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Question

我有一个章节表，其中大约有17个条目都属于一本书。

chapters:
 --------------------------
|  id  | book_id | content |
 --------------------------
|   1  |    1    |   ...   |
|   2  |    1    |   ...   |
|   3  |    1    |   ...   |
|   4  |    1    |   ...   |
|   5  |    1    |   ...   |
|  ... |    1    |   ...   |
 --------------------------

还有一个图书馆表，可以将用户链接到他们保存的图书：

library:
 ----------------
| user | book_id |
 ----------------
|   1  |    1    |
|   1  |    2    |
|   2  |    1    |
|   3  |    1    |
|   4  |    1    |
|  ... |    1    |
 ----------------

我可以使用：

SELECT COUNT(*) FROM chapters WHERE book_id = 1

并且表正确输出17因为书1有17章

如果我这样做：

SELECT COUNT(*) FROM library WHERE book_id = 1

我也得到5的正确结果，因为5个用户已将书1添加到他们的库中。

但是，我似乎无法使用JOIN将两个查询组合在一起。我想要输出类似这样的SQL代码：

output:
 -----------------------------
|  book  | chapters | readers |
 -----------------------------
|    1   |    17    |    5    |
 -----------------------------

Answer 1

select c.book_id, count(c.id) as chapters, count(l.user) as readers
from chapters c
join library l on l.book_id = c.book_id
where c.book_id = 1
group by c.book_id

Answer 2

//检查以下内容：

select "1" as book,
    (SELECT COUNT( * ) FROM chapters WHERE book_id = 1) as chapters, 
    (SELECT COUNT( * ) FROM library WHERE book_id = 1) as readers
from dual;

Answer 3

如果尝试下面的查询，结果集将包括libary中的所有行，本章中的章节为85行（17x5）：

select c.book_id, c.id as chapters, l.user as readers
from chapters c
inner join library l on l.book_id = c.book_id
where c.book_id = 1;

  

只要两个表中的列之间存在匹配，INNER JOIN关键字就会选择两个表中的所有行。

您可以阅读更多Here。

然而，为了获得唯一的行数，即。来自85记录结果集的17章和5位读者必须使用如下所示的计数：

select c.book_id, count(distinct(c.id)) as chapters, count(distinct(l.user)) as readers
from chapters c
inner join library l on l.book_id = c.book_id
where c.book_id = 1;

您可以阅读有关count distinct here的更多信息。

Answer 4

有一个答案：

select c.book_id, count(distinct c.id) as chapters, count(distinct l.user) as readers
from chapters c
left join library l on l.book_id = c.book_id
where c.book_id = 1

或所有书籍：

select c.book_id, count(distinct c.id) as chapters, coalesce(count(distinct l.user), 0) as readers
from chapters c
left join library l on l.book_id = c.book_id
group by c.book_id