sqlite删除列a和列b不在前n项中的所有结果

假设我有下表

a    b    c
-----------
1    1    5
1    2    3
4    1    2
1    2    4
4    2    10

我想删除前n行中没有一行在a和b中具有与该行相同的值的所有行

例如，各种n的结果表是

n=1

n=2

n=3

n=4

---

我一直在尝试使用not in或not exists来解决这一问题，但由于我对两个列的匹配感兴趣，而不仅仅是1或整个记录，因此我很难做到这一点。

由于您没有定义特定的顺序，因此结果没有完全定义，但这取决于实现的任意选择，即在limit子句中首先计算哪些行。例如，不同的SQLite版本可能会给出不同的结果。话虽如此，我相信您需要以下查询：

select t1.* from table1 t1, 
(select distinct t2.a, t2.b from table1 t2 limit N) tabledist 
where t1.a=tabledist.a and t1.b=tabledist.b;

其中，应将N替换为所需的行数

编辑：因此，要直接从现有表中删除，您需要以下内容：

with toremove(a, b, c) as 
    (select * from table1 tt 
    EXCEPT select t1.* from table1 t1, 
    (select distinct t2.a, t2.b from table1 t2 limit N) tabledist 
    where t1.a=tabledist.a and t1.b=tabledist.b) 
delete from table1 where exists 
(select * from toremove 
where table1.a=toremove.a and table1.b=toremove.b and table1.c=toremove.c);

---

标准的顺序是什么？是n=ID吗？这是不相关的-我只是想得到一个较小的数据集合，使其更可行，但我需要将任何给定a的所有数据与b一起，以使其有效。非常感谢。不过，我认为在第二部分中包含c是没有必要的

a    b    c
-----------
1    1    5
1    2    3
4    1    2
1    2    4

// The fourth row has the same values in a and b as the second, so it is not deleted. The first 3 rows of course match themselves so are not deleted

a    b    c
-----------
1    1    5
1    2    3
4    1    2
1    2    4 

// The first 4 rows of course match themselves so are not deleted. The fifth row does not have the same value in both a and b as any of the first 4 rows, so is deleted.

select t1.* from table1 t1, 
(select distinct t2.a, t2.b from table1 t2 limit N) tabledist 
where t1.a=tabledist.a and t1.b=tabledist.b;

with toremove(a, b, c) as 
    (select * from table1 tt 
    EXCEPT select t1.* from table1 t1, 
    (select distinct t2.a, t2.b from table1 t2 limit N) tabledist 
    where t1.a=tabledist.a and t1.b=tabledist.b) 
delete from table1 where exists 
(select * from toremove 
where table1.a=toremove.a and table1.b=toremove.b and table1.c=toremove.c);