Question

我有以下查询：

select "houses"."id", 
"houses"."uuid", 
COUNT(1) OVER() as full_count from "houses" 
CROSS JOIN LATERAL jsonb_array_elements(houses.types) house_types 
inner join "hoods" on "hoods"."id" = "houses"."hood_id" and "hoods"."owner_id" = 2 
inner join "groups" on "groups"."hood_id" = "hoods"."id" and "groups"."manager_id" = 54 
where house_types->>'type' = 'big' 
group by "houses"."id", "houses"."uuid" 
order by lower(houses.name) asc 
limit 20

哪位正确地给我提供了头类型为“ big”的前20栋房屋，这些房屋位于Hood_id为2且与之关联的管理者为54个房屋的房屋中。

现在，问题是，有时我会拥有名称相同的房屋，而我只想保留其中一间并除去其余的。例如：

如果我的房屋桌子如下：

id, types, name
1, [{ type: 'rating' }], 'white house'
2, [{ type: 'rating' }], 'white house'
3, [{ type: 'rating' }], 'red house'

我只会得到ID为1和3的行。

在PostgreSQL中做到这一点的一种好方法是，假设我既可以将偏移量和限制都应用于查询，又希望删除重复项。

Answer 1

使用group by代替distinct on：

select distinct on (lower(h.name)) h.id, h.uuid
      COUNT(*) OVER() as full_count
from houses h cross join lateral
     jsonb_array_elements(h.types) ht inner join
     "hoods" ho
     on ho.id = h.hood_id and
        ho.owner_id = 2 inner join
     "groups" g
     on g.hood_id = ho.id and
        g.manager_id = 54 
where house_types->>'type' = 'big' 
order by lower(houses.name) asc  
limit 20;

编辑：

select h.*, count(*) over ()  as full_count
from (select distinct on (lower(h.name)) h.id, h.uuid
      from houses h cross join lateral
           jsonb_array_elements(h.types) ht inner join
           "hoods" ho
           on ho.id = h.hood_id and
              ho.owner_id = 2 inner join
           "groups" g
           on g.hood_id = ho.id and
              g.manager_id = 54 
      where house_types->>'type' = 'big' 
      order by lower(houses.name) asc  
     ) h
limit 20

选择行并根据列的值删除重复项

1 个答案: