php pdo lastInsertId（）作为第二个表的输入

我感谢所有的帮助！！我是PDO新手，在对两个不同的表执行插入时遇到问题。我在第二个表（fk_employee）中有一个字段，我想要第一个查询（insert）中最后记录的id。运行查询时，我收到以下错误：

查询失败：SQLSTATE[42000]：语法错误或访问冲突：1064您的SQL语法有错误；请查看与MySQL服务器版本对应的手册，以了解在“；”附近使用的正确语法第10行的“地址1”、“地址2”、“城市”、“TN”、“77777”

以下是我的方法：

        function completeCallback()
    {
        // finally, all form data is valid and the user has confirmed they
        // want to proceed. Now we do all the final processing here.

        $name_title = $this->getValue('name_title');
        $emp_lastname = $this->getValue('emp_lastname');
        $emp_suffix = $this->getValue('emp_suffix');
        $emp_firstname = $this->getValue('emp_firstname');
        $emp_middlename = $this->getValue('emp_middlename');
        $emp_prefername = $this->getValue('emp_prefername');
        $address1 = $this->getValue('address1');
        $address2 = $this->getValue('address2');
        $city = $this->getValue('city');
        $state = $this->getValue('state');
        $postal_code = $this->getValue('postal_code');
        $country = $this->getValue('country');

      $conn = parent::connect();
      $sql = "INSERT INTO " . TBL_EMPLOYEES . " (
                name_title,
                emp_lastname,
                emp_firstname,
                emp_suffix,
                emp_prefername,
                emp_middlename
              ) VALUES (
                :name_title,
                :emp_lastname,
                :emp_firstname,
                :emp_suffix,
                :emp_prefername,
                :emp_middlename
              )";

      try {
        $st = $conn->prepare( $sql );
        $st->bindValue( ":name_title", $name_title, PDO::PARAM_STR );
        $st->bindValue( ":emp_lastname", $emp_lastname, PDO::PARAM_STR );
        $st->bindValue( ":emp_firstname", $emp_firstname, PDO::PARAM_STR );
        $st->bindValue( ":emp_suffix", $emp_suffix, PDO::PARAM_STR );
        $st->bindValue( ":emp_prefername", $emp_prefername, PDO::PARAM_STR );
        $st->bindValue( ":emp_middlename", $emp_middlename, PDO::PARAM_STR );
        if($st->execute()){
            $lastInsertID = $conn->lastInsertId();
        }
      $sql = "INSERT INTO " . TBL_EMPLOYEEADD . " (
                fk_employee,
                address1,
                address2,
                city,
                state,
                postal_code,
                country
              ) VALUES (
                :fk_employee;
                :address1,
                :address2,
                :city,
                :state,
                :postal_code,
                :country
              )";

        $st = $conn->prepare( $sql );
        $st->bindValue( ':fk_employee', $lastInsertID, PDO::PARAM_INT );
        $st->bindValue( ':address1', $address1, PDO::PARAM_STR );
        $st->bindValue( ':address2', $address2, PDO::PARAM_STR );
        $st->bindValue( ':city', $city, PDO::PARAM_STR );
        $st->bindValue( ':state', $state, PDO::PARAM_STR );
        $st->bindValue( ':postal_code', $postal_code, PDO::PARAM_STR );
        $st->bindValue( ':country', $country, PDO::PARAM_STR );
        if($st->execute()){
            $lastInsertID = $conn->lastInsertId();
        }
      } catch ( PDOException $e ) {
        parent::disconnect( $conn );
        die( "Query failed: " . $e->getMessage() );
      }
    }

这是我的两张桌子：

CREATE TABLE IF NOT EXISTS `ds_employee` (
  `id_employee` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `name_title` varchar(20) DEFAULT NULL,
  `emp_lastname` varchar(100) NOT NULL DEFAULT '',
  `emp_suffix` varchar(50) DEFAULT NULL,
  `emp_firstname` varchar(100) NOT NULL DEFAULT '',
  `emp_prefername` varchar(100) NOT NULL DEFAULT '',
  `emp_middlename` varchar(100) NOT NULL DEFAULT '',
  PRIMARY KEY (`id_employee`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 AUTO_INCREMENT=19 ;

表2

我相信这很简单，但我一直在敲我的头，想不出来

谢谢，
REF
将该查询更改为以下内容

$sql = "INSERT INTO " . TBL_EMPLOYEEADD . " (
                fk_employee,
                address1,
                address2,
                city,
                state,
                postal_code,
                country
              ) VALUES (
                :fk_employee,
                :address1,
                :address2,
                :city,
                :state,
                :postal_code,
                :country
              )";

我很抱歉用这么愚蠢的问题来打扰你！！非常感谢你。就这样！！无关：fk_员工是否建议我认为它的功能？
$sql = "INSERT INTO " . TBL_EMPLOYEEADD . " (
                fk_employee,
                address1,
                address2,
                city,
                state,
                postal_code,
                country
              ) VALUES (
                :fk_employee,
                :address1,
                :address2,
                :city,
                :state,
                :postal_code,
                :country
              )";