Php mysqli prepare语句使用select由于未知原因失败

为什么会失败？我连接到了正确的数据库，而且钥匙肯定在那里

//verify key against key list in database
if ($prep_statement = $db->prepare("SELECT id FROM key_list WHERE keys=?"))
{
    $prep_statement->bind_param('s',$key);
    $prep_statement->execute();
    echo $prep_statement->num_rows;
     $prep_statement->close();
} else {
    printf("Prepared Statement Error: %s\n", $db->error); 
    die();      
}

我得到以下错误，if语句返回false

Prepared Statement Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'key=?' at line 1 

以下是我的表格结构：

-- MySQL dump 10.11
--
-- Host: localhost    Database: gameserverdev
-- ------------------------------------------------------
-- Server version   5.0.51a-24+lenny4

/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;
/*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;
/*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;
/*!40101 SET NAMES utf8 */;
/*!40103 SET @OLD_TIME_ZONE=@@TIME_ZONE */;
/*!40103 SET TIME_ZONE='+00:00' */;
/*!40014 SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0 */;
/*!40014 SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0 */;
/*!40101 SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='NO_AUTO_VALUE_ON_ZERO' */;
/*!40111 SET @OLD_SQL_NOTES=@@SQL_NOTES, SQL_NOTES=0 */;

--
-- Table structure for table `key_list`
--

DROP TABLE IF EXISTS `key_list`;
SET @saved_cs_client     = @@character_set_client;
SET character_set_client = utf8;
CREATE TABLE `key_list` (
  `id` bigint(20) unsigned NOT NULL auto_increment,
  `keys` char(16) NOT NULL,
  `datetime_registered` datetime NOT NULL,
  `banned` tinyint(1) NOT NULL,
  `comment` char(255) default NULL,
  PRIMARY KEY  (`id`)
) ENGINE=MyISAM AUTO_INCREMENT=3 DEFAULT CHARSET=latin1;
SET character_set_client = @saved_cs_client;

--
-- Dumping data for table `key_list`
--

LOCK TABLES `key_list` WRITE;
/*!40000 ALTER TABLE `key_list` DISABLE KEYS */;
INSERT INTO `key_list` VALUES (2,'1111222233334444','2010-07-31 16:04:30',0,NULL);
/*!40000 ALTER TABLE `key_list` ENABLE KEYS */;
UNLOCK TABLES;

--
-- Dumping routines for database 'gameserverdev'
--
DELIMITER ;;
DELIMITER ;
/*!40103 SET TIME_ZONE=@OLD_TIME_ZONE */;

/*!40101 SET SQL_MODE=@OLD_SQL_MODE */;
/*!40014 SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS */;
/*!40014 SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS */;
/*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT */;
/*!40101 SET CHARACTER_SET_RESULTS=@OLD_CHARACTER_SET_RESULTS */;
/*!40101 SET COLLATION_CONNECTION=@OLD_COLLATION_CONNECTION */;
/*!40111 SET SQL_NOTES=@OLD_SQL_NOTES */;

-- Dump completed on 2010-07-31 23:00:56

您正在绑定结果，而不是绑定。将bind_result更改为bind_param，它应该可以正常工作

密钥是MySQL。您需要将列名更改为非保留字，或者需要引用它。在MySQL中，是反引号，`，但如果启用了ANSI引号，则可以使用双引号

因此，你的声明应该是：

SELECT id FROM key_list WHERE `key` = ?

---

哈哈，我以前没注意到，但你用的是保留字。答案已更新。现在可以工作了。我试了两把钥匙和两把钥匙，但显然它们都是保留的。谢谢