Question

我有下面的架构

CREATE TABLE `vocabulary` (
`vid` int(10) unsigned NOT NULL auto_increment,
`name` varchar(255),
PRIMARY KEY vid (`vid`)   
);

CREATE TABLE `term` (
`tid` int(10) unsigned NOT NULL auto_increment,
`vid` int(10) unsigned NOT NULL default '0',
`name` varchar(255),
PRIMARY KEY tid (`tid`)   
);

CREATE TABLE `article` (
`aid` int(10) unsigned NOT NULL auto_increment,
`body` text,
PRIMARY KEY aid (`aid`)     
);

CREATE TABLE `article_index` (
`aid` int(10) unsigned NOT NULL default '0',
`tid` int(10) unsigned NOT NULL default '0'
) 

INSERT INTO `vocabulary` values (1, 'vocabulary 1');
INSERT INTO `vocabulary` values (2, 'vocabulary 2');

INSERT INTO `term` values (1, 1, 'term v1 t1');
INSERT INTO `term` values (2, 1, 'term v1 t2 ');
INSERT INTO `term` values (3, 2, 'term v2 t3');
INSERT INTO `term` values (4, 2, 'term v2 t4');
INSERT INTO `term` values (5, 2, 'term v2 t5');

INSERT INTO `article` values (1, "");
INSERT INTO `article` values (2, "");
INSERT INTO `article` values (3, "");
INSERT INTO `article` values (4, "");
INSERT INTO `article` values (5, "");

INSERT INTO `article_index` values (1, 1);
INSERT INTO `article_index` values (1, 3); 
INSERT INTO `article_index` values (2, 2);
INSERT INTO `article_index` values (3, 1);
INSERT INTO `article_index` values (3, 3); 
INSERT INTO `article_index` values (4, 3); 
INSERT INTO `article_index` values (5, 1);
INSERT INTO `article_index` values (5, 4);

实施例。选择一个defiend词汇表的术语（具有非零文章索引），例如VID = 2

 select a.tid, count(*) as article_count  from term t JOIN article_index a   
 ON t.tid = a.tid where t.vid = 2 group by t.tid;

   
    +-----+---------------+
    | tid | article_count |
    +-----+---------------+
    |   3 |             3 |
    |   4 |             1 |
    +-----+---------------+

问题：

选择条款
一个。一个defiend词汇表（非零文章索引，例如vid = 1 =＆gt; term {1,2}）
湾过滤器（a），只需要那些与特定术语的文章相关联的术语，例如： tid = 3，结果=＆gt; {1}，因为tid = 2的术语被排除，因为没有与tid = 3

SQL：不正确，因为article_count错误

SELECT t.tid, count(*) as c FROM term t JOIN article_index i ON t.tid = i.tid WHERE t.vid = 1 AND i.aid IN 

    ( SELECT i.aid FROM term t JOIN article_index i ON t.tid = i.tid WHERE i.tid = 3) 

GROUP BY t.tid ;

结果：

+-----+---------------+
| tid | article_count |
+-----+---------------+
|   1 |             2 |
+-----+---------------+

预期结果：由于只有一篇文章与tid = 3和tid = 1

+-----+---------------+
| tid | article_count |
+-----+---------------+
|   1 |             1 |
+-----+---------------+

Answer 1

它不是很好但很容易到达

SELECT tid, count(*)
FROM 

(


SELECT distinct t.tid
FROM term t 
JOIN article_index i 
ON t.tid = i.tid 
WHERE t.vid = 1 
AND i.aid IN 

    ( SELECT i.aid 
      FROM term t 
          JOIN article_index i 
          ON t.tid = i.tid 
      WHERE i.tid = 3) 

) t
GROUP BY tid

Answer 2

您只需将count(*)替换为count(distinct t.vid) ...

具有前置条件的复杂MySQL表选择/连接

2 个答案: