我想使用这个类查询:
class DB {
private static $_instance = null;
private $_pdo,
$_query,
$_error = false,
$_results,
$_count = 0;
private function __construct() {
try {
$this->_pdo = new PDO('mysql:host=' . Config::get('mysql/host') . ';dbname=' . Config::get('mysql/db'), Config::get('mysql/username'), Config::get('mysql/password'));
} catch(PDOException $e) {
die($e->getMessage());
}
}
public static function getInstance() {
if(!isset(self::$_instance)) {
self::$_instance = new DB();
}
return self::$_instance;
}
public function query($sql, $params = array()) {
$this->_error = false;
if($this->_query = $this->_pdo->prepare($sql)) {
$x = 1;
if(count($params)) {
foreach ($params as $param) {
$this->_query->bindValue($x, $param);
$x++;
}
}
if($this->_query->execute()) {
$this->_results = $this->_query->fetchAll(PDO::FETCH_OBJ);
$this->_count = $this->_query->rowCount();
}else {
$this->_error = true;
}
}
return $this;
}
我正试图达到这样的查询方法:
$sql = "SELECT * FROM name
WHERE namel LIKE :seek";
$db = DB::getInstance();
$db->query($sql, array(':seek' => $seek));
我收到这个警告: 警告:PDOStatement :: execute():SQLSTATE [HY093]:参数号无效:参数未在第36行的C:\ xampp \ htdocs \ blabla \ classes \ DB.php中定义
我甚至不确定我的方式达到DB类。
答案 0 :(得分:1)
更新你的查询fn() - 问题在你的bindValue中你需要传递param和值来绑定
public function query($sql, $params = array()) {
$this->_error = false;
if($this->_query = $this->_pdo->prepare($sql)) {
if(count($params)) {
foreach ($params as $param => $value) {
$this->_query->bindValue($param, $value);
}
}
if($this->_query->execute()) {
$this->_results = $this->_query->fetchAll(PDO::FETCH_OBJ);
$this->_count = $this->_query->rowCount();
}else {
$this->_error = true;
}
}
return $this;
}
这取决于您的喜欢结果,如果您希望它显示开始,之后您必须像这样更新您的值:
$this->_query->bindValue($param, '%' . $value . '%');
我不知道你为什么要回来这个。
此外,如果您试图从课外获取结果,则需要公开才能访问该结果。
public $_results;