mysql笔试题(1)

有一张数据表(grade)如下:

-- ----------------------------
-- Table structure for `grade`
-- ----------------------------
DROP TABLE IF EXISTS `grade`;
CREATE TABLE `grade` (
  `name` varchar(64) COLLATE utf8_bin DEFAULT NULL,
  `course` varchar(64) COLLATE utf8_bin DEFAULT NULL,
  `grade` int(11) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_bin;

-- ----------------------------
-- Records of grade
-- ----------------------------
INSERT INTO `grade` VALUES ('张三', '语文', '81');
INSERT INTO `grade` VALUES ('张三', '数学', '88');
INSERT INTO `grade` VALUES ('张三', '英语', '78');
INSERT INTO `grade` VALUES ('李四', '语文', '77');
INSERT INTO `grade` VALUES ('李四', '数学', '88');
INSERT INTO `grade` VALUES ('李四', '英语', '68');
INSERT INTO `grade` VALUES ('王二', '数学', '100');

问题1:请查询出所有课程成绩均大于80分的学生姓名

答案：

select name from grade where grade>80 and name not in(select name from grade where grade<=80) group by name;