我在MySQL数据库中有这些表用于简单的大学图书馆应用程序（由于显而易见的原因，我将它们简化了一点）：
SELECT * FROM language;
╔═════════════╦══════════╗
║ language_id ║ language ║
╠═════════════╬══════════╣
║ 11 ║ English ║
║ 12 ║ Russian ║
║ 13 ║ German ║
╚═════════════╩══════════╝
SELECT * FROM subject;
╔════════════╦═════════════╗
║ subject_id ║ subject ║
╠════════════╬═════════════╣
║ 21 ║ Mathematics ║
║ 22 ║ History ║
║ 23 ║ Chemistry ║
║ 24 ║ Physics ║
╚════════════╩═════════════╝
SELECT publisher_id, publisher FROM publisher;
╔══════════════╦═══════════════╗
║ publisher_id ║ publisher ║
╠══════════════╬═══════════════╣
║ 31 ║ Berkley Books ║
║ 32 ║ Penguin ║
╚══════════════╩═══════════════╝
SELECT author_id, first_name, last_name FROM author;
╔═══════════╦════════════╦═══════════╗
║ author_id ║ first_name ║ last_name ║
╠═══════════╬════════════╬═══════════╣
║ 41 ║ Joe ║ Schmoe ║
║ 42 ║ John ║ Smith ║
╚═══════════╩════════════╩═══════════╝
SELECT book_id, title, language_id, subject_id, publisher_id FROM book;
╔═════════╦═══════╦═════════════╦════════════╦══════════════╗
║ book_id ║ title ║ language_id ║ subject_id ║ publisher_id ║
╠═════════╬═══════╬═════════════╬════════════╬══════════════╣
║ 1 ║ A ║ 11 ║ 22 ║ 31 ║
║ 2 ║ B ║ 13 ║ 24 ║ 32 ║
╚═════════╩═══════╩═════════════╩════════════╩══════════════╝
SELECT book_id, author_id FROM book_author;
╔═════════╦════════════╗
║ book_id ║ author_id ║
╠═════════╬════════════╣
║ 1 ║ 41 ║
║ 1 ║ 42 ║
║ 2 ║ 41 ║
╚═════════╩════════════╝
SELECT book_copy_id, book_id FROM book_copy;
╔══════════════╦═════════╗
║ book_copy_id ║ book_id ║
╠══════════════╬═════════╣
║ 1 ║ 1 ║
║ 2 ║ 1 ║
║ 3 ║ 1 ║
║ 4 ║ 2 ║
║ 5 ║ 2 ║
╚══════════════╩═════════╝
我有一个看起来像这样的查询（我为了以防万一而向你展示）：
SELECT b.book_id,
b.title,
GROUP_CONCAT(CONCAT_WS(' ', a.last_name,
a.first_name)
ORDER BY a.last_name
SEPARATOR ', ') AS authors,
l.language,
s.subject,
p.publisher
FROM book AS b
LEFT JOIN book_author AS ba
ON b.book_id = ba.book_id
LEFT JOIN author AS a
ON ba.author_id = a.author_id
LEFT JOIN language AS l
ON b.language_id = l.language_id
LEFT JOIN subject AS s
ON b.subject_id = s.subject_id
LEFT JOIN publisher AS p
ON b.publisher_id = p.publisher_id
GROUP BY b.title ASC;
╔═════════╦═══════╦════════════════════════╦══════════╦═════════╦═══════════════╗
║ book_id ║ title ║ authors ║ language ║ subject ║ publisher ║
╠═════════╬═══════╬════════════════════════╬══════════╬═════════╬═══════════════╣
║ 1 ║ A ║ Joe Schmoe, John Smith ║ English ║ History ║ Berkley Books ║
║ 2 ║ B ║ John Smith ║ German ║ Physics ║ Penguin ║
╚═════════╩═══════╩════════════════════════╩══════════╩═════════╩═══════════════╝
现在，我想有这个选择：
╔══════════════╦═══════╦════════════════════════╦══════════╦═════════╦═══════════════╗
║ book_copy_id ║ title ║ authors ║ language ║ subject ║ publisher ║
╠══════════════╬═══════╬════════════════════════╬══════════╬═════════╬═══════════════╣
║ 1 ║ A ║ Joe Schmoe, John Smith ║ English ║ History ║ Berkley Books ║
║ 2 ║ A ║ Joe Schmoe, John Smith ║ English ║ History ║ Berkley Books ║
║ 3 ║ A ║ Joe Schmoe, John Smith ║ English ║ History ║ Berkley Books ║
║ 4 ║ B ║ John Smith ║ German ║ Physics ║ Penguin ║
║ 5 ║ B ║ John Smith ║ German ║ Physics ║ Penguin ║
╚══════════════╩═══════╩════════════════════════╩══════════╩═════════╩═══════════════╝
我也有这两个表：
SELECT student_id, first_name, last_name FROM student;
╔════════════╦════════════╦═══════════╗
║ student_id ║ first_name ║ last_name ║
╠════════════╬════════════╬═══════════╣
║ 81 ║ Bob ║ Dylan ║
║ 82 ║ Jim ║ Carrey ║
╚════════════╩════════════╩═══════════╝
SELECT student_id, book_copy_id FROM loan;
╔════════════╦══════════════╗
║ student_id ║ book_copy_id ║
╠════════════╬══════════════╣
║ 81 ║ 1 ║
║ 81 ║ 4 ║
║ 82 ║ 5 ║
╚════════════╩══════════════╝
我需要编写一个产生这个选择的查询：
