使用cte构建连续状态的运行：

with recursive cte(s_date, date, item, status, s_count, result) as (
    select e.date, e.date, e.item, e.status, 1, '[]'::jsonb from entries e 
    left join entries e1 on e1.item = e.item and e.date - interval '1 day' = e1.date where e1.date is null
    union all
    select c.s_date, e.date, c.item, e.status, 
      case when e.status = c.status then c.s_count + 1 else 1 end, 
      case when e.status = c.status then c.result else c.result || jsonb_build_object('s', c.status, 'c', c.s_count) end 
    from cte c join entries e on e.item = c.item and c.date + interval '1 day' = e.date
)
select date(t1.md), t1.item, e.status, (select max(((v -> 'c')#>>'{}')::int) 
      from jsonb_array_elements(r::jsonb) v where (v -> 's')#>>'{}' = e.status) 
    from (select t.s_date, t.item, max(t.date) md, max(t.result::text) r 
 from (select c.s_date, c.date, c.item, c.result || jsonb_build_object('s', c.status, 'c', c.s_count) result from cte c) t
group by t.s_date, t.item) t1
join entries e on e.item = t1.item and date(e.date) = date(t1.md)

See fiddle.

根据您的评论，您希望找到连续状态where status = last status value的最大计数，这变成了一个间隙和岛问题。这可以使用两个row_numbers之间的差和last_value函数来解决，如下所示：

with last_status as
(
  select *, 
      last_value(status) over (partition by item order by date_ 
        range between unbounded preceding and unbounded following) current_status,
      max(date_) over (partition by item) latest_date,
      row_number() over (partition by item order by date_) -
      row_number() over (partition by item, status order by date_) grp
  from table_name
),
consecutive_status_counts as
(
  select latest_date, item, current_status, status, count(*) cnt
  from last_status 
  where current_status = status
  group by latest_date, item, current_status, status, grp 
)
select latest_date, 
       item,
       current_status, 
       max(cnt) number_of_days_on_current
from consecutive_status_counts
group by latest_date, item, current_status
order by item

See demo