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用一條SQL語句查出每門課都大於80分的學生的姓名

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用一條SQL語句查出每門課都大於80分的學生的姓名,數據表結構如下:

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建表SQL如下:

SET FOREIGN_KEY_CHECKS=0;

-- ----------------------------
-- Table structure for grade
-- ----------------------------
DROP TABLE IF EXISTS `grade`;
CREATE TABLE `grade` (
  `name` varchar(255) NOT NULL,
  `class` varchar(255) NOT NULL,
  `score` tinyint(4
) NOT NULL ) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4; -- ---------------------------- -- Records of grade -- ---------------------------- INSERT INTO `grade` VALUES (張三, 語文, 81); INSERT INTO `grade` VALUES (張三, 數學, 75); INSERT INTO `grade` VALUES (李四, 語文, 76); INSERT INTO `grade` VALUES
(李四, 數學, 90); INSERT INTO `grade` VALUES (王五, 語文, 81); INSERT INTO `grade` VALUES (王五, 數學, 100); INSERT INTO `grade` VALUES (王五, 英語, 90); SET FOREIGN_KEY_CHECKS=1;

查詢每門課都大於80分的同學的姓名:

SELECT DISTINCT name FROM grade WHERE name NOT IN(SELECT DISTINCT name FROM grade WHERE score <=
80);

查詢平均分大於80的學生的姓名:

SELECT name FROM (SELECT COUNT(*) AS t,SUM(score) AS num,name FROM `grade` GROUP BY name) AS a WHERE a.num > 80*t;

用一條SQL語句查出每門課都大於80分的學生的姓名