在oracle中要求分割功能

Question

现在有一个像这样的字符串:
789 + 456-239

我想得到一个这样的列表:

sign | num
+      789
+      456
-      239

Answer 1

这可能是一种方法,通过使用常用的分裂字符串方法,只适合您处理标志的需要.

-- test case
with yourString(str) as 
        (
        select '+789+456-239 ' str
        from dual
        )
-- query
SELECT regexp_substr(str, '[+-]', 1, level) sign,
       regexp_substr(str, '[^+-]+', 1, level) num
  FROM ( select case                                  
                  when substr(str, 1, 1) in ('+','-') then str
                  -- I add a plus sign if the first char of the string is not a sign
                  else '+' || str 
                end as str
         from yourString
       )
CONNECT BY regexp_instr(str, '[+-]', 1, level ) > 0

这给出了:

SIGN           NUM
-------------- --------------
+              789
+              456
-              239

在这里,我假设如果在该字符串的开头没有给出符号,则第一个符号将是+.

如果需要处理多个字符串,则需要在表中标识一个ID,这CONNECT BY会变得有点复杂:

-- test case
with yourString(id, str) as 
        (
        select 1, '+789+456-239' from dual union all
        select 2, '789+456-239'  from dual union all
        select 3, '-789+456-239' from dual
        )
-- query
SELECT id,
       regexp_substr(str, '[+-]', 1, level) sign,
       regexp_substr(str, '[^+-]+', 1, level) num
  FROM ( select id,
                case                                  
                  when substr(str, 1, 1) in ('+','-') then str
                  -- I add a plus sign if the first char of the string is not a sign
                  else '+' || str 
                end as str
         from yourString
       )
CONNECT BY regexp_instr(str, '[+-]', 1, level ) > 0
  and prior id = id
  and prior sys_guid() is not null 

结果:

       ID SIGN  NUM
---------- ----- --------------------------------------------
         1 +     789
         1 +     456
         1 -     239
         2 +     789
         2 +     456
         2 -     239
         3 -     789
         3 +     456
         3 -     239