多對多MySQL表中的計數和多個分組

Question

我想知道整天，我無法完成它。

我有一個簡單的測試表，以news系統爲例。我們有news,tags和categories。 News可以有許多tags和一個category。我需要的是統計每個category中每個tag下有多少個news。例如，我們可以在政治類別下帶有通用標籤的4 news，在科學類別下帶有通用標籤的2 news。

我的表是這樣的：

news: 
    - news_id 
    - category_id 
    - title 

categories: 
    - category_id 
    - category_name 

tags: 
    - tag_id 
    - tag_name 

news_tags: 
    - news_id 
    - tag_id 

下面是一個簡單的腦圖，以澄清什麼，我需要： http://i.stack.imgur.com/6ySiJ.png

下面是一個查詢，我沒有成功嘗試：

SELECT *, COUNT(n.news_id) AS news_count FROM news AS n 
LEFT JOIN categories AS c ON n.category_id = c.category_id 
LEFT JOIN news_tags AS tn ON n.news_id = tn.news_id 
LEFT JOIN tags AS t ON tn.tag_id = t.tag_id 
GROUP BY t.tag_id, c.category_id; 

Answer 1

您的查詢無誤我的行爲。 運行查詢時會出現什麼錯誤/爲什麼它不成功？

這是我嘗試設置您的情況：

mysql> select * from news; 
+---------+-------------+------------------------+ 
| news_id | category_ID | title     | 
+---------+-------------+------------------------+ 
|  1 |   1 | politics and general 1 | 
|  2 |   1 | politics and general 2 | 
|  3 |   1 | politics and general 3 | 
|  4 |   1 | politics and general 4 | 
|  5 |   2 | science and general 1 | 
|  6 |   2 | science and general 2 | 
|  7 |   2 | science and funny 1 | 
+---------+-------------+------------------------+ 

mysql> select * from tags; 
+--------+----------+ 
| tag_id | tag_name | 
+--------+----------+ 
|  1 | general | 
|  2 | funny | 
+--------+----------+ 

mysql> select * from news_tags; 
+---------+--------+ 
| news_id | tag_id | 
+---------+--------+ 
|  1 |  1 | 
|  2 |  1 | 
|  3 |  1 | 
|  4 |  1 | 
|  5 |  1 | 
|  6 |  1 | 
|  7 |  2 | 
+---------+--------+ 

mysql> select * from categories; 
+-------------+---------------+ 
| category_id | category_name | 
+-------------+---------------+ 
|   1 | politics  | 
|   2 | science  | 
+-------------+---------------+ 

查詢的結果：

+---------+-------------+------------------------+-------------+---------------+---------+--------+--------+----------+------------+ 
| news_id | category_ID | title     | category_id | category_name | news_id | tag_id | tag_id | tag_name | news_count | 
+---------+-------------+------------------------+-------------+---------------+---------+--------+--------+----------+------------+ 
|  1 |   1 | politics and general 1 |   1 | politics  |  1 |  1 |  1 | general |   4 | 
|  5 |   2 | science and general 1 |   2 | science  |  5 |  1 |  1 | general |   2 | 
|  7 |   2 | science and funny 1 |   2 | science  |  7 |  2 |  2 | funny |   1 | 
+---------+-------------+------------------------+-------------+---------------+---------+--------+--------+----------+------------+ 

然而，它沒有任何意義SELECT *因爲你的標籤總計數/類別，像title這樣的東西在聚合時沒有意義。

您可以試試：

SELECT c.category_name, t.tag_name, COUNT(n.news_id) AS news_count FROM news AS n 
LEFT JOIN categories AS c ON n.category_id = c.category_id 
LEFT JOIN news_tags AS tn ON n.news_id = tn.news_id 
LEFT JOIN tags AS t ON tn.tag_id = t.tag_id 
GROUP BY t.tag_id, c.category_id; 

要獲取：

+---------------+----------+------------+ 
| category_name | tag_name | news_count | 
+---------------+----------+------------+ 
| politics  | general |   4 | 
| science  | general |   2 | 
| science  | funny |   1 | 
+---------------+----------+------------+