2

出于某种原因，下面的查询允许重复名称。这是为什么？

SELECT id, name_without_variants, SUM(relevance) as total_relevance FROM (
    SELECT 
        card_definitions.id, 
            card_definitions.name_without_variants,
        (MATCH(card_definitions.name_without_variants) AGAINST ('lost soul site discard')) * 0.40 AS relevance
        FROM card_definitions
        GROUP BY name_without_variants, id
    UNION
    SELECT 
        card_definitions.id,
            card_definitions.name_without_variants,
        (MATCH(card_def_identities.special_ability_text) AGAINST ('lost soul site discard')) * 0.05 AS relevance
        FROM card_def_identities 
        INNER JOIN card_definitions ON card_def_identities.card_def_sid = card_definitions.id 
        GROUP BY name_without_variants, id
    UNION
    SELECT 
        card_definitions.id,
            card_definitions.name_without_variants,
        (MATCH(brigades.brigade_color) AGAINST ('lost soul site discard')) * 0.30 AS relevance
        FROM brigades 
        INNER JOIN card_def_brigades ON brigades.id = card_def_brigades.brigade_sid
        INNER JOIN card_definitions ON card_def_brigades.card_def_sid = card_definitions.id 
        GROUP BY name_without_variants, id
    UNION
    SELECT 
        card_definitions.id,
            card_definitions.name_without_variants,
        (MATCH(identifiers.identifier) AGAINST ('lost soul site discard')) * 0.20 AS relevance
        FROM identifiers
        INNER JOIN card_def_identifiers ON identifiers.id = card_def_identifiers.identifier_sid
        INNER JOIN card_definitions on card_def_identifiers.card_def_sid = card_definitions.id 
        GROUP BY name_without_variants, id
    UNION
    SELECT 
        card_definitions.id,
            card_definitions.name_without_variants,
        (MATCH(card_effects.effect) AGAINST ('lost soul site discard')) * 0.05 AS relevance
        FROM card_effects
        INNER JOIN card_def_effects ON card_effects.id = card_def_effects.effect_sid
        INNER JOIN card_definitions on card_def_effects.card_def_sid = card_definitions.id 
        GROUP BY name_without_variants, id
    ) AS combined_search 
GROUP BY name_without_variants, id
HAVING total_relevance > 0
ORDER BY total_relevance DESC
LIMIT 10;

这是我得到的结果。注意两个Lost Soul [Site Doubler]

2623    Lost Soul [Deck Discard]    6.35151714086533
1410    Lost Soul [Hand Discard]    6.29273346662521
1495    Lost Soul [Discard Card]    5.93360201716423
1442    Lost Soul [Demon Discard]   5.91308708190918
1497    Lost Soul [Site Doubler]    5.05888686180115
1498    Lost Soul [Site Doubler]    5.05888686180115
2572    Lost Soul [Site Guard]  4.82421946525574
2774    Lost Soul [Far Country] 3.39325473308563
2891    Fortify Site [RoA2] 2.77084048986435
1418    Lost Soul [Hopper]  2.63041100502014
4

2 回答 2

2

因为 ID 不同，并且您按 ID 分组，所以每个 ID 都有多行，这就是这样GROUP BY做的。如果您将顶层更改SELECT为

SELECT name_without_variants, SUM(relevance) as total_relevance

和外部 GROUP BY 到：

GROUP BY name_without_variants

您应该会看到不同的名称，但不会再有 id。

于 2012-04-04T17:18:27.070 回答
0 
GROUP BY name_without_variants, id

您按 name_without_variants，id 分组。两条记录的 id 不同：

1497    Lost Soul [Site Doubler]    5.05888686180115
1498    Lost Soul [Site Doubler]    5.05888686180115

您需要决定如何管理 id。

从 group by 中删除 id 并将聚合函数添加到您选择的 id 列中。或者只是一起删除列。

这是一个简化为单个查询的示例。请理解我对您的架构或数据没有完整的看法，也没有经过测试。我在这里也做了一些假设。但是，如果架构是关系的，这应该会带回您正在寻找的内容：

SELECT cd.id, cd.name_without_variants, (((MATCH(cd.name_without_variants) AGAINST ('lost soul site discard')) * 0.40)+
                                   ((MATCH(cdi.special_ability_text) AGAINST ('lost soul site discard')) * 0.05)+
                                   ((MATCH(b.brigade_color) AGAINST ('lost soul site discard')) * 0.30)+
                                   ((MATCH(i.identifier) AGAINST ('lost soul site discard')) * 0.20)+
                                   ((MATCH(ce.effect) AGAINST ('lost soul site discard')) * 0.05)
                                  ) as total_relevance 
FROM card_definitions cd 
 LEFT OUTER JOIN card_def_identities cdi ON cd.id=cdi.card_def_sid
 LEFT OUTER JOIN brigades b ON cd.id=b.card_def_sid
 LEFT OUTER JOIN identifiers i ON i.id=cdi.identifier_sid
 LEFT OUTER JOIN card_def_effects cde ON cde.card_def_sid=cd.id
 LEFT OUTER JOIN card_effects ce ON ce.id=cde.effect_sid
GROUP BY cd.id, cd.name_without_variants
HAVING total_relevance > 0
ORDER BY total_relevance DESC
LIMIT 10;
于 2012-04-04T17:23:26.573 回答