Question

解析字符串“projects / project / 23”时可能会将其扩展为此结果：

projects
projects/project
projects/project/23

我想我可以使用滞后/引导窗口函数并进行此SQL查询：

select  lag(unnest, 1, '') OVER () || '/' || unnest AS unnest
from unnest(string_to_array('projects/project/23', '/'))

但它不起作用。如何制作？

谢谢。

一个解决方案：

好的我找到了一个解决方案（但我觉得它不是最优的）

WITH t AS (
        SELECT  row_number() OVER (),
                string_to_array('projects/project/23', '/')
        FROM unnest(string_to_array('projects/project/23', '/'))
)
SELECT  array_to_string(string_to_array[1:t.row_number], '/')
FROM t

Answer 1

答案的主要部分 - 您需要使用string_agg滚动总计：

with cte as (
    select a, row_number() over() as rn
    from unnest(string_to_array('projects/project/23', '/')) as a
)
select string_agg(a, '/') over(order by rn)
from cte

sql fiddle demo

它会起作用，但不保证元素的顺序（尽管我不能使查询工作错误）。要 100％保证元素顺序，您可以使用generate_subscripts函数：

with cte as (
    select string_to_array('projects/project/23', '/') as a
), cte2 as (
    select a, generate_subscripts(a, 1) as i from cte
)
select string_agg(a[i], '/') over(order by i)
from cte2

sql fiddle demo

将url的部分扩展为行

1 个答案: