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_id

1 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_idtrigram_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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