我有一张桌子partenaire。合伙人可能有一个或多个地址。当然，一个地址可能“属于”多个合伙人。所以我有 3 个表partenaire：partenaire_adresse和adresse。该地址只有一个城镇（ville法语），所以我id_ville在表中有一个外键addresse。
SELECT
p.nom,
v.nom, v.id_region as id_r, v.id_departement as id_p,
r.description as region
FROM partenaire p
JOIN partenaire_adresse pa
ON pa.id_partenaire=p.id
JOIN adresse a
ON a.id=pa.id_adresse
JOIN ville v
ON v.id=a.id_ville
JOIN region r
ON v.id_region=r.id
LIMIT 4;
这给了我这些结果：
+----------------------------+-------------+------+------+--------+
| nom | nom | id_r | id_p | region |
+----------------------------+-------------+------+------+--------+
| Ferme Auberge Christlesgut | Breitenbach | 1 | 2 | Alsace |
| Alice Pizza | Strasbourg | 1 | 1 | Alsace |
| Au Vieux Cellier | Strasbourg | 1 | 1 | Alsace |
| Auberge du 7Eme Art | Strasbourg | 1 | 1 | Alsace |
+----------------------------+-------------+------+------+--------+
现在，如果我LEFT OUTER JOIN在最后一张表 ( region) 上执行 a ，结果将不一样：
SELECT
p.nom,
v.nom, v.id_region as id_r, v.id_departement as id_p,
r.description as region
FROM partenaire p
JOIN partenaire_adresse pa
ON pa.id_partenaire=p.id
JOIN adresse a
ON a.id=pa.id_adresse
JOIN ville v
ON v.id=a.id_ville
LEFT OUTER JOIN region r
ON v.id_region=r.id
LIMIT 4;
不一样的结果。看：
+---------------------+----------+------+------+----------------+
| nom | nom | id_r | id_p | region |
+---------------------+----------+------+------+----------------+
| 'Le 144' Petrossian | Paris 18 | 12 | 43 | Île-de-France |
| 'Le 144' Petrossian | Paris 08 | 12 | 43 | Île-de-France |
| 'O'Quai' | Vouvray | 7 | 26 | Centre |
| 'O'Quai' | Tours | 7 | 26 | Centre |
+---------------------+----------+------+------+----------------+
我认为不应该，因为在第一个查询中，id_region 和 id_departement 不为空，所以如果你做一个“JOIN”或“LEFT OUTER JOIN”，结果应该是一样的。还是我错过了什么？