これが私のデータベース構造の例です。
uid email name
-------------------------------
1 john@gmail.com John
2 peter@gmail.com Peter
3 colin@gmail.com Colin
本のテーブル
book_id book_name
-------------------
1 PHP
2 MySQL
3 Javascripts
4 CSS
履歴表
history_uid history_bookid
---------------------------
1 1
1 2
1 3
2 1
2 3
3 3
複数選択クエリを含む現在のステートメントhistory.history_bookid
SELECT users.uid,
users.email,
users.`name`,
history.history_uid,
history.history_bookid,
GROUP_CONCAT(history.history_bookid) AS BookID,
GROUP_CONCAT(book_name) AS bookTitles
FROM users
INNER JOIN history
ON users.uid = history.history_uid
INNER JOIN book
ON history.history_bookid = book.book_id
WHERE history.history_bookid = 1
OR history.history_bookid = 3
GROUP BY users.uid
結果
uid email name history_uid history_bookid BookID bookTitles
---------------------------------------------------------------------------------
1 john@gmail.com John 1 3 3,1 Javascripts,PHP
2 peter@gmail.com Peter 2 1 1,3 PHP,Javascripts
3 colin@gmail.com Colin 3 3 3 Javascripts
質問 :
読んだ本の完全なリストを取得するにはどうすればよいuid = 1
ですか?私が欲しい出力例は、すべての本を含めることですhistory table
uid email name history_uid history_bookid BookID bookTitles
---------------------------------------------------------------------------------
1 john@gmail.com John 1 3 1,2,3 PHP,MySQL,Javascripts
2 peter@gmail.com Peter 2 1 1,3 PHP,Javascripts
3 colin@gmail.com Colin 3 3 3 Javascripts