使用PL/SQL分析来依赖子字符串？

Question

如何编写查询以返回以K开头的农场的计数

为什么(partition by id,substr(farm,1))计算为1

with tree_harvest
as ( 
select 1 as id, 'PINE' as tree, 'K001' as farm from dual union all
select 1 as id, 'PINE' as tree, '0003' as farm from dual union all
select 1 as id, 'PINE' as tree, 'K002' as farm from dual union all
select 1 as id, 'PINE' as tree, 'K003' as farm from dual
)
select id, tree,farm,
       count(*) over (partition by id) as id_count,
       case
       when regexp_like(farm,'^K','i') 
       then count(*) over (partition by id,substr(farm,1))
       else 0
       end as k_count
       from tree_harvest;

期望结果

 ID TREE FARM   ID_COUNT  K_COUNT  
  1 PINE  0003  4         0
  1 PINE  K001  4         3
  1 PINE  K002  4         3
  1 PINE  K003  4         3

Answer 1

下面是解决问题的解决方案，它应该比当前的方法更快(更有效)。请注意，这里这两个解析函数只由id分区；条件计数是在count()调用本身中单独处理的。而且，对K或k的比较都是不区分大小写的；在您尝试的查询中，有一个比较不敏感。我也避免正则表达式(更慢)，因为这里不需要正则表达式。

with tree_harvest
as ( 
select 1 as id, 'PINE' as tree, 'K001' as farm from dual union all
select 1 as id, 'PINE' as tree, '0003' as farm from dual union all
select 1 as id, 'PINE' as tree, 'K002' as farm from dual union all
select 1 as id, 'PINE' as tree, 'K003' as farm from dual
)
select id, tree,farm,
       count(*) over (partition by id) as id_count,
       case when lower(farm) like 'k%' then
           count(case when lower(farm) like 'k%' then 1 end) 
                over (partition by id) else 0 end as k_count
       from tree_harvest;
       
        ID TREE FARM   ID_COUNT    K_COUNT
---------- ---- ---- ---------- ----------
         1 PINE K001          4          3
         1 PINE K003          4          3
         1 PINE K002          4          3
         1 PINE 0003          4          0