SQL：如何根据类别进行选择？

Question

有两个表格，categories和books，我想根据给定的类别选择所有书籍。

分类表：

cat_id | book_id 
---------------- 
1  | 1 
2  | 1 
3  | 1 
3  | 2 

书籍表：

id | name 
---------------- 
1 | abc 
2 | def 

我试过SELECT * FROM categories WHERE cat_id IN(1,3)但随后返回包含给定类别中的至少一个书。我想要的是它只返回包含所有类别的书籍，所以它应该只返回book_id = 1的所有（或一个）行，因为它是唯一具有所有给定类别的书籍。

Answer 1

尝试：

select book_id 
from categories 
group by book_id 
having sum((cat_id in (1,3))::int) = 2 

或者，如果你打算从支持直接传递数组给它（这样的：http://fxjr.blogspot.com/2009/05/npgsql-tips-using-in-queries-with.html）语言传递一个数组Postgres的，使用这样的：

select book_id 
from categories 
group by book_id 
having sum((cat_id = ANY(ARRAY[1,3]))::int) = 2 

如果你想得到书名：

select categories.book_id, books.name 
from categories 
join books on books.id = categories.book_id 
group by categories.book_id 
    ,books.name 
having sum((categories.cat_id in (1,3))::int) = 2 

@Evan Carroll，修改查询：

ANSI SQL的方式：

select categories.book_id, books.name 
from categories 
join books on books.id = categories.book_id 
group by categories.book_id 
    ,books.name 
having count(case when categories.cat_id in (1,3) then 1 end) = 2 

三世书名称：

select book_id 
from categories 
group by book_id 
having count(case when cat_id in (1,3) then 1 end) = 2 

---

什么是内联的条件和相同的条款中（即其计数值的优势。having），而不是单独把条件where条款及其having条款计数？...

select book_id 
from categories 
where category_id in (1,3) 
group by book_id 
having count(*) = 2 

...如果我们两个内嵌的条件和条款having其计数值，我们可以很方便的我们可以通过查询所有分类为1和3的书籍，或者分类为2和3和4的。面向未来的FTW！此外，对组合类别和数量的测试彼此相邻，再加上可读性因素。

为了方便那种查询：

select book_id 
from categories 
group by book_id 
having 
    count(case when cat_id in (1,3) then 1 end) = 2 
    or count(case when cat_id in (2,3,4) then 1 end) = 3 

要达到的性能（有时，实现了性能和可读性;不要拌匀），必须复制having子句，其中的元素测试条款：

select book_id 
from categories 
where cat_id in (1,2,3,4) 
group by book_id 
having 
    count(case when cat_id in (1,3) then 1 end) = 2 
    or count(case when cat_id in (2,3,4) then 1 end) = 3 

[编辑]

顺便说一句，这里的惯用MySQL的：

select book_id 
from categories 
group by book_id 
having sum(cat_id in (1,3)) = 2 

Answer 2

实际上，您每本书都会获得多个条目。如果n的n类别被分配给这本书，你会得到n这本书的条目。所以，你可以组你的查询，只选择那些谁拥有ñ点击：如说我编辑查询，以便它返回一个包含所有种类的书籍：

SELECT T.cat_id, count(*) hits FROM 
(
    SELECT * FROM categories WHERE cat_id IN(1,3) 
) T 
GROUP BY T.cat_id 
HAVING hits = 2 

Answer 3

试试这个：

SELECT * FROM books WHERE id IN 
(SELECT book_id 
FROM categories 
GROUP BY book_id 
HAVING COUNT(distinct cat_id) = (select count(distinct cat_id) from categories)) 

编辑在这个问题

Answer 4

加入对你要求每个类别：

SELECT books.* 
FROM books 
    JOIN categories cat1 ON cat1.book_id = books.book_id 
    JOIN categories cat3 ON cat3.book_id = books.book_id 
WHERE cat1.cat_id = 1 
     AND cat3.cat_id = 3 

或者，如果您不想添加内部连接，则可以等效使用WHERE EXISTS（半连接）。

Answer 5

另一种替代方法：

SELECT book_id FROM categories WHERE cat_id = 1 
INTERSECT 
SELECT book_id FROM categories WHERE cat_id = 3; 

您可以继续链相交，如果你有两个以上的类别相匹配。

Answer 6

SELECT * FROM 
(
SELECT b.id, count(c.cat_id) as cat_count 
FROM books AS b 
JOIN cats AS c 
    ON (b.id = c.book_id) 
GROUP BY b.id 
) AS t 
WHERE t.cat_count = (SELECT DISTINCT count(cat_id) FROM cat); 

这是假设一本书不能同日而语的两倍。这将选择任一类别中的所有图书，对类别进行计数，并确保类别数量是类别的最大数量。