我有一张桌子来存储有关我的兔子的信息。看起来像这样:
create table rabbits (rabbit_id bigserial primary key, info json not null); insert into rabbits (info) values ('{"name":"Henry", "food":["lettuce","carrots"]}'), ('{"name":"Herald","food":["carrots","zucchini"]}'), ('{"name":"Helen", "food":["lettuce","cheese"]}');
我该如何找到喜欢胡萝卜的兔子?我想出了这个:
select info->>'name' from rabbits where exists ( select 1 from json_array_elements(info->'food') as food where food::text = '"carrots"' );
我不喜欢那个查询。一团糟。
作为专职兔子管理员,我没有时间更改数据库架构。我只想适当地喂兔子。有没有更可读的方法来执行该查询?
从PostgreSQL 9.4开始,您可以使用?operator:
?
select info->>'name' from rabbits where (info->'food')::jsonb ? 'carrots';
如果改用 jsonb 类型,甚至可以?在"food"键上为查询建立索引: __
"food"
alter table rabbits alter info type jsonb using info::jsonb; create index on rabbits using gin ((info->'food')); select info->>'name' from rabbits where info->'food' ? 'carrots';
当然,作为专职兔子饲养员,您可能没有时间这样做。
更新: 这是在一张1,000,000只兔子的桌子上性能改进的演示,其中每只兔子喜欢两种食物,其中10%像胡萝卜:
d=# -- Postgres 9.3 solution d=# explain analyze select info->>'name' from rabbits where exists ( d(# select 1 from json_array_elements(info->'food') as food d(# where food::text = '"carrots"' d(# ); Execution time: 3084.927 ms d=# -- Postgres 9.4+ solution d=# explain analyze select info->'name' from rabbits where (info->'food')::jsonb ? 'carrots'; Execution time: 1255.501 ms d=# alter table rabbits alter info type jsonb using info::jsonb; d=# explain analyze select info->'name' from rabbits where info->'food' ? 'carrots'; Execution time: 465.919 ms d=# create index on rabbits using gin ((info->'food')); d=# explain analyze select info->'name' from rabbits where info->'food' ? 'carrots'; Execution time: 256.478 ms