这是php数据库中创建表的正确函数吗？

初始化变量

<? php
$host = 'localhost';
$user = 'root';
$pword = '';
$dbname = 'mydb';
$tablename = 'userdata';
$con = mysqli_connect($host, $user, $pword); 

---

？>

变化不大。建议将

$con

&

$tablename

参数传递到

createtable（）

，最后调用

createtable（）

函数

$host = 'localhost';
$user = 'root';
$pword = '';
$dbname = 'mydb';
$tablename = 'userdata';
$con = mysqli_connect($host, $user, $pword, $dbname);
function createtable($con, $tablename) {
//mysqli_connect($host, $user, $pword);
$sql = "CREATE TABLE IF NOT EXISTS
  $tablename(uid int(10) unsigned NOT NULL AUTO_INCREMENT,
  firstname char(60),
  lastname char(60),
  username varchar(60),
  password varchar(60),
  gender enum('male','female') NOT NULL,
  course set('PHP','HTML','CSS','Javascript'),
  comments longtext,
  PRIMARY KEY(uid)
  )";
$result = mysqli_query($con,$sql);
if($result){echo "Table created.";}
else{echo mysqli_error($con);}
}
createtable($con, $tablename);

---

也许最好是“否”，以及为什么需要一个函数在每个表上创建相同的表call@FerozAkbar因为如果它不存在，我只想创建同一个表一次，但使用我自己的函数，这个问题似乎离题了，因为它询问的是个人辅导和代码检查。

$host = 'localhost';
$user = 'root';
$pword = '';
$dbname = 'mydb';
$tablename = 'userdata';
$con = mysqli_connect($host, $user, $pword, $dbname);
function createtable($con, $tablename) {
//mysqli_connect($host, $user, $pword);
$sql = "CREATE TABLE IF NOT EXISTS
  $tablename(uid int(10) unsigned NOT NULL AUTO_INCREMENT,
  firstname char(60),
  lastname char(60),
  username varchar(60),
  password varchar(60),
  gender enum('male','female') NOT NULL,
  course set('PHP','HTML','CSS','Javascript'),
  comments longtext,
  PRIMARY KEY(uid)
  )";
$result = mysqli_query($con,$sql);
if($result){echo "Table created.";}
else{echo mysqli_error($con);}
}
createtable($con, $tablename);