示例
';
$sql = "create table emp5(id INT AUTO_INCREMENT,name VARCHAr(20) NOT NULL,
emp_salary INT NOT NULL,primary key (id))";
if(mysqli_query($conn, $sql)){
echo "Table emp5 created successfully";
}else{
echo "Could not create table: ". mysqli_error($conn);
}
mysqli_close($conn);
?>
执行上面代码得到以下结果 -
Connected successfully
Table emp5 created successfully



