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MySQL的左连接与计数(在右表属性)

[英]mysql left join with count(on right table attribute) group by

I have two tables, one with tags and another one with the actually selected tags of an article (table relation). 我有两个表,一个带有标签,另一个带有标签(表关系)的实际选择的标签。 I want to add the condition to select the amount of the actually selected tags of the current article. 我想添加条件以选择当前文章的实际选择的标签数量。 example selecting article 1: 选择文章1的示例:

--------------------------
| id | name   | selected |
--------------------------
| 0  | this   | 1        |
| 1  | is     | 0        | 
| 2  | sparta | 1        |
--------------------------

Ive got to this far: 我到目前为止:

SELECT t.*, count(t.id) as `selected`
FROM tag t LEFT JOIN relation r ON t.id = r.tid
GROUP BY t.id

table tag: 表格标签:

---------------
| id | name   |
---------------
| 0  | this   |
| 1  | is     |
| 2  | sparta |
---------------

table relation: 表关系:

-------------
| tid | aid |
-------------
| 0   | 2   |
| 0   | 1   |
| 2   | 1   |
-------------

EDIT : The first query returns the selected tags for an article and that's not the desired behaviour 编辑:第一个查询返回所选的文章标签,这不是所需的行为

Is that the query you're looking for? 您正在寻找的查询吗?

SELECT A.id
      , COUNT(T.id) AS [nb selected tags]
FROM article A
LEFT OUTER JOIN relation R ON R.aid = A.id
LEFT OUTER JOIN tag T ON T.id = R.tid
                         AND T.selected = 1
GROUP BY A.id

Hope this will help you. 希望这会帮助你。

PS: If you just want the result of a specific Article, you juste have to add a WHERE clause before the GROUP BY . PS:如果只需要特定文章的结果,则只需在GROUP BY之前添加WHERE子句。


Query returning the desired data: 查询返回所需数据:

SELECT T.id
     ,T.name
     , CASE
        WHEN R.tid IS NOT NULL THEN 1
        ELSE 0
      END AS [selected]
FROM tag T
LEFT OUTER JOIN relation R ON R.tid = T.id
                            AND R.selected = 1
                            AND R.aid = ...

I assume that the table relation has a unicity on tid and aid. 我假设表relation在提示和帮助上具有统一性。

Let me know if the query returns the expected result. 让我知道查询是否返回预期结果。

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