sql查询未正确分组

Question

由于某种原因，以下查询允许重复的名称。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 

Answer 1

因为ID是不同的，你是ID分组，你多行的每个，这就是GROUP BY一样。如果您改变了顶层SELECT到

SELECT name_without_variants, SUM(relevance) as total_relevance 

和外GROUP BY到：

GROUP BY name_without_variants 

你应该看到不同的名称，但将不再有ID。

Answer 2

GROUP BY name_without_variants, id 

您正在按name_without_variants，id进行分组。该id在两条记录上有所不同：

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

您需要决定如何管理id。

从组中删除id，然后将聚合函数添加到select中的id列。或者只是一起删除列。

下面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全面的看法，也没有经过测试。我也在这里做一些假设。但是，如果架构是关系型的，这应该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;