Oracle Regexp根据首次出现的分隔符来拆分字符串

Question

我有以下数据

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使用oracle regexpr的预期输出

1241 - ASGEF - AFZHE
1242 - ASGEF1
1243 - UJHAHAHG - AGHBNG

Answer 1

SUBSTR + INSTR可能会运行得更快（即不需要正则表达式）。 result显示，而result_2显示一个正则表达式选项。

SQL> with test (col) as
  2    (select '1241 - ASGEF - AFZHE'     from dual union all
  3     select '1242 - ASGEF1'            from dual union all
  4     select '1243 - UJHAHAHG - AGHBNG' from dual
  5    )
  6  select col,
  7         trim(substr(col, instr(col, '-') + 1)) result,
  8         ltrim(regexp_substr(col, '-.+'), '- ') result_2
  9  from test;

COL                      RESULT                   RESULