Question

我是PHP中使用OOP模型的新手。我一直在尝试从数据库中检索数据，但与私人相关的东西让我陷入困境。这是我的代码。

<?php 

require ("UserData.php");

class Database{

    public function getUser($sql){

    include ("includes/connect.php");
        $statement = $conn->prepare ($sql);
        $statement->execute();

        while ($row = $statement->fetch()) {

            $dataSet[] = new UserData($row);
        }

        if (!empty($dataSet)) {
            return $dataSet;

        }else{
            die();
        }
    }
}
?>

第二个文件

<?php 

class UserData
{
    private $user_id, $phone,$name,$address;

    public function _construct($dbrow){ 

        $this->user_id = $dbrow['user_id'];
        $this->name = $dbrow['name'];
        $this->phone = $dbrow['phone'];
        $this->address = $dbrow['address'];
    }
    public function getUserId(){
        return $this->user_id;
    }
    public function getUserName(){
        return $this->name;
    }
    public function getUserPhone(){
        return $this->phone;
    }
    public function getUserAddress(){
        return $this->address;
    }
}
?>

和最后一个

<?php require ("Database.php"); ?>

<html>
<head>
    <title>OOP</title>
</head>
<body>
    <?php 
    include("includes/connect.php");

    $db = new Database();

    $dataSet = $db -> getUser ("SELECT * from user");


    if ($dataSet) {

        foreach ($dataSet as $data) {   
            echo "<p>";
            echo "ID" .$data->getUserId()."<br />";
            echo "Name" .$data->getUserName()."<br />";
            echo "Phone" .$data->getUserPhone()."<br />";
            echo "Address" .$data->getUserAddress()."<br />";
            echo "</p>";
        }

    }else{
        echo "no result found";
    }
    ?>
</body>
</html>

好吧，我尝试var_dump dataSet，但错误显示出来。

array（2）{[0] =＆gt; object（UserData）＃5（4）{ [＆＃34; USER_ID＆＃34;：＆＃34;的UserData＆＃34;：私人] =＆GT; NULL [＆＃34; phone＆＃34;：＆＃34; UserData＆＃34;：private] =＆gt; NULL [＆＃34; name＆＃34;：＆＃34; UserData＆＃34;：private] =＆gt;空值 [＆＃34;地址＆＃34;：＆＃34;的UserData＆＃34;：私人] =＆GT; NULL} [1] =＆gt; object（UserData）＃6（4）{ [＆＃34; USER_ID＆＃34;：＆＃34;的UserData＆＃34;：私人] =＆GT; NULL [＆＃34; phone＆＃34;：＆＃34; UserData＆＃34;：private] =＆gt; NULL [＆＃34; name＆＃34;：＆＃34; UserData＆＃34;：private] =＆gt;空值 [＆＃34;地址＆＃34;：＆＃34;的UserData＆＃34;：私人] =＆GT; NULL}}

那么有人能告诉我哪些地方会使代码转储？

Answer 1

除了这一行，一切都没问题，

public function _construct($dbrow){ ...
                ^^^^^^^^^^ it should be double underscore, not single

UserData类中的构造函数方法是错误的。它应该是，

public function __construct($dbrow){ ...

Answer 2

在PHP 5.6.0+中，您可以使用magic mathod __debugInfo（）

class UserData
{
    public function __debugInfo() {
        return ['user_id' => $this->user_id, 'phone' => $this->phone, 'name' => $this->name, 'address' => $this->address];
    }
}

或使变量公开，当你有像getUserId

这样的方法时，这是不可取的

或创建转储方法，如：

class UserData
{
    public function dump() {
        var_dump($this);
    }
}

btw在课程Database方法getUser中，您一直包括调用方法的时间，快速修复：使用require_once代替include，最好是调用{课堂外的{1}} / include / require

OOP-php - 无法从数据库中检索数据

2 个答案: