不知道我的查询是否正在做它应该做的事

Question

我不确定我的查询是否正在做问题的要求。我要么理解错误的问题，要么我对这个问题想要什么感到困惑。但是，从问题中我得到的是找到所有会员的所有memberID，firstname和lastname，这些会员在“Harry Potter”这本书的过去已经签出/或已经签出。

这里有一个问题：

列出MEMBERID，名字，和姓氏谁借当前或过去在冠军“哈利·波特”的库中的所有书籍的成员。如果任何此类书籍具有多份副本，则该成员必须至少借用一本此类书籍的副本。

代码：

CREATE TABLE Book 
    (bookID INT, 
    ISBN INT, 
    title varchar (60), 
    author varchar (20), 
    publish_year INT, 
    category varchar(20), 
    PRIMARY KEY (bookID)); 

CREATE TABLE Member 
    (memberID INT, 
    lastname varchar (20), 
    firstname varchar (20), 
    address varchar(20), 
    phone_number INT, 
    limit_ INT, 
    PRIMARY KEY (memberID)); 

CREATE TABLE CurrentLoan 
    (memberID INT , 
    bookID INT, 
    loan_date DATE, 
    due_date DATE, 
    PRIMARY KEY (memberID, bookID), 
    FOREIGN KEY (memberID) REFERENCES Member(memberID), 
    FOREIGN KEY (bookID) REFERENCES Book(bookID)); 

CREATE TABLE History 
    (memberID INT, 
    bookID INT, 
    loan_date DATE, 
    return_date DATE, 
    PRIMARY KEY (memberID, bookID, loan_date), 
    FOREIGN KEY (memberID) REFERENCES Member(memberID), 
    FOREIGN KEY (bookID) REFERENCES Book(bookID)); 

INSERT INTO Book VALUES (10, 7771452369, 'XML and XQuery', 'Author Le', 2017, 'reference'); 
INSERT INTO Book VALUES (11, 8881245525, 'XQuery: The XML Query Language', 'Jack Se', 2017, 'reference'); 
INSERT INTO Book VALUES (12, 9991123546, 'Yellow Bird', 'Jake Red', 2014, 'reference'); 
INSERT INTO BOOK VALUES (13, 1212121212, 'The Giving Tree', 'Shel Silverstein', 1964, 'fiction'); 
INSERT INTO BOOK VALUES (14, 2121212121, 'Gone Fishing', 'Shel Silverstein', 1964, 'reference'); 
INSERT INTO BOOK VALUES (15, 1313131313, 'The Lazy Dog', 'Jake Red', 2016, 'childrens'); 
INSERT INTO BOOK VALUES (16, 3131313131, 'The Red Bird', 'Jake Red', 2016, 'childrens'); 
INSERT INTO BOOK VALUES (17, 1414141414, 'The Very Blue Boy', 'Ben Jen', 2006, 'fiction'); 
INSERT INTO Book VALUES (18, 1113312336, 'Harry Potter 1', 'J. K. Rowling', 2000, 'fiction'); 
Insert INTO Book VALUES (19, 1113331142, 'Harry Potter 1', 'J. K. Rowling', 2000, 'fiction'); 
INSERT INTO Book VALUES (20, 2221257787, 'The Real Harry Potter 2', 'J. K. Rowling', 2009, 'fiction'); 
INSERT INTO Book VALUES (21, 2221254896, 'The Fake Harry Potter 3', 'J. K. Rowling', 2010, 'fiction'); 
INSERT INTO Book VALUES (22, 2221254896, 'The Fake Harry Potter 3', 'J. K. Rowling', 2010, 'fiction'); 

INSERT INTO Member VALUES (001, 'Lee', 'Nancy', 'Brownlea Drive', 1254896325, 10); 
INSERT INTO Member VALUES (002, 'Le', 'Ray', '10th Street', 1234561256, 2); 
INSERT INTO Member VALUES (003, 'Kan', 'Charlie', '5th Street', 1234567236, 8); 
INSERT INTO Member VALUES (004, 'Brown', 'Joe', 'Elm Street', 1234567845, 9); 
INSERT INTO Member VALUES (005, 'Smith', 'John', '33 East', 1234567890, 3); 
INSERT INTO Member VALUES (006, 'Kope', 'NON', '358 spence', 2145345625, 5); 

