Question

当我尝试这条线时：

unserialize($_SESSION['user']);

我收到此错误：类用户的对象无法转换为字符串

当我回应会话时，我得到：

O:4:"user":7:{s:9:"accountId";s:1:"2";s:9:"firstName";s:12:"Ronald";s:8:"lastName";s:5:"Test";s:5:"email";s:22:"blabla@hotmail.com";s:10:"postalCode";s:7:"J0L 8G0";s:8:"username";s:12:"testuser";s:4:"role";s:1:"0";}

这是用户的课程：

class user {
var $accountId;
var $firstName;
var $lastName;
var $email;
var $postalCode;
var $username;
var $role;

/**
 * @return mixed
 */
public function getAccountId()
{
    return $this->accountId;
}

/**
 * @param mixed $accountId
 */
public function setAccountId($accountId)
{
    $this->accountId = $accountId;
}

/**
 * @return mixed
 */
public function getFirstName()
{
    return $this->firstName;
}

/**
 * @param mixed $firstName
 */
public function setFirstName($firstName)
{
    $this->firstName = $firstName;
}

/**
 * @return mixed
 */
public function getLastName()
{
    return $this->lastName;
}

/**
 * @param mixed $lastName
 */
public function setLastName($lastName)
{
    $this->lastName = $lastName;
}

/**
 * @return mixed
 */
public function getEmail()
{
    return $this->email;
}

/**
 * @param mixed $email
 */
public function setEmail($email)
{
    $this->email = $email;
}

/**
 * @return mixed
 */
public function getPostalCode()
{
    return $this->postalCode;
}

/**
 * @param mixed $postalCode
 */
public function setPostalCode($postalCode)
{
    $this->postalCode = $postalCode;
}

/**
 * @return mixed
 */
public function getUsername()
{
    return $this->username;
}

/**
 * @param mixed $username
 */
public function setUsername($username)
{
    $this->username = $username;
}

/**
 * @return mixed
 */
public function getRole()
{
    return $this->role;
}

/**
 * @param mixed $role
 */
public function setRole($role)
{
    $this->role = $role;
}

function __construct($accountId,$firstName,$lastName,$email,$postalCode,$username,$role){
    $this->setAccountId($accountId);
    $this->setFirstName($firstName);
    $this->setLastName($lastName);
    $this->setEmail($email);
    $this->setPostalCode($postalCode);
    $this->setUsername($username);
    $this->setRole($role);
}

}

这是我创建对象的地方：

public function createUserObject($username){
    session_start();
    require("user.php");
    include("connection.php");
    $username = mysql_real_escape_string($username);

    mysql_connect("$SQLhost", "$SQLusername", "$SQLpassword") or die("cannot connect");
    mysql_select_db("$SQLdb_name") or die("cannot select DB");

    $username = mysql_real_escape_string($username);
    $sql = "SELECT * FROM Account WHERE username='$username'";
    $result = mysql_query($sql);

    $rows = mysql_fetch_assoc($result);
    $user = new user($rows['accountId'],$rows['firstName'],$rows['lastName'],$rows['email'],$rows['postalCode'],$rows['username'],$rows['role']);
    $_SESSION['user'] = serialize($user);

    if($result){
        return true;
    }else{
        return false;
    }
}

谢谢，如果有人能解释我的问题。

Answer 1

Mathieu，这是一个众所周知的问题。因此，如果您可以稍微修改您的代码，那将是一件好事。

请参阅以下帖子，了解一些想法：http://blog.pagalhost.com/codeigniter-php-serialize-unserialize-issues/

因此，您需要做的事情是以下方式

//To serialize
$result_string = base64_encode(serialize($array));

//To unserialize
$restored_array = unserialize(base64_decode($result_string ));

有关更多信息和想法，请查看PHP Manual - Unserialize Function

Unserialize（）错误无法转换为字符串

1 个答案: