqli_error ( );

after you run the INSERT-query to check if there are any errors?

[vytas]

isn't it called mysql_error() ?

[Mr_CHISOL]

isn't it called mysql_error() ?

Yes it is, but there's an mysqli_error() too, but that requires a handler/link to the database (i as in Improved).

[justsomeguy]

I think it's the query:mysql_query("INSERT INTO users_online (rand-num) VALUES ('".$user_number."')");You are saying the field is called "rand minus num". Either don't use hyphens in field names, because in SQL and PHP it is an arithmetic operator, or put the fieldname in backquotes.mysql_query("INSERT INTO users_online (`rand-num`) VALUES ('".$user_number."')");

[duncan_cowan]

dont worry i tried using the mysql error function but to do this I set the query as a variable and just setting it as a variable seemed to work.Thanks to all those who suggested!