0

我的查询：

SELECT CONCAT(f.name, ' ', f.parent_names) AS FullName,
       stts.name AS 'Status',
       u.name AS Unit,
       city.name AS City,
       hus.mobile1 AS HusbandPhone,
       wife.mobile1 AS WifePhone,
       f.phone AS HomePhone,
       f.contact_initiation_date AS InitDate,
       fh.created_at AS StatusChangeDate,
       cmt.created_at AS CommentDate,
       cmt.comment AS LastComment
FROM families f
JOIN categories stts ON f.family_status_cat_id = stts.id
JOIN units u ON f.unit_id = u.id
JOIN categories city ON f.main_city_cat_id = city.id
JOIN contacts hus ON f.husband_id = hus.id
JOIN contacts wife ON f.wife_id = wife.id
JOIN comments cmt ON f.id = cmt.commentable_id
AND cmt.created_at =
    (SELECT MAX(created_at)
     FROM comments
     WHERE commentable_id = f.id)
JOIN family_histories fh ON f.id = fh.family_id
AND fh.created_at =
    (SELECT MAX(created_at)
     FROM family_histories
     WHERE family_id = f.id
         AND family_history_cat_id = 1422)
WHERE f.id = 17883

问题：结果是 2 行 - 但它们是相同的。为什么我得到 2 个结果而不是一个？

4

2 回答 2

1

可能评论或family_histories有双重关系，您可以通过返回*结果来检查它。某处应该有区别。

提供完整的结果以找到问题。

您可以（但不推荐）通过设置来解决它DISTINCT。

SELECT DISTINCT CONCAT(f.name, ' ', f.parent_names) AS FullName, stts.name AS 'Status', u.name AS Unit, city.name AS City, hus.mobile1 AS HusbandPhone, wife.mobile1 AS WifePhone, f.phone AS HomePhone, f.contact_initiation_date AS InitDate, fh.created_at AS StatusChangeDate, cmt.created_at AS CommentDate, cmt.comment AS LastComment
FROM families f JOIN categories stts ON f.family_status_cat_id = stts.id
JOIN units u ON f.unit_id = u.id
JOIN categories city ON f.main_city_cat_id = city.id
JOIN contacts hus ON f.husband_id = hus.id
JOIN contacts wife ON f.wife_id = wife.id
JOIN comments cmt ON f.id = cmt.commentable_id AND cmt.created_at = (SELECT MAX(created_at) FROM comments WHERE commentable_id = f.id)
JOIN family_histories fh ON f.id = fh.family_id AND fh.created_at = (SELECT MAX(created_at) FROM family_histories WHERE family_id = f.id AND family_history_cat_id = 1422)
WHERE f.id = 17883
于 2013-03-04T20:47:02.933 回答
1

当您连接存在一对多关系的表时，您似乎会返回多行，而实际上该行在主表中仅存在一次，但其中一个子表中有多行相关数据.

另请注意，您期望的数据可能会丢失，因为子表中没有匹配的记录，在这种情况下使用LEFT JOIN。

于 2013-03-04T20:50:34.777 回答