在 mysql 中加入四个表。语法错误答案

【问题标题】：JOIN FOUR TABLES in mysql. Error in syntax在 mysql 中加入四个表。语法错误
【发布时间】：2015-09-19 07:24:38
【问题描述】：

我需要在mysql中加入四个表。

我的数据库结构：

DROP DATABASE IF EXISTS db_applicant;

CREATE DATABASE db_applicant
  DEFAULT CHARACTER SET 'utf8'
  DEFAULT COLLATE 'utf8_unicode_ci';

USE db_applicant;

--
-- TABLE: PROFESSION
--
CREATE TABLE PROFESSION (
  PROFESSION_ID   INT         NOT NULL AUTO_INCREMENT,
  PROFESSION_NAME VARCHAR(50) NOT NULL,
  PRIMARY KEY (PROFESSION_ID)
);

--
-- TABLE: SUBJECT
--
CREATE TABLE SUBJECT (
  SUBJECT_ID   INT         NOT NULL AUTO_INCREMENT,
  SUBJECT_NAME VARCHAR(50) NOT NULL,
  PRIMARY KEY (SUBJECT_ID)
);

--
-- TABLE: APPLICANT
--
CREATE TABLE APPLICANT (
  APPLICANT_ID  INT         NOT NULL AUTO_INCREMENT,
  PROFESSION_ID INT         NOT NULL,
  LAST_NAME     VARCHAR(30) NOT NULL,
  FIRST_NAME    VARCHAR(30) NOT NULL,
  ENTRANCE_YEAR INT         NOT NULL,
  PRIMARY KEY (APPLICANT_ID),
  FOREIGN KEY (PROFESSION_ID) REFERENCES PROFESSION (PROFESSION_ID)
);

--
-- TABLE: APPLICANT_RESULT
--
CREATE TABLE APPLICANT_RESULT (
  APPLICANT_RESULT_ID INT NOT NULL AUTO_INCREMENT,
  APPLICANT_ID        INT NOT NULL,
  SUBJECT_ID          INT NOT NULL,
  MARK                INT,
  PRIMARY KEY (APPLICANT_RESULT_ID),
  FOREIGN KEY (SUBJECT_ID)
  REFERENCES SUBJECT (SUBJECT_ID),
  FOREIGN KEY (APPLICANT_ID)
  REFERENCES APPLICANT (APPLICANT_ID)
);


--
-- TABLE: SPECIALITY_SUBJECT
--
CREATE TABLE SPECIALITY_SUBJECT (
  SP_SB_ID      INT NOT NULL AUTO_INCREMENT,
  PROFESSION_ID INT NOT NULL,
  SUBJECT_ID    INT NOT NULL,
  PRIMARY KEY (SP_SB_ID),
  FOREIGN KEY (PROFESSION_ID)
  REFERENCES PROFESSION (PROFESSION_ID),
  FOREIGN KEY (PROFESSION_ID)
  REFERENCES PROFESSION (PROFESSION_ID),
  FOREIGN KEY (SUBJECT_ID)
  REFERENCES SUBJECT (SUBJECT_ID)
);

我需要这样的输出：

first_name（申请表中的这一列），last_name（申请表中的这一列），entry_year（申请表中的这一列），professional_name（表中的这一列），subject_name（表中的这一列），mark (此列来自表申请者_结果）。

你可以看到，我有相关的领域。但我需要强大的内部查询。为此，我创建了具有结构的新表：

CREATE TABLE APP(
  ALL_ID              INT NOT NULL AUTO_INCREMENT,
  APPLICANT_ID        INT NOT NULL,
  SUBJECT_ID          INT NOT NULL,
  PROFESSION_ID       INT NOT NULL,
  APPLICANT_RESULT_ID INT NOT NULL,
  PRIMARY KEY (ALL_ID),
  FOREIGN KEY (SUBJECT_ID)
  REFERENCES SUBJECT (SUBJECT_ID),
  FOREIGN KEY (APPLICANT_ID)
  REFERENCES APPLICANT (APPLICANT_ID),
  FOREIGN KEY (PROFESSION_ID)
  REFERENCES PROFESSION (PROFESSION_ID),
  FOREIGN KEY (APPLICANT_RESULT_ID)
  REFERENCES APPLICANT_RESULT (APPLICANT_RESULT_ID)
);

还有我的内心：

SELECT ap.ALL_ID, a.FIRST_NAME, a.LAST_NAME,
       a.ENTRANCE_YEAR, p.PROFESSION_NAME s.SUBJECT_NAME, ar.MARK
FROM app ap
JOIN (applicant a, profession p, subject s, applicant_result ar)
 ON ap.APPLICANT_ID = a.APPLICANT_ID
AND ap.SUBJECT_ID = s.SUBJECT_ID
AND ap.PROFESSION_ID = p.PROFESSION_ID
AND ap.APPLICANT_RESULT_ID = ar.APPLICANT_RESULT_ID;

但我有错误：

[2015-09-19 10:08:52] [42000][1064] 你的 SQL 有错误句法;检查与您的 MySQL 服务器版本相对应的手册在 '.SUBJECT_NAME, ar.MARK FROM app ap 附近使用正确的语法 JOIN（申请人 a，职业 p，学科 s，a' 在第 1 行

【问题讨论】：

你少了一个逗号。

标签： mysql sql join

【解决方案1】：

您不认为，在 s.SUBJECT_NAME 之前的 select 语句中这里缺少逗号“，”吗？

SELECT ap.ALL_ID, a.FIRST_NAME, a.LAST_NAME,
       a.ENTRANCE_YEAR, p.PROFESSION_NAME s.SUBJECT_NAME, ar.MARK

【讨论】：