SQL：如何根據類別進行選擇？

Question

有兩個表格，categories和books，我想根據給定的類別選擇所有書籍。

分類表：

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

書籍表：

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

我試過SELECT * FROM categories WHERE cat_id IN(1,3)但隨後返回包含給定類別中的至少一個書。我想要的是它只返回包含所有類別的書籍，所以它應該只返回book_id = 1的所有（或一個）行，因爲它是唯一具有所有給定類別的書籍。

Answer 1

嘗試：

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

或者，如果你打算從支持直接傳遞數組給它（這樣的：http://fxjr.blogspot.com/2009/05/npgsql-tips-using-in-queries-with.html）語言傳遞一個數組Postgres的，使用這樣的：

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

如果你想得到書名：

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，修改查詢：

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 

三世書名稱：

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

---

什麼是內聯的條件和相同的條款中（即其計數值的優勢。having），而不是單獨把條件where條款及其having條款計數？...

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

...如果我們兩個內嵌的條件和條款having其計數值，我們可以很方便的我們可以通過查詢所有分類爲1和3的書籍，或者分類爲2和3和4的。面向未來的FTW！此外，對組合類別和數量的測試彼此相鄰，再加上可讀性因素。

爲了方便那種查詢：

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 

要達到的性能（有時，實現了性能和可讀性;不要拌勻），必須複製having子句，其中的元素測試條款：

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 

[編輯]

順便說一句，這裏的慣用MySQL的：

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

Answer 2

實際上，您每本書都會獲得多個條目。如果n的n類別被分配給這本書，你會得到n這本書的條目。所以，你可以組你的查詢，只選擇那些誰擁有ñ點擊：如說我編輯查詢，以便它返回一個包含所有種類的書籍：

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

試試這個：

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)) 

編輯在這個問題

Answer 4

加入對你要求每個類別：

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，如果您不想添加內部連接，則可以等效使用WHERE EXISTS（半連接）。

Answer 5

另一種替代方法：

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

您可以繼續鏈相交，如果你有兩個以上的類別相匹配。

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); 

這是假設一本書不能同日而語的兩倍。這將選擇任一類別中的所有圖書，對類別進行計數，並確保類別數量是類別的最大數量。