INSERT INTO CurrentLoan VALUES (001, 10, '13-SEP-17', '14-NOV-17'); 
INSERT INTO CurrentLoan VALUES (001, 11, '13-SEP-17', '14-NOV-17'); 
INSERT INTO CurrentLoan VALUES (001, 18, '13-SEP-17', '14-NOV-17'); 
INSERT INTO CurrentLoan VALUES (001, 19, '13-SEP-17', '14-NOV-17'); 
INSERT INTO CurrentLoan VALUES (001, 21, '13-SEP-17', '14-NOV-17'); 
INSERT INTO CurrentLoan VALUES (001, 22, '13-SEP-17', '14-NOV-17'); 
INSERT INTO CurrentLoan VALUES (002, 11, '14-FEB-17', '12-MAR-17'); 
INSERT INTO CurrentLoan VALUES (003, 19, '12-OCT-17', '09-NOV-17'); 
INSERT INTO CurrentLoan VALUES (004, 13, '12-OCT-17', '09-NOV-17'); 
INSERT INTO CurrentLoan VALUES (004, 17, '12-OCT-17', '09-NOV-17'); 

INSERT INTO History VALUES (001, 15, '03-JAN-16', '25-MAY-16'); 
INSERT INTO History VALUES (001, 21, '03-JAN-16', '25-MAY-16'); 
INSERT INTO History VALUES (002, 21, '03-JAN-16', '25-MAY-16'); 
INSERT INTO History VALUES (002, 18, '03-JAN-16', '25-MAY-16'); 
INSERT INTO History VALUES (002, 15, '03-JAN-16', '25-MAY-16'); 
INSERT INTO History VALUES (002, 10, '03-JAN-16', '25-MAY-16'); 
INSERT INTO History VALUES (003, 10, '12-FEB-16', '05-MAY-16'); 
INSERT INTO History VALUES (004, 13, '12-JUN-16', '05-AUG-16'); 
COMMIT; 

查询：

SELECT memberID, firstname, lastname 
FROM (SELECT Member.memberID, firstname, lastname 
     FROM Member, Book, CurrentLoan 
     WHERE Member.memberID = CurrentLoan.memberID 
     AND Book.bookID = CurrentLoan.bookID 
     AND Book.title LIKE '%Harry Potter%') 
UNION 
SELECT memberID, firstname, lastname 
FROM (SELECT Member.memberID, firstname, lastname 
     FROM Member, Book, History 
     WHERE Member.memberID = History.memberID 
     AND Book.bookID = History.bookID 
     AND Book.title LIKE '%Harry Potter%'); 

Answer 1

你的代码是正确的，在提供正确答案的条款。然而，明确的JOIN声明更好，并且不太容易产生误解。此外，寻找到oracle结构的INSERT ALL ...

SELECT 
    Member.memberID AS MemberID 
    , Member.firstname AS FirstName 
    , Member.lastname AS LastName 
FROM 
    Member 
     INNER JOIN 
      CurrentLoan ON CurrentLoan.memberID = Member.memberID 
     INNER JOIN 
      Book ON CurrentLoan.bookID = Book.bookID 
      AND Book.title LIKE '%Harry Potter%' 
UNION SELECT 
     Member.memberID 
    , Member.firstname 
    , Member.lastname 
FROM 
    Member 
     INNER JOIN 
      History ON History.memberID = Member.memberID 
     INNER JOIN 
      Book ON History.bookID = Book.bookID 
      AND Book.title LIKE '%Harry Potter%'; 

UPDATE

如果你正在寻找的只是谁借了所有三个不同的HP本本的成员，那么下面的查询是一个你要。请注意，使用现有数据，您根本得不到任何结果。将计数更改为2以查找已阅读本系列中至少2本书的成员。今天的

SELECT 
     MemberID 
    , FirstName 
    , LastName 
FROM (
    SELECT 
      m.memberID AS MemberID 
     , m.firstname AS FirstName 
     , m.lastname AS LastName 
     , b.Title AS BookTitle 
    FROM 
     Member m 
      INNER JOIN CurrentLoan c ON c.memberID = m.memberID 
      INNER JOIN Book b ON c.bookID = b.bookID 
       AND b.title LIKE '%Harry Potter%' 
    UNION SELECT 
      m.memberID 
     , m.firstname 
     , m.lastname 
     , b.Title 
    FROM 
     Member m 
      INNER JOIN History h ON h.memberID = m.memberID 
      INNER JOIN Book b ON h.bookID = b.bookID 
       AND b.title LIKE '%Harry Potter%' 
) ResultSet 
GROUP BY MemberID, FirstName, LastName 
HAVING COUNT(DISTINCT BookTitle) = 3;