如何在oracle中获取给定字符串之间的字符串

Question

这就是我的数据的样子。

APP_APX_PLM_PostCategory~~~pavan anand~~~2019-09-26 15:03:39@@@MF_APX_PLM_PostBuyProgram.msgflow***MF_APX_PLM_PostBuyProgram1.msgflow***MF_APX_PLM_PostBuyProgram2.msgflow^^^APP_APX_PLM_Promo~~~skola2
~~~2019-09-30 14:34:11@@@MF_APX_PLM_Promo1.msgflow***MF_APX_PLM_Promo2.msgflow^^^APP_APX_PLM_Santosh~~~skola2~~~2019-09-30 14:39:26@@@MF_PKMS_DA_TransferPickInboundPreProcessor.msgflow***MF_PKMS_DA_TransferPickInboundPreProcessor.msgflow

我想获取从第一个字符到第一次出现 ^^^ 的数据，以及从第一个 ^^^ 到第二个 ^^^ 的数据。

谁能帮我在oracle中得到这个？

Answer 1

使用INSTR查找分隔符的位置，然后使用SUBSTR提取子字符串：

Oracle 设置：

CREATE TABLE test_data ( value ) AS
  SELECT 'APP_APX_PLM_PostCategory~~~pavan anand~~~2019-09-26 15:03:39@@@MF_APX_PLM_PostBuyProgram.msgflow***MF_APX_PLM_PostBuyProgram1.msgflow***MF_APX_PLM_PostBuyProgram2.msgflow^^^APP_APX_PLM_Promo~~~skola2
~~~2019-09-30 14:34:11@@@MF_APX_PLM_Promo1.msgflow***MF_APX_PLM_Promo2.msgflow^^^APP_APX_PLM_Santosh~~~skola2~~~2019-09-30 14:39:26@@@MF_PKMS_DA_TransferPickInboundPreProcessor.msgflow***MF_PKMS_DA_TransferPickInboundPreProcessor.msgflow' FROM DUAL;

查询：

SELECT SUBSTR( value, 1, first_position - 1 ) AS first_substr,
       SUBSTR( value, first_position + 3, second_position - first_position - 3 ) AS second_substr
FROM   (
  SELECT value,
         INSTR( value, '^^^', 1, 1 ) AS first_position,
         INSTR( value, '^^^', 1, 2 ) AS second_position
  FROM   test_data
)

输出：

FIRST_SUBSTR | SECOND_SUBSTR :------------------------------------------------- -------------------------------------------------- -------------------------------------------------- -------------------- | :------------------------------------------------- -------------------------------------------------- -------- APP_APX_PLM_PostCategory~~~pavan anand~~~2019-09-26 15:03:39@@@MF_APX_PLM_PostBuyProgram.msgflow***MF_APX_PLM_PostBuyProgram1.msgflow***MF_APX_PLM_PostBuyProgram2.msgflow | APP_APX_PLM_Promo~~~skola2
~~~2019-09-30 14:34:11@@@MF_APX_PLM_Promo1.msgflow***MF_APX_PLM_Promo2.msgflow

db小提琴here

Answer 2

使用 REGEXP_SUBSTR

https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=0981e7232b873934c5d4e833c141f47a

with tbl as 
(select 'APP_APX_PLM_PostCategory~~~pavan anand~~~2019-09-26 15:03:39@@@MF_APX_PLM_PostBuyProgram.msgflow***MF_APX_PLM_PostBuyProgram1.msgflow***MF_APX_PLM_PostBuyProgram2.msgflow^^^APP_APX_PLM_Promo~~~skola2~~~2019-09-30 14:34:11@@@MF_APX_PLM_Promo1.msgflow***MF_APX_PLM_Promo2.msgflow^^^APP_APX_PLM_Santosh~~~skola2~~~2019-09-30 14:39:26@@@MF_PKMS_DA_TransferPickInboundPreProcessor.msgflow***MF_PKMS_DA_TransferPickInboundPreProcessor.msgflow' str 
from dual)
select TRIM( '^' FROM REGEXP_SUBSTR(str, '.+?\^{3}')),
TRIM( '^' FROM REGEXP_SUBSTR(str, '\^{3}.+?\^{3}',1,1,'n'))  from tbl

输出

APP_APX_PLM_PostCategory~~~pavan anand~~~2019-09-26 15:03:39@@@MF_APX_PLM_PostBuyProgram.msgflow***MF_APX_PLM_PostBuyProgram1.msgflow***MF_APX_PLM_PostBuyProgram2.msgflow  APP_APX_PLM_Promo~~~skola2~~~2019-09-30 14:34:11@@@MF_APX_PLM_Promo1.msgflow***MF_APX_PLM_Promo2.msgflow

Answer 3

当窗口弹出时复制你的字符串或替换为 :str

select 
substr(:str,
instr(:str,'^^^',1,1) +3,
instr(:str,'^^^',1,2) - instr(:str,'^^^',1,1) - 3)
from dual;

Answer 4

我将以不同的方式处理它，并假设您正准备解析这个多分隔字符串。由于第一级分隔似乎是'^^^'，让我们使用connect by 遍历字符串并拆分这些元素，只返回前2 个，因为这是您的要求。这将返回由 '^^^' 分隔的前 2 个元素：

with tbl(str) as (
  select 'APP_APX_PLM_PostCategory~~~pavan anand~~~2019-09-26 15:03:39@@@MF_APX_PLM_PostBuyProgram.msgflow***MF_APX_PLM_PostBuyProgram1.msgflow***MF_APX_PLM_PostBuyProgram2.msgflow^^^APP_APX_PLM_Promo~~~skola2~~~2019-09-30 14:34:11@@@MF_APX_PLM_Promo1.msgflow***MF_APX_PLM_Promo2.msgflow^^^APP_APX_PLM_Santosh~~~skola2~~~2019-09-30 14:39:26@@@MF_PKMS_DA_TransferPickInboundPreProcessor.msgflow***MF_PKMS_DA_TransferPickInboundPreProcessor.msgflow' str 
  from dual
)
select regexp_substr(str, '(.*?)(\^\^\^|$)', 1, level, NULL, 1) element
from tbl
connect by level <= 2; --regexp_count(str, '\^\^\^')+1;

请注意，如果您只需要特定元素，请将 regexp_substr 调用中的“级别”替换为元素的编号。