-- 'a' stands for 'the quantity of book copies that are available for a loan'
-- 't' stands for 'the total quantity of book copies'
╔═══╦═══╦═══════╦════════════════════════╦══════════╦═════════╦═══════════════╗
║ a ║ t ║ title ║ authors ║ language ║ subject ║ publisher ║
╠═══╬═══╬═══════╬════════════════════════╬══════════╬═════════╬═══════════════╣
║ 2 ║ 3 ║ A ║ Joe Schmoe, John Smith ║ English ║ History ║ Berkley Books ║
║ 0 ║ 2 ║ B ║ John Smith ║ German ║ Physics ║ Penguin ║
╚═══╩═══╩═══════╩════════════════════════╩══════════╩═════════╩═══════════════╝
我希望我已经非常明确地用图表概述了它。请把这两个问题的问题写下来。
答案 0 :(得分：1)
您需要加入book_copy并更改group by条件。这解决了您的第一个问题：
SELECT bc.book_copy_id,
b.title,
GROUP_CONCAT(CONCAT_WS(' ', a.last_name,
a.first_name)
ORDER BY a.last_name
SEPARATOR ', ') AS authors,
l.language,
s.subject,
p.publisher
FROM book AS b
LEFT JOIN book_author AS ba
ON b.book_id = ba.book_id
LEFT JOIN author AS a
ON ba.author_id = a.author_id
LEFT JOIN language AS l
ON b.language_id = l.language_id
LEFT JOIN subject AS s
ON b.subject_id = s.subject_id
LEFT JOIN publisher AS p
ON b.publisher_id = p.publisher_id
left join book_copy bc
on b.book_id = bc.book_id
GROUP BY bc.book_copy_id ASC;
答案 1 :(得分：0)
看起来我解决了第一个问题。请看看它是否合法（但感觉它完美无缺）：
SELECT bc.book_copy_id,
subquery.title,
subquery.authors,
subquery.subject,
subquery.language,
subquery.publisher
FROM book_copy AS bc
LEFT JOIN (SELECT b.book_id,
b.title,
GROUP_CONCAT(CONCAT_WS(' ', a.last_name,
a.first_name)
ORDER BY a.last_name
SEPARATOR ', ') AS authors,
l.language,
s.subject,
p.publisher
FROM book AS b
LEFT JOIN book_author AS ba
ON b.book_id = ba.book_id
LEFT JOIN author AS a
ON ba.author_id = a.author_id
LEFT JOIN language AS l
ON b.language_id = l.language_id
LEFT JOIN subject AS s
ON b.subject_id = s.subject_id
LEFT JOIN publisher AS p
ON b.publisher_id = p.publisher_id
GROUP BY b.title) AS subquery
ON bc.book_id = subquery.book_id
ORDER BY subquery.title
现在我已经完成了第二个问题。这是解决方案：
SELECT subquery.book_id,
(COUNT(bc.book_copy_id) - (SELECT COUNT(l.book_copy_id)
FROM loan AS l
WHERE l.checked_in IS NULL AND
l.book_copy_id IN (SELECT bc.book_copy_id
FROM book_copy AS bc
WHERE bc.book_id = subquery.book_id))) AS available,
COUNT(bc.book_copy_id) AS total,
subquery.title,
subquery.authors,
subquery.subject,
subquery.language,
subquery.publisher
FROM book_copy AS bc
LEFT JOIN (SELECT b.book_id,
b.title,
GROUP_CONCAT(CONCAT_WS(' ', a.last_name,
a.first_name)
ORDER BY a.last_name
SEPARATOR ', ') AS authors,
l.language,
s.subject,
p.publisher
FROM book AS b
LEFT JOIN book_author AS ba
ON b.book_id = ba.book_id
LEFT JOIN author AS a
ON ba.author_id = a.author_id
LEFT JOIN language AS l
ON b.language_id = l.language_id
LEFT JOIN subject AS s
ON b.subject_id = s.subject_id
LEFT JOIN publisher AS p
ON b.publisher_id = p.publisher_id
GROUP BY b.title) AS subquery
ON bc.book_id = subquery.book_id
GROUP BY subquery.book_id
ORDER BY subquery.title;
答案 2 :(得分：0)
这应该可以解决问题，但不确定它是如何有效的。
请注意，我将GROUP BY更改为book_id而不是title。如果两本书具有相同的标题，它们仍然是不同的书籍（不确定是否会发生这种情况）。
SELECT COUNT(DISTINCT bc.book_copy_id) - COUNT(DISTINCT lo.book_copy_id) AS a,
COUNT(DISTINCT bc.book_copy_id) AS t,
b.book_id,
b.title,
GROUP_CONCAT(DISTINCT CONCAT_WS(' ', a.last_name, a.first_name) ORDER BY a.last_name SEPARATOR ', ') AS authors,
l.language,
s.subject,
p.publisher
FROM book AS b
LEFT JOIN book_author AS ba ON b.book_id = ba.book_id
LEFT JOIN author AS a ON ba.author_id = a.author_id
LEFT JOIN language AS l ON b.language_id = l.language_id
LEFT JOIN subject AS s ON b.subject_id = s.subject_id
LEFT JOIN publisher AS p ON b.publisher_id = p.publisher_id
LEFT JOIN book_copy AS bc ON (bc.book_id = b.book_id)
LEFT JOIN loan AS lo ON (lo.book_copy_id = bc.book_copy_id)
GROUP BY b.book_id;
- 编辑
请注意，GROUP_CONCAT中有DISTINCT。如果两位作者具有相同的名称，则只会列出一次。不确定这是否重要。