Question

What’s wrong with the following code:

$table = $email . "Entries";
$query = "CREATE TABLE ".$table."(FirstName varchar(15),LastName varchar(15),Age int)";

I want to create a table whose name is dynamic based on a posted variable. The above doesn’t work however. I have also tried:

$query = "CREATE TABLE $table(FirstName varchar(15),LastName varchar(15),Age int)";

Any Ideas?

Answer 1

I cannot find the relevant documentation, but would assume that if $email contains an @ this would be an invalid table name in MySQL.

CREATE TABLE [e-mail] (FirstName varchar(15),LastName varchar(15),Age int)

If this is the case, try enclosing the table name in backticks:

$query = "CREATE TABLE `$table` (FirstName varchar(15),LastName varchar(15),Age int)";