Question

我有一个字符串：'&articleNumber=370878&variantNumber=&bundleNumber=3&supplierNumber=20064&discountType=13'（不带撇号，我想将其拆分为以下值：

articleNumber variantNumber bundleNumber supplierNumber discountType 
370878        (null)        3            20064          13

我能够基于'='之前的字符串提取所需的值，但是我希望有一个更优化的解决方案，一次提取所有内容，第一行是列名，第二行是值。< / p>

有可能吗？注意：字符串可以有1、2、3、4或5对。

Answer 1

类似这样的作品。可以做得更优雅。

select regexp_substr(regexp_substr('articleNumber=370878&variantNumber=&bundleNumber=3&supplierNumber=20064&discountType=13'
                                  ,'(.*?)(&|$)'
                                  ,1
                                  ,1
                                  ,null
                                  ,1)
                    ,'(.*?)(=|$)'
                    ,2
                    ,2
                    ,null
                    ,1) as articlenumber
      ,regexp_substr(regexp_substr('articleNumber=370878&variantNumber=&bundleNumber=3&supplierNumber=20064&discountType=13'
                                  ,'(.*?)(&|$)'
                                  ,2
                                  ,2
                                  ,null
                                  ,1)
                    ,'(.*?)(=|$)'
                    ,2
                    ,2
                    ,null
                    ,1) as variantnumber
      ,regexp_substr(regexp_substr('articleNumber=370878&variantNumber=&bundleNumber=3&supplierNumber=20064&discountType=13'
                                  ,'(.*?)(&|$)'
                                  ,3
                                  ,3
                                  ,null
                                  ,1)
                    ,'(.*?)(=|$)'
                    ,2
                    ,2
                    ,null
                    ,1) as bundlenumber
      ,regexp_substr(regexp_substr('articleNumber=370878&variantNumber=&bundleNumber=3&supplierNumber=20064&discountType=13'
                                  ,'(.*?)(&|$)'
                                  ,4
                                  ,4
                                  ,null
                                  ,1)
                    ,'(.*?)(=|$)'
                    ,2
                    ,2
                    ,null
                    ,1) as suppliernumber
      ,regexp_substr(regexp_substr('articleNumber=370878&variantNumber=&bundleNumber=3&supplierNumber=20064&discountType=13'
                                  ,'(.*?)(&|$)'
                                  ,5
                                  ,5
                                  ,null
                                  ,1)
                    ,'(.*?)(=|$)'
                    ,2
                    ,2
                    ,null
                    ,1) as discounttype
  from dual

Answer 2

尝试一下：

SET DEFINE OFF
WITH DATAA(D) AS 
(SELECT '&articleNumber=370878&variantNumber=&bundleNumber=3&supplierNumber=20064&discountType=13' 
FROM DUAL)
SELECT SUBSTR(D,INSTR(D,'=',1,1)+1,INSTR(D,'&',1,2)-INSTR(D,'=',1,1) - 1) AS articleNumber,
SUBSTR(D,INSTR(D,'=',1,2)+1,INSTR(D,'&',1,3)-INSTR(D,'=',1,2) - 1) AS variantNumber,
SUBSTR(D,INSTR(D,'=',1,3)+1,INSTR(D,'&',1,4)-INSTR(D,'=',1,3) - 1) AS bundleNumber,
SUBSTR(D,INSTR(D,'=',1,4)+1,INSTR(D,'&',1,5)-INSTR(D,'=',1,4) - 1) AS supplierNumber,
SUBSTR(D,INSTR(D,'=',1,5)+1) AS supplierNumber
FROM DATAA;

--
Output:
--
370878 | (null) | 3 | 20064 | 13

干杯！

Answer 3

SQL> with input as
  2  ( select '&articleNumber=370878&variantNumber=&bundleNumber=3&supplierNumber=20064&discountType=13' text
  3      from dual
  4  )
  5  , splitted_by_ampersand as
  6  ( select regexp_substr(text,'[^&]+',1,level) text
  7      from input
  8   connect by level <= regexp_count(text,'&')
  9  )
 10  , splitted_by_equal_sign as
 11  ( select substr(text,1,instr(text,'=')-1) name
 12         , substr(text,instr(text,'=')+1) value
 13      from splitted_by_ampersand
 14  )
 15  select *
 16    from splitted_by_equal_sign
 17   pivot ( max(value)
 18           for name in
 19           ( 'articleNumber'  as "articleNumber"
 20           , 'variantNumber'  as "variantNumber"
 21           , 'bundleNumber'   as "bundleNumber"
 22           , 'supplierNumber' as "supplierNumber"
 23           , 'discountType'   as "discountType"
 24           )
 25         )
 26  /

articleNumber        variantNumber        bundleNumber         supplierNumber       discountType
-------------------- -------------------- -------------------- -------------------- --------------------
370878                                    3                    20064                13

1 rij is geselecteerd.

使用Oracle SQL分割带有正则表达式的字符串

3 个答案: