MySQL - 關系數據庫和標記

Question

我一直在摸索這個問題，所以我想我會問堆棧溢出（注意：我是 SQL 新手，試圖了解更多 SQL，所以請尊重和解釋）：

我有一張 sql 表，看起來像這樣，稱為“帖子”：

id | user
--------------------------------
0  | tim
1  | tim
2  | bob

另一個稱為“標簽”的標簽以文本形式存儲在帖子（在“帖子”表中）上：

id | postID | tag
--------------------------------
0  | 0      | php
1  | 2      | php
2  | 0      | mysql
3  | 1      | mysql
4  | 1      | sql
5  | 3      | perl

(To clarify, the concept where: id=0 is tagged php,mysql; id=1 is tagged sql,mysql; id=2 is tagged php; id=3 is tagged perl.)

我如何編寫 WHERE 語句來獲取標記為 x 而不是 y 的帖子（x 和 y 將由 php 定義）？

例如，我怎樣才能獲得所有標記為 mysql 而不是 php 的帖子？

編輯

您能否解釋一下如何添加多個標簽進行搜索（例如獲取所有標記的 mysql 和遞歸但不是 php）

Answer 1

select *
from
    (select distinct postID
    from tags
    where tag = "mysql") as t1
left join
    (select distinct postID
     from tags
     where tag = "php") as t2
using (postID)
where t2.postID is NULL

示例 2：獲取所有標記的 mysql 和遞歸但不是 php：

select *
from
    ((select distinct postID
    from tags
    where tag = "mysql") as t1
join
    (select distinct postID
    from tags
    where tag = "recursion") as t3
using (postID))
left join
    (select distinct postID
     from tags
     where tag = "php") as t2
using (postID)
where t2.postID is NULL

Answer 2

我對 EXISTS 的建議：

SELECT DISTINCT postID
FROM tags t1
WHERE EXISTS(SELECT NULL
             FROM tag t2
             WHERE t2.id = t1.id
               AND t2.tag = 'x')
  AND NOT EXISTS(SELECT NULL
                   FROM tag t2
                  WHERE t2.id = t1.id
                    AND t2.tag = 'y')

Answer 3

天真、簡單、便攜：

SELECT *
FROM posts
WHERE EXISTS (SELECT * FROM tags WHERE tags.postid = posts.id AND tags.tag = 'x')
AND NOT EXISTS (SELECT * FROM tags WHERE tags.postid = posts.id AND tags.tag = 'y')

現在根據執行計划，您可以做其他事情來優化它。