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我有一个简单的表格来存储主题的结果，所以一个用户可能做了很少的主题，他们可能有不同的分数。我想创建一个存储过程来检索给定学生的最小和最大主题详细信息：

DELIMITER $$

CREATE DEFINER=`root`@`localhost`
  PROCEDURE `Search_Min_Marks_For_Student`(IN Student_code SMALLINT)
BEGIN

SELECT Subject_Subject_code AS `Minimum Scored Subject`,
       Subject_title AS `Subject Title`,
       Min(Total_mk) AS Marks 
FROM result,subject 
WHERE result.Student_Student_code = Student_code AND
      Subject_Subject_code=Subject_code;

END

我为最低分数写了一个，它工作正常，但是有没有办法可以将最高结果添加到同一个查询中？

-- Table structure for table `result`
--

DROP TABLE IF EXISTS `result`;
/*!40101 SET @saved_cs_client     = @@character_set_client */;
/*!40101 SET character_set_client = utf8 */;
CREATE TABLE `result` (
  `Result_code` smallint(6) NOT NULL AUTO_INCREMENT,
  `Student_Student_code` smallint(6) NOT NULL,
  `Subject_Subject_code` smallint(6) NOT NULL,
  `Practical_mk` smallint(6) NOT NULL,
  `Assignment_mk` smallint(6) NOT NULL,
  `Exam_mk` smallint(6) NOT NULL,
  `Total_mk` smallint(6) NOT NULL,
  `Grade` varchar(20) NOT NULL,
  PRIMARY KEY (`Result_code`),
  KEY `fk_Result_Subject1_idx` (`Subject_Subject_code`),
  KEY `fk_Result_Student1_idx` (`Student_Student_code`),
  CONSTRAINT `fk_Result_Student1` FOREIGN KEY (`Student_Student_code`) REFERENCES `student` (`Student_code`) ON DELETE NO ACTION ON UPDATE NO ACTION,
  CONSTRAINT `fk_Result_Subject1` FOREIGN KEY (`Subject_Subject_code`) REFERENCES `subject` (`Subject_code`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB AUTO_INCREMENT=26 DEFAULT CHARSET=utf8;
/*!40101 SET character_set_client = @saved_cs_client */;

--
-- Table structure for table `subject`
--

DROP TABLE IF EXISTS `subject`;
/*!40101 SET @saved_cs_client     = @@character_set_client */;
/*!40101 SET character_set_client = utf8 */;
CREATE TABLE `subject` (
  `Subject_code` smallint(6) NOT NULL AUTO_INCREMENT,
  `Subject_title` varchar(50) NOT NULL,
  `Num_of_credits` smallint(6) NOT NULL,
  `Description` varchar(100) DEFAULT NULL,
  `Course_Course_code` smallint(6) NOT NULL,
  `Department_Dep_code` smallint(6) NOT NULL,
  PRIMARY KEY (`Subject_code`),
  KEY `fk_Subject_Course1_idx` (`Course_Course_code`),
  KEY `fk_Subject_Department1_idx` (`Department_Dep_code`),
  CONSTRAINT `fk_Subject_Course1` FOREIGN KEY (`Course_Course_code`) REFERENCES `course` (`Course_code`) ON DELETE NO ACTION ON UPDATE NO ACTION,
  CONSTRAINT `fk_Subject_Department1` FOREIGN KEY (`Department_Dep_code`) REFERENCES `department` (`Dep_code`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB AUTO_INCREMENT=8004 DEFAULT CHARSET=utf8;
/*!40101 SET character_set_client = @saved_cs_client */;
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1 回答 1

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您可以使用（有点冗长的）连接以及相应的代码和标题来获得两者。此查询将最佳和最差主题作为两个单独的行返回；

SELECT DISTINCT s.Subject_Subject_code, s.Subject_title, r1.Total_mk
FROM result r1
LEFT JOIN result r2 
  ON r1.Total_mk > r2.Total_mk
 AND r1.Student_Student_code = r2.Student_Student_code
LEFT JOIN result r3 
  ON r1.Total_mk < r3.Total_mk
 AND r1.Student_Student_code = r3.Student_Student_code
JOIN Subject s
  ON r1.Subject_code=s.Subject_Subject_code
WHERE (r2.Total_mk IS NULL OR r3.Total_mk IS NULL)
 AND r1.Student_Student_code = 1;

它所做的基本上是为学生选择不存在较低分数（r2 连接）或不存在更好分数（r3 连接）的行，然后与主题连接以获取要显示的数据。

一个用于测试的 SQLfiddle。

于 2013-05-11T15:08:01.340 回答