Reputation: 139
I am passing a query variable in PHP. I would like to make it the table name, but I know there is probably a SQL syntax error. When I print the statement, the variable is passed meaning it works, but the database simply isn't created.
Here is my code for the creation of the database:
$DBName = "database_name";
$sql = "CREATE TABLE '$DBName'.'$login' (
ClientID int NOT NULL AUTO_INCREMENT,
AgentClients varchar(15),
ClientTotal int
)";
mysql_query($sql,$link);
where `$login = $_POST['login'];
Also, I'm not worried about security breaches at the moment, so don't worry about that.
Any insight would be greatly appreciated.
Upvotes: 0
Views: 1220
Reputation: 21866
You must use backticks for your tablename, and not quotes:
$sql = "CREATE TABLE `$DBName`.`$login` (
ClientID int NOT NULL AUTO_INCREMENT,
AgentClients varchar(15),
ClientTotal int,
PRIMARY KEY (`ClientID`)
)";
Upvotes: 6