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，一个名为“出发”，一个名为“到达”。要让UNION使用不同的列名，我必须给另一列NULL，以便它不会忽略它，也不会将其包含在查询中。

我的问题是我的结果如下：

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和efg根据ID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（min / max）应用于结果集。由于空值不包含在汇总中，您只需获得组合结果：

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

Answer 2

我不确定你为什么试图建立联盟。

尝试这种方式：

SELECT journey.id, 
    TIME_FORMAT(ADDTIME(journey.departure, SEC_TO_TIME(SUM(l1.elapsed))), '%H:%i') AS departure, 
    TIME_FORMAT(ADDTIME(journey.departure, SEC_TO_TIME(SUM(l2.elapsed))), '%H:%i') AS arrival
FROM journey
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.stop = "1000DEHS7812"
    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

合并UNION ALL行以删除NULL值

2 个答案: