你也可以使用regex_replace来实现：

select id, "SLA text",
    case when "SLA text" = 'Yes' then
        trim(leading '/' from regexp_replace(text,'(?:.*The average processing time for this type of request is (\d+) days\.)(?:
This delay is reduced to (\d+) day for requests with high or critical priority.)?.*' , '\2/\1', 'g'))
    else '' end "SLA2 text"
from(
  SELECT 
      id,
      CASE 
          WHEN regexp_replace(description, ' ', ' ', 'g') ILIKE '%The average processing%' 
          THEN 'Yes'
          ELSE 'No'
      END AS "SLA text",
      regexp_replace(description, ' ', ' ', 'g') text
  FROM 
      test
) t

在这里，我们用从这个模式中找到的数字替换你的模式。当第二个数字丢失时，修剪第一个斜线。

可以将substirng与正则表达式一起使用

SELECT 
    id,
    CASE 
        WHEN regexp_replace(description, ' ', ' ', 'g') ILIKE '%The average processing%' 
        THEN 'Yes'
        ELSE 'No'
    END AS "SLA text",
  substring(description, '([0-9]*) day') days
FROM 
    test
where substring(description, '([0-9]*) day') IS NULL

SELECT 1

SELECT 
    id,
    CASE 
        WHEN regexp_replace(description, ' ', ' ', 'g') ILIKE '%The average processing%' 
        THEN 'Yes'
        ELSE 'No'
    END AS "SLA text",
  substring(description, '([0-9]*) day') days
FROM 
    test
where substring(description, '([0-9]*) day') IS NULL
UNION ALL
select 
  id,
    MAX(CASE 
        WHEN regexp_replace(description, ' ', ' ', 'g') ILIKE '%The average processing%' 
        THEN 'Yes'
        ELSE 'No'
    END) AS "SLA text",  
  STRING_AGG(match[1], '/' ORDER BY match[1]) as days
from test
cross join lateral regexp_matches(description, '([0-9]*) day', 'g') as match
Group by id

SELECT 5

fiddle