用PHP实现DB拉取功能

Question

我是PHP编程的新手，正在开发一个有趣的小游戏来帮助自己学习。我从其他人那里得到了一些代码帮助，从数据库中提取角色的统计数据，但我很难让它发挥作用。当我现在尝试运行它时，我只收到了"服务器错误"。数据库信息很好，我以前有一个从数据库中提取的工作函数，但希望通过类函数将其通用化。这是我迄今为止所拥有的。

DB类：

<?php
class db_class
{
    //db connection portion
    protected $mysqli;
    private $db_host = 'XXXXXXX';
    private $db_user = 'Filler';
    private $db_password = 'Filler';
    protected $db_name = 'Filler';
    //db connection portion    
    public function __construct($db_host = null, $db_user = null, $db_password = null, $db_name = null) {
        if (!empty($db_host)) {
            $this->db_host = $db_host;
        }
        // validate other parameters similarly
        //database connection object
        $mysqli = new mysqli($this->db_host, $this->db_user, $this->db_password, $this->db_name);
        if ($mysqli->connect_error) {
            throw new Exception('Connect Error: ' . $mysqli->connect_errno . ', ' . $mysqli->connect_error);
        } else {
            $this->mysqli = $mysqli;
        }
    }
    public function getPlayerStats($id) {    
        if (empty($id)) {
            throw new Exception ('An empty value was passed for id');
        }
        // verify this is integer-like value
        $id      = (string) $id;
        $pattern = '/^d+$/';
        if (!preg_match($pattern, $id) !== 1) {
            throw new Exception ('A non-integer value was passed for id');
        }
        $id = (int) $id;
        $query = "SELECT id, name, strength, defense, level, health, type, experience FROM characters WHERE id = :id";
        $stmt  = $this->mysqli->prepare($query);
        $stmt->bind_param('i', $id);
        $result = $stmt->execute();
        if (false === $result) {
            throw new Exception('Query error: ' . $stmt->error);
        } else {
            $obj = new stdClass();
            $stmt->bind_result($obj->id, $obj->name, $obj->strength, $obj->defense, $obj->level, $obj, health, $obj->type, $obj->experience);
            $stmt->fetch();
            $stmt->close();
            return $obj;
        }
    }
}
?>

DB类函数调用：

<?php
include "db_class.php";
echo "made it out here1";
$classobject = new db_class();
echo "made it out here2";
$results = $classobject->getPlayerStats('1');
print_r($results);
echo "made it out here3";
$id         = "id: " . $results['id'];
$name       = "name: " . $results['charname'];
$strength   = "strength: " . $results['strength'];
$defense    = "defense: " . $results['defense'];
$health     = "health: " . $results['health'];
$level      = "level: " . $results['level'];
$type       = "type: " . $results['type'];
$experience = "experience: " . $results['experience'];
echo "<br/>";
echo "made it out here4";
?>

调试这段代码很困难，因为我习惯于在编译器中输入特征线并处理VBA之类的代码错误，所以任何调试技巧都会非常有用。我在这里做错了什么？提前感谢！

Answer 1

您编写了

public __construct($db_host = NULL, ...

但是构造函数是函数。你需要

public function __construct($db_host = NULL, ...

db_class构造函数接受四个参数。此实例化不通过。

$classobject = new db_class();

所以你的连接字符串中出现了垃圾。把它整理好，你就可以上路了。

---

您可以通过构建最低版本来避免大量调试。例如，你可以先写这个。

<?php
class db_class{
  public function __construct($db_host = NULL, $db_user = NULL, $db_password = NULL, $db_name = NULL) {
  }
} 
?>

如果有效，请将其签入版本控制，然后添加一些代码。（你怎么知道它是否有效？测试它。）