I have a table with addresses, dates and some other info.
我有一张包含地址，日期和其他信息的表格。
I want to select an N unique quantity of addresses, and of that N records, I want to select everything from all the rows where that address is, because the address is not a unique value in that column.
我想选择N个唯一数量的地址，并且在N个记录中，我想从该地址的所有行中选择所有内容，因为该地址不是该列中的唯一值。
Date | Address | MoreInfo
1/2/2000 | 123 SUPER LN | SOME MORE BORING INFO
2/3/2000 | 123 SUPER LN | SOME MORE BORING INFO
5/3/2000 | 567 AWESOME ST | SOME MORE BORING INFO
5/26/2000 | 567 AWESOME ST | SOME MORE BORING INFO
7/2/2000 | 987 SUPER LN | SOME MORE BORING INFO
8/3/2000 | 1100 BORING DR | SOME MORE BORING INFO
11/7/2000 | 1100 BORING DR | SOME MORE BORING INFO
1/5/2001 | 564 SUPER LN | SOME MORE BORING INFO
So for that data, I would like to get, say 3 different addresses, (123 SUPER LN, 567 AWESOME ST, 987 SUPER LN) and get everything from all the rows those 3 addresses are at.
所以对于那些数据，我想得到3个不同的地址（123 SUPER LN，567 AWESOME ST，987 SUPER LN）并从这3个地址所在的所有行中获取所有内容。
I had a query:
我有一个问题：
SELECT *
FROM my_table
WHERE my_condition = true
GROUP BY address
But this brings only 1 row per address, and I would like to get all the rows where those N addresses are at. Thanks!
但是这每个地址只带一行，我想获得那些N个地址所在的所有行。谢谢！
1 个解决方案
#1
3
I think this is what you want:
我想这就是你想要的：
select t.*
from (select distinct address
from my_table
where my_condition = true
limit 100 -- this is the N
) a join
my_table t
on a.address = t.address;
The first subquery gets the list of addresses. The join brings back the original rows.
第一个子查询获取地址列表。连接将返回原始行。
#1
3
I think this is what you want:
我想这就是你想要的：
select t.*
from (select distinct address
from my_table
where my_condition = true
limit 100 -- this is the N
) a join
my_table t
on a.address = t.address;
The first subquery gets the list of addresses. The join brings back the original rows.
第一个子查询获取地址列表。连接将返回原始行。