这是一个 SQL 问题,不知道要使用哪种类型的 JOIN、GROUP BY 等,它用于聊天程序,其中消息与房间相关,并且房间中的每一天都链接到抄本等。
基本上,在输出我的成绩单时,我需要显示哪些用户在该成绩单上聊天。目前我通过消息链接它们,如下所示:
SELECT rooms.id, rooms.name, niceDate, room_transcripts.date, long
FROM room_transcripts
JOIN rooms ON room_transcripts.room=rooms.id
JOIN transcript_users ON transcript_users.room=rooms.id AND transcript_users.date=room_transcripts.date
JOIN users ON transcript_users.user=users.id
WHERE room_transcripts.deleted=0 AND rooms.id IN (1,2)
ORDER BY room_transcripts.id DESC, long ASC
结果集如下所示:
Array
(
[0] => Array
(
[id] => 2
[name] => Room 2
[niceDate] => Wednesday, April 14
[date] => 2010-04-14
[long] => Jerry Seinfeld
)
[1] => Array
(
[id] => 1
[name] => Room 1
[niceDate] => Wednesday, April 14
[date] => 2010-04-14
[long] => Jerry Seinfeld
)
[2] => Array
(
[id] => 1
[name] => Room 1
[niceDate] => Wednesday, April 14
[date] => 2010-04-14
[long] => Test Users
)
)
我希望数组中的每个元素都代表一个成绩单条目,并将用户分组在一个数组中作为条目的元素。所以“long”将是一个列出所有名称的数组。这能做到吗?
目前我只是附加名称,当成绩单日期和房间更改时,我会回溯它们,但我会对文件和突出显示的消息执行相同的操作,但它很困惑。
谢谢。
最佳答案
您不能将数组设为 long
结果中的字段,因为每个数组元素代表表中的一行。但是您可以使用连接的字符串,并在需要时拆分字符串。查询将如下所示:
SELECT rooms.id, rooms.name, niceDate, room_transcripts.date, group_concat(long) as long
FROM room_transcripts
JOIN rooms ON room_transcripts.room=rooms.id
JOIN transcript_users ON transcript_users.room=rooms.id AND transcript_users.date=room_transcripts.date
JOIN users ON transcript_users.user=users.id
WHERE room_transcripts.deleted=0 AND rooms.id IN (1,2)
GROUP BY room_transcripts.id
ORDER BY room_transcripts.id DESC
关于SQL 组结果作为列数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/2640293/