在 Class 中创建数据库 Table 作为函数

Creating a DataBase Table inside a Class as a Function

我正在尝试缩短我的代码,这不仅是为了可读性,也是为了自定义我正在处理的项目。

我创建了一个连接到数据库的 class,但我正在努力使用一个函数来创建一个带有列的 table。

目前 class 看起来像这样:

class DataBase {

    private $link;
    private $host, $username, $password, $database;

    public function __construct($host, $username, $password, $database){
        $this->host        = $host;
        $this->username    = $username;
        $this->password    = $password;
        $this->database    = $database;

        $this->link = mysql_connect($this->host, $this->username, $this->password)
            OR die("There was a problem connecting to the database.");

        mysql_select_db($this->database, $this->link)
            OR die("There was a problem selecting the database.");

        return true;
    }

    public function query($query) {
        $result = mysql_query($query);
        if (!$result) die('Invalid query: ' . mysql_error());
        return $result;
    }

    public function __destruct() {
        mysql_close($this->link)
            OR die("There was a problem disconnecting from the database.");
    }
}

可以看到查询的方法已经添加了。其 运行 的示例:

$db = new DataBase('localhost',$user,$pass,$name);
$db->query('SELECT * FROM table WHERE id="0"');

任何人都可以给我发送一些代码来添加添加插入 table 的功能吗?我试过这个:

public function create_table($t_data) {
    $result = $t_data;
    if (!$result) die('Invalid query: ' . mysql_error());
    return $result;
}

用法:

$t_data = 'CREATE TABLE log_users(
     uid VARCHAR(1024) NOT NULL,
     username VARCHAR(33) NOT NULL,
     password VARCHAR(18) NOT NULL,
     admin VARCHAR(1) DEFAULT 0,
     key VARCHAR(18) NOT NULL,
     constant VARCHAR(1) DEFAULT 0)';

 $db->create_table($t_data);

我建议您查看 MySQLiPDO,因为您使用的是已弃用的函数 mysql,它现在很容易受到攻击。我已经更新了您的 class(未测试)以帮助您入门。这也解决了您最初无法创建 table.

的问题
class DataBase {

    private $link;
    // May not need these, see updated __construct method
    private $host, $username, $password, $database;

    public function __construct($host, $username, $password, $database){
        // Unless you need them elsewhere, no reason to set $this->host, $this->username, etc...you can just access directly like below
        $this->link = new mysqli($host, $username, $password, $database);

        // Check connection (which also checks selection of database)
        if ($this->link->connect_error) {
            die("Connection failed: " . $this->link->connect_error);
        }
    }

    // You will need to research and update this to work with mysqli (right now it's ripe for SQL injection)!
    public function query($query) {
        $result = mysql_query($query);
        if (!$result) die('Invalid query: ' . mysql_error());
        return $result;
    }

    // This method will create a table based on the SQL you send it
    public function create_table($sql) {
        if ($this->link->query($sql) === TRUE) {
            return "Table created successfully";
        } else {
            return "Error creating table: " . $this->link->error;
        }
    }

    // Close connection
    public function __destruct() {
        $this->link->close();
    }
}