MySQL检查从最后到开始的条件是否为真，如果停止则返回

Question

I really love to know how this is possible without doing a for loop.My way is lame,I check the last then take the previous check again and add this values to an array and when the condition wasn't true anymore I return the array.This is not pure sql and uses php for loop so it's not very performant,I need to do this with sql.Example:

chatTable:
from_id    to_id     msg                  time
2          1         Hi                   2 days ago(as timestamp)
1          2         Hello                yesterday
1          2         How you doing?       today

So here we are going to take all the messages that id 1 sent from the latest until It's not him anymore.

1->2 How you doing  = (true)
1->2 Hello  = (true)
2->1 Hi = (false)

so

return
1->2 How you doing
1->2 Hello

Answer 1

First way is to order the recrods in desc order of timestamp. Then look for patterns in your data such that the next_from_id is same as that of the current_from_id. If they are different then ignore.

If this looks correct the following can be done

select * from (
    select from_id
           ,to_id
           ,timestamp
           ,lead(from_id) over(order by timestamp desc) as next_frm_id
      from chatTable
      )x
 where x.from_id = x.next_frm_id

Answer 2

If I understand correctly, you want the last set of messages that are for from_id = 1 . That would be:

select ct.*
from chatTable ct
where ct.from_id = 1 and
      ct.timestamp > (select max(ct2.timestamp)
                      from chatTable ct2
                      where ct2.from_id <> 1
                     );

This assumes that the most recent message if from id 1.