我在列中有一个文本,例如
Hello World %UC#abc#UC%. How are you %UC#def#UC%. Have a nice day %UC#ghi#UC%.
我想使用REGEXP_REPLACE
(或任何)函数将%UC#< value >#UC%
替换为UNISTR(< value >)
。从上面的例子中,结果应该是
Hello World (UNISTR of abc). How are you (UNISTR of def). Have a nice day (UNISTR of ghi).
基本上它应该剥离%UC#
并将其中的值替换为值的UNISTR
。
有没有办法实现这个目标?
答案 0 :(得分:0)
这可能是11g以及之中的一种方式:
with test(s) as ( select 'Hello World %UC#abc#UC%. How are you %UC#def#UC%. Have a nice day %UC#ghi#UC%.' || '%UC##UC%' from dual)
select listagg (str) within group ( order by lev)
from (
select regexp_substr(s, '(^|#UC%)(.*?)(%UC#)', 1, level, '', 2) ||
UPPER(regexp_substr(s, '(%UC#)(.*?)(#UC%)', 1, level, '', 2)) as str,
level as lev
from test
connect by instr(s, '%UC#', 1, level ) > 0
)
给出了(我使用UPPER
代替UNISTR
来清除结果):
Hello World ABC. How are you DEF. Have a nice day GHI.
这里的想法是使用常用的分割字符串技术,将'%UC#...#UC%'
包裹的部分视为分隔符;请注意,我在输入字符串中添加了一个小字符串('%UC##UC%'
)来处理输入字符串的最后一部分,使查询将字符串视为以及(空)'%UC#...#UC%'
序列结束处理。
在Oracle 10g中,我不能像listagg
那样使用regexp_substr
,因此,解决方案有点复杂。
这里我根本不使用正则表达式,并通过SYS_CONNECT_BY_PATH
计算聚合;要做到这一点,我需要确定一个永远不会出现在输入文本中的字符串,比如说'@@'
:
with test as ( select 'Hello World %UC#abc#UC%. How are you %UC#def#UC%. Have a nice day %UC#ghi#UC%.' || '%UC##UC%' as s from dual)
测试为(选择&#39; Hello World%UC#abc#UC%。你好%UC#def#UC%。祝你有个美好的一天%UC#ghi#UC%。&#39; | |&#39;%UC ## UC%&#39; as s from dual)
select replace ( sys_connect_by_path (
substr(s, case when level = 1 then 1 else instr(s,'#UC%', 1, level-1) +4 end, instr(s, '%UC#', 1, level) -case when level = 1 then 1 else instr(s,'#UC%', 1, level-1) +4 end ) ||
UPPER(substr(s, instr(s, '%UC#', 1, level) + 4, instr(s,'#UC%', 1, level) - (instr(s, '%UC#', 1, level) + 4)) )
, '@@'
),
'@@') str
from test
where connect_by_isleaf = 1
connect by instr(s, '%UC#', 1, level ) > 0