我需要恢復某個查詢的行數,但它使我這個錯誤試圖讓非對象錯誤的性質
試圖讓非對象的屬性在C:\ XAMPP \ htdocs中\樣本\包括\ CheckUsername.php上線18
這裏是我的代碼:
<?php
class Db_CheckUsername{
protected $_conn;
protected $_username;
protected $_errors = array();
public function __construct($username,$conn){
$this->_username = $username;
$this->_conn = $conn;
}
public function isUsernameAvailable(){
$sql = "SELECT ";
$sql .= "FROM accounts ";
$sql .= "WHERE username = {$this->_username}";
$result = $this->_conn->query($sql);
$numRows = $result->num_rows;
if($numRows == 0){
return true;
}else{
return false;
}
}
}
?>
我該如何解決這個問題?
哪一個正好是'18號線'? – Havelock
確保'_con'實際上包含有效的連接對象。你可以通過type-hinting構造函數來實現:'public function __construct($ username,MySQLi $ conn){' –
@Havelock:'$ numRows = $ result-> num_rows;' –