Question

我正在做一个测试，朋友派我去测试我的PHP技能 - 但我已经碰壁了。我需要能够在插入记录后使用新的详细信息更新数据库。这完全是通过脚本完成的 - 没有GUI。

以下是代码：

<?php
class UserModel {
public $name = null, $occupation = null, $email = null, $oldname = null,     $oldoccupation = null, $oldemail = null, $me, $handler, $result;

public function __construct(){
    try{
        $this->handler = new PDO('mysql:host=127.0.0.1;dbname=lab19', 'root', 'root');
        $this->handler->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    } catch (PDOException $e) {
        echo $e->getMessage();
        die();
    }
}

public function create($fields = array()){
    if(count($fields) == 3){
        $this->name = $fields['name'];
        $this->occupation = $fields['occupation'];
        $this->email = $fields['email'];
    }

}

public function _save(){
    if($this->oldname == null && $this->oldoccupation == null && $this->oldemail == null){
        $sql = "INSERT INTO users (name, occupation, email) VALUES (:name, :occupation, :email)";
        $this->result = $this->handler->prepare($sql);
        $this->result->execute(array(
            ':name'=>$this->name,
            ':occupation'=>$this->occupation,
            ':email'=>$this->email
        ));
    } else {
        $sql = "UPDATE users SET name = :name, occupation = :occupation, email = :email WHERE name = :oldname";
        $this->result = $this->handler->prepare($sql);
        $this->result->execute(array(
            ':name'=>$this->name,
            ':occupation'=>$this->occupation,
            ':email'=>$this->email,
            ':oldname'=>$this->oldname
        ));
    }
}

public function name($given_name = null) {
    if($this->name == null){
        if($given_name != null){
            $this->name = $given_name;
        }
    } else {
        if($given_name != null){
            $this->oldname = $this->name;
            $this->name = $given_name;
        }
    }
    return $this->name;
}

public function occupation($given_occupation = null) {
    if($this->occupation == null){
        if($given_occupation != null){
            $this->occupation = $given_occupation;
        }
    } else {
        if($given_occupation != null){
            $this->oldoccupation = $this->occupation;
            $this->occupation = $given_occupation;
        }
    }
    return $this->occupation;
}

public function email($given_email = null){
    $this->verifyEmail($given_email);
    if($given_email != null){
        // $this->oldemail = $this->email; // THIS LINE IS THE ISSUE
        $this->email = $given_email;
    }
    return $this->email;
}

public function verifyEmail($givenemail = null){
    if($givenemail == null){
        $email = $this->email;
    } else {
        $email = $givenemail;

    }
    if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
        throw new Exception('Email not valid.');
        die();
    }

}
}

$user = new UserModel();
$user->create(array(
'name' => 'Luke',
'occupation' => 'Programmer',
'email' => 'luke@gmail.com'
));
$user->_save();
// $user->name('Jack');
// $user->occupation();
try {
$user->email('demo@example.co.za');
} catch (Exception $e) {
echo $e->getMessage();
}
$user->_save();

但是当我有这行$this->oldemail = $this->email;时，电子邮件不会在数据库中发生变化 - 但是当我把它取出时一切正常。可能是什么问题??

Answer 1

哪里你有oldemail设置？是在你尝试做_save之前吗？如果是这样，您最终可能会在不存在的记录上执行UPDATE，而不是INSERT。代码不是很好 - 也许不是试图弄清楚是做INSERT还是UPDATE，你可以简单地做一个REPLACE INTO。

Answer 2

当前拥有$user->name('Jack');和$user->occupation();注释掉$this->oldname和$this->oldoccupation的方式永远不会从null的初始设置更改。

在UPDATE声明中，您的条件为WHERE name = :oldname。由于$this->oldname为null，因此您不会更新任何内容。

数据库中的PHP更新字段 - 1行停止字段更新

2 个答案: