I have a problem that is quite simple to understand, but for me it is not that simple to implement.
我有一个很容易理解的问题,但对我而言,实现起来并不那么简单。
I have a table named Time:
我有一个名为Time的表:
-----------------------------------
DAY TIME1 TIME2
-----------------------------------
1 08.00 09.40
1 09.40 10.00
1 10.00 11.40
2 08.00 08.50
2 08.50 10.40
----------------------------------
What I want to get is like:
我想得到的是:
-------------------
DAY TIME
-------------------
1 08.00
1 09.40
1 10.00
1 11.00
2 08.00
2 08.50
2 10.40
------------------
I have tried this code:
我试过这段代码:
SELECT DISTINCT sub.val FROM (
SELECT Time1 AS val FROM Time
UNION ALL
SELECT Time2 AS val FROM Time
) AS sub
And it is only return the Time column. I have no idea about joining the "Day" column. While I am trying to add the "Day" after the "DISTINCT sub.val", it got error. How to do it?
它只返回时间列。我不知道加入“日”专栏。当我试图在“DISTINCT sub.val”之后添加“Day”时,它出错了。怎么做?
Thank you very much.
非常感谢你。
4 个解决方案
#1
5
select distinct day, time1 from mytab
union
select distinct day, time2 from mytab
#2
3
SELECT DISTINCT sub.Day, sub.val as Time FROM (
SELECT Day, Time1 AS val FROM Time
UNION ALL
SELECT Day, Time2 AS val FROM Time
) AS sub
#3
3
try this
尝试这个
SELECT DISTINCT *
FROM (
SELECT day, time1 FROM Time
UNION ALL SELECT day, time2 FROM Time
) AS x
ORDER BY x.day
#4
2
This might help:
这可能有所帮助:
SELECT
DAY,
TIME1
FROM
YourTable AS T
UNION
SELECT
DAY,
TIME2
FROM
YourTable AS T
ORDER BY DAY,TIME1
#1
5
select distinct day, time1 from mytab
union
select distinct day, time2 from mytab
#2
3
SELECT DISTINCT sub.Day, sub.val as Time FROM (
SELECT Day, Time1 AS val FROM Time
UNION ALL
SELECT Day, Time2 AS val FROM Time
) AS sub
#3
3
try this
尝试这个
SELECT DISTINCT *
FROM (
SELECT day, time1 FROM Time
UNION ALL SELECT day, time2 FROM Time
) AS x
ORDER BY x.day
#4
2
This might help:
这可能有所帮助:
SELECT
DAY,
TIME1
FROM
YourTable AS T
UNION
SELECT
DAY,
TIME2
FROM
YourTable AS T
ORDER BY DAY,TIME1