Question

select id,rules from links where id=2;
 id |                                                                                   rules                                                                                   
----+---------------------------------------------------------------------------------------------------------------------------------------------------------------------------
  2 | {"{\"id\": \"61979e81-823b-419b-a577-e2acb34a2f40\", \"url\": \"https://www.wikijob.co.uk/about-us\", \"what\": \"country\", \"matches\": \"GB\", \"percentage\": null}"}

我正在尝试使用此处的https://www.postgresql.org/docs/9.6/functions-json.html

运算符来获取jsonb的元素

无论我使用“ url”还是下面的整数，我都会得到类似的结果。

select id,rules->1 from links where id=2;
ERROR:  operator does not exist: jsonb[] -> integer
LINE 1: select id,rules->1 from links where id=2;
                       ^
HINT:  No operator matches the given name and argument type(s). You might need to add explicit type casts.

我在做什么错？

PS Postgres版本9.6.12。

Answer 1

该列是一个数组，您可以使用索引访问第一个元素：

select id, rules[1]
from links
where id = 2

请务必同时检查this answer.

在横向联接中使用jsonb_each()可以在单独的行中查看所有规则：

select id, key, value
from links
cross join jsonb_each(rules[1]) as rule(key, value)
where id = 2

您可以通过以下方式获得一条规则：

select id, value as url
from links
cross join jsonb_each(rules[1]) as rule(key, value)
where id = 2 and key = 'url'

使用unnest()在数组的所有元素中查找url，例如：

select id, unnest(rules)->'url' as url
from links
where id = 2;

错误：运算符不存在：jsonb []->整数

1 个答案: