Question

需要帮助将以下字符串转换为所需的格式。我将具有以下几个值。是否可以使用REGEXP或更好的方法轻松实现此目的？

当前格式来自A列

Region[Envionment Lead|||OTC|||06340|||List Program|||TX|||Z3452|||Souther Region 05|||M7894|||California Divison|||Beginning]

Region[Coding Analyst|||BA|||04561|||Water Bridge|||CA|||M8459|||West Region 09|||K04956|||East Division|||Supreme]

A列的必需格式

Region[actingname=Envionment Lead,commonid=OTC,insturmentid=06340,commonname=List Program]

Region[actingname=Coding Analyst,commonid=BA,insturmentid=04561,commonname=Water Bridge]

修订后的数据

**Column data**  
Region[Coding Analyst|||BA|||reg pro|||04561|||08/16/2011|||Board member|||AZ|||06340|||Whiter Bridge|||CA|||M0673|||West Region 09|||K04956|||East Division|||Supreme]

**required Data** 
{actingname=06340, actingid=M0673, insturmentid=BA, insturmentname=Coding Analyst, commonname=West Region 09, stdate=08/16/2011, linnumber=04561, linstate=CA, linname=Supreme}

问题在于获取字符串的10、11、12和15位置。我可以得到低于10位的任何东西，但不能达到10位或更多的字符串位置。你能指导我在这里我想念的吗

'{actingname=\8,actingid=\11,insturmentid=\2,insturmentname=\1,commonname=\12, stdate=\5,linnumber=4,linstate=10,linname=15}'--Here 10,11,12 and 15 posistion are not being fethched

Answer 1

我使用了REGEXP_REPLACE

import requests
from bs4 import BeautifulSoup

query = "deep"
yahoo = "https://search.yahoo.com/search?q=" + query + "&n=" + str(10)
raw_page = requests.get(yahoo)

soup = BeautifulSoup(raw_page.text)

for link in soup.find_all(attrs={"class": "ac-algo fz-l ac-21th lh-24"}):
    print (link.text, link.get('href'))

或者像更新一样

SELECT REGEXP_REPLACE(
    'Region[Envionment Lead|||OTC|||06340|||List Program|||TX|||Z3452|||Souther Region 05|||M7894|||California Divison|||Beginning]',
    '^Region\[([[:alpha:][:space:][:digit:]]*)\|\|\|([[:alpha:]]*)\|\|\|([[:digit:]]*)\|\|\|([[:alpha:][:space:][:digit:]]*).*',
    'Region[actingname=\1,commonid=\2,instrumentid=\3,commonname=\4]') as replaced
FROM dual

Answer 2

您可以连续使用regexp_substr和listagg

with t1(str1) as
(
 select 'Region[Coding Analyst|||BA|||04561|||Water Bridge]' from dual   
),   t2(str2) as
(
 select 'actingname,commonid,insturmentid,commonname' from dual   
),   t3 as
(
select regexp_substr(str1, '[^|||]+', 1, level) str1,
       regexp_substr(str2, '[^,]+', 1, level)||'=' str2,
       level as lvl
  from t1
  cross join t2 
connect by level <= regexp_count(str1, '[^|||]+')
),   t4 as
(
select case when lvl = 1 then 
                 replace(str1,'[','['||str2)
       else
                 str2||str1
       end as str, lvl
  from t3
)    
select listagg(str,',') within group (order by lvl) as "Result String" from t4;

Result String
----------------------------------------------------------------------------------------
Region[actingname=Coding Analyst,commonid=BA,insturmentid=04561,commonname=Water Bridge]

P.S。我将第二个作为样本，由于第三个字符串以等号结尾的元组标签为4，因此将第一个字符串取为4，这是因为三串分隔的子字符串的数量。 Demo

Answer 3

这将起作用：

select substr(regexp_replace(regexp_replace(regexp_replace
(regexp_replace(regexp_replace("col1",'\[','[actingname='),
                     '\|\|\|',',commonid=',1,1,'i'),
'\|\|\|',',insturmentid=',1,1,'i'),
'\|\|\|',',commonname=',1,1,'i'),
    '\|',']',1,1,'i'),
             1,regexp_instr(regexp_replace(regexp_replace(regexp_replace
(regexp_replace(regexp_replace("col1",'\[','[actingname='),
                     '\|\|\|',',commonid=',1,1,'i'),
'\|\|\|',',insturmentid=',1,1,'i'),
'\|\|\|',',commonname=',1,1,'i'),
    '\|',']',1,1,'i'),'\]')-1            )||']'
from Table1;

检查： http://sqlfiddle.com/#!4/3ddfa0/11

谢谢!!!!!!

Oracle字符串转换

3 个答案: