如何使用“ in”运算符返回0而不是null
我有三张桌子:
当我执行此命令时:
select count(coalesce(text_id,0)), text_id from text_trigram
where text_id in (1, 2, 3)
and trigram_id = 1
group by text_id;
结果出来了,却没有null
我想要的结果0
:
count|text_id1 1
1 2
这是我除了拥有的东西:
count|text_id 1 1
1 2
0 3
此外,我想执行以下操作:
select count(coalesce(text_id,0)), text_id from text_trigram
where text_id in (1, 2, 3)
and trigram_id in (1, 2, 3)
group by text_id;
count|text_id|trigram_id
1 1 1
1 1 2
0 1 3
1 2 1
1 2 2
1 2 3
0 3 1
有可能的?还是使用in
运算符是错误的?
回答:
我认为这里的困惑是您假设的值为null text_id=3
,但实际上 没有匹配的行 。考虑以下简化版本:
select *from text_trigram
where text_id in (3)
如果没有带有的条目,则不会返回任何行text_id=3
。它不会伪造一行带有一串空值的行。
为了即使没有匹配的数据也强制存在一行,您可以创建一个包含这些ID的表表达式,例如
select * from(
values (1), (2), (3)
) as required_ids ( text_id );
然后是LEFT JOIN
您的数据,因此您可以NULL
找到没有匹配数据的地方:
select *from
(
values (1), (2), (3)
) as required_ids ( text_id )
left join text_trigram
on text_trigram.text_id = required_ids.text_id;
要进行第一个查询,请注意两件事:
count
忽略空值,因此count(text_trigram.text_id)
将仅对在text_trigram
表中找到匹配项的行进行计数任何额外条件都需要放在的
on
子句中left join
,以便它们从中消除行text_trigram
,而不是整个查询中的行select count(text_trigram.text_id), required_ids.text_id
from
(
values (1), (2), (3)
) as required_ids ( text_id )
left join text_trigram
on text_trigram.text_id = required_ids.text_id
and text_trigram.trigram_id = 1
group by text_id
order by text_id;
此更改为每个排列text_id
并trigram_id
只想涉及额外的表表达式和CROSS JOIN
:
select required_text_ids.text_id, required_trigram_ids.trigram_id, count(text_trigram.text_id)from
(
values (1), (2), (3)
) as required_text_ids( text_id )
cross join
(
values (1), (2), (3)
) as required_trigram_ids( trigram_id )
left join text_trigram
on text_trigram.text_id = required_text_ids.text_id
and text_trigram.trigram_id = required_trigram_ids.trigram_id
group by required_text_ids.text_id, required_trigram_ids.trigram_id
order by required_text_ids.text_id, required_trigram_ids.trigram_id;
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