合并 UNION ALL 行以删除 NULL 值

Question

SELECT journey.id, TIME_FORMAT(ADDTIME(journey.departure, SEC_TO_TIME(SUM(l1.elapsed))), '%H:%i') AS departure, null AS arrival
FROM journey
INNER JOIN journey_day ON journey_day.journey = journey.code
INNER JOIN pattern ON pattern.code = journey.pattern
INNER JOIN link l1 ON l1.section = pattern.section AND l1.stop = "370023139"
INNER JOIN link l2 ON l2.section = pattern.section AND l2.id <= l1.id
WHERE journey.service = "11-252-_-y08-1" AND journey_day.day = 1 AND journey.code NOT IN (SELECT journey
                                                                                          FROM journey_non_operation
                                                                                          WHERE "2015-03-01" BETWEEN date_start AND date_end) AND pattern.direction = "outbound"
GROUP BY journey.id

UNION ALL

SELECT journey.id, null AS departure, TIME_FORMAT(ADDTIME(journey.departure, SEC_TO_TIME(SUM(l1.elapsed))), '%H:%i') AS arrival
FROM journey
INNER JOIN journey_day ON journey_day.journey = journey.code
INNER JOIN pattern ON pattern.code = journey.pattern
INNER JOIN link l1 ON l1.section = pattern.section AND l1.stop = "1000DEHS7812"
INNER JOIN link l2 ON l2.section = pattern.section AND l2.id <= l1.id
WHERE journey.service = "11-252-_-y08-1" AND journey_day.day = 1 AND journey.code NOT IN (SELECT journey
                                                                                          FROM journey_non_operation
                                                                                          WHERE "2015-03-01" BETWEEN date_start AND date_end) AND pattern.direction = "outbound"
GROUP BY journey.id

上面是两个查询，其结果由 UNION ALL 子句合并。您会注意到查询返回不同的列，一列称为“出发”，一列称为“到达”。为了使用UNION不同的列名，我必须给出另一列，NULL这样它就不会忽略它，也不会将其包含在查询中。

我的问题是我的结果如下所示：

id | departure | arrival
1 asd NULL
2 asd NULL
3 asd NULL
4 asd NULL
5 NULL efg
6 NULL efg
7 NULL efg
8 NULL efg

如何合并行以便asd根据efgID 进行匹配？

期望的结果：

id | departure | arrival
1 asd efg
2 asd efg
3 asd efg
4 asd efg

Answer 1

只需将聚合函数（最小/最大）应用于结果集即可。由于空值不包含在聚合中，您将只得到组合结果：

select id, min(departure), min(arrival)
from (your query) as q
group by id