使用類方法時未創建表

嗨，我創建了以下作業。我對此很陌生，不明白為什么第二個代碼不起作用。任何建議將受到極大歡迎。

 
 
class Dbcon {
    private $severname;
    private $username;
    private $password;
    private $dbname;
    private $charset;
 
 
public function connect () {
 
$this->servername ='localhost';
$this->username = 'ucmsadmin';
$this->password = 'P@xxwxrdxxx';
$this->dbname = 'umcs';
$this->charset = 'utmf8mb4';
 
 
 try {
 
 
$dsn = "mysql:host=".$this->severname.";dbname=".$this->dbname.";charset".$this->charset;
$pdo = new PDO($dsn , $this->username, $this->password);
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO:: ERRMODE_EXCEPTION);
echo "Connected";
 
 
$queryCreateUsersTable = "CREATE TABLE IF NOT EXISTS `USERS` (
    `ID` int(11) unsigned NOT NULL auto_increment,
    `USER_NAME` varchar(255) NOT NULL default '',
    `EMAIL` varchar(255) NOT NULL default '',
    `PASSWORD` varchar(255) NOT NULL default '',
    `ACCESS_LEVEL` tinyint(1) unsigned NOT NULL default '1',
    `ACCOUNT_APPROVED` boolean NOT NULL default '0',
    
    PRIMARY KEY  (`ID`))";
 
if(!$pdo->query($queryCreateUsersTable)){
    echo "Table creation failed: (" . $pdo->errno . ") " . $pdo->error;
 
} else {
    echo' table created';
}
 
return $pdo;
 
 } catch (PDOException $e) {
 echo "connection failed: ".$e->getMessage();
 
 }
}
}

但是，如果我將創建資料庫表的部分移動到公共函式中，我不會收到任何錯誤，但不會創建表。這是我所做的；

 
 
class Dbcon {
    private $severname;
    private $username;
    private $password;
    private $dbname;
    private $charset;
 
 
public function connect () {
 
$this->servername ='localhost';
$this->username = 'ucmsadmin';
$this->password = 'P@33w0rd101';
$this->dbname = 'umcs';
$this->charset = 'utmf8mb4';
 
 
 try {
 
 
$dsn = "mysql:host=".$this->severname.";dbname=".$this->dbname.";charset".$this->charset;
$pdo = new PDO($dsn , $this->username, $this->password);
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO:: ERRMODE_EXCEPTION);
echo "Connected";
 
 
return $pdo;
 
 } catch (PDOException $e) {
 echo "connection failed: ".$e->getMessage();
 
 }
}
 
public function createtable() {
    $this->connect();
    $queryCreateUsersTable = "CREATE TABLE IF NOT EXISTS `USERS` (
    `ID` int(11) unsigned NOT NULL auto_increment,
    `USER_NAME` varchar(255) NOT NULL default '',
    `EMAIL` varchar(255) NOT NULL default '',
    `PASSWORD` varchar(255) NOT NULL default '',
    `ACCESS_LEVEL` tinyint(1) unsigned NOT NULL default '1',
    `ACCOUNT_APPROVED` boolean NOT NULL default '0',
    
    PRIMARY KEY  (`ID`))";
    return $queryCreateUsersTable;
 
}
}

在 index.php 我運行

$object = new Dbcon;
$object->connect();
$object2 = new Dbcon;
$object2->createTable();?> 

uj5u.com熱心網友回復：

該createTable()函式只是定義了一個包含一些 SQL 的字串。它永遠不會將該 SQL 發送到要執行的資料庫。

基于原始代碼，您可以更像這樣撰寫它：

public function createtable() {
    $pdo = $this->connect();
    $queryCreateUsersTable = "CREATE TABLE IF NOT EXISTS `USERS` (
    `ID` int(11) unsigned NOT NULL auto_increment,
    `USER_NAME` varchar(255) NOT NULL default '',
    `EMAIL` varchar(255) NOT NULL default '',
    `PASSWORD` varchar(255) NOT NULL default '',
    `ACCESS_LEVEL` tinyint(1) unsigned NOT NULL default '1',
    `ACCOUNT_APPROVED` boolean NOT NULL default '0',
    PRIMARY KEY  (`ID`))";

    if(!$pdo->query($queryCreateUsersTable)){
      echo "Table creation failed: (" . $pdo->errno . ") " . $pdo->error;
    } else {
      echo' table created';
    }
}

但是我要指出，每次運行查詢時都創建一個新連接是非常低效的。相反，您應該嘗試在 PHP 腳本的生命周期內創建一次連接，并繼續重復使用它。

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