我在终端中创建了一个名为“somedb”的新数据库

CREATE DATABASE somedb

在我的桌面上,我有从PHPMyadmin下载的sql转储：somedb.sql

我试过了：

somedb < /Users/myname/Desktop/somedb.sql

结果：错误1064(42000)：您的sql语法中有错误

MysqL -u myname -p -h localhost somedb </Users/myname/Desktop/somedb.sql

结果：错误1064(42000)：您的sql语法有错误;

我是sql新手(导入此数据库的目的是进行教科书练习)

我已授予自己所有权限,但没有密码.

知道我做错了什么吗？

这是sql转储文件的顶部：

-- PHPMyAdmin sql Dump
-- version 4.0.2
-- http://www.PHPmyadmin.net
--
-- Host: localhost
-- Generation Time: Jun 18,2013 at 02:22 PM
-- Server version: 5.5.31-30.3
-- PHP Version: 5.2.17

SET sql_MODE = "NO_AUTO_VALUE_ON_ZERO";
SET time_zone = "+00:00";


/*!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 */;

--
-- Database: `somedb`
--
CREATE DATABASE IF NOT EXISTS `somedb` DEFAULT CHARACTER SET utf8 COLLATE utf8_general_ci;
USE `somedb`;

-- --------------------------------------------------------

--
-- Table structure for table `actions`
--

CREATE TABLE IF NOT EXISTS `actions` (
  `action_id` int(11) NOT NULL AUTO_INCREMENT,`action` varchar(75) NOT NULL,`qualifiers` text NOT NULL,`response` varchar(75) NOT NULL,`response_vars` text NOT NULL,`active` tinyint(4) NOT NULL,PRIMARY KEY (`action_id`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 COMMENT='Stores user defined actions triggered by certain events' AUTO_INCREMENT=3 ;

-- --------------------------------------------------------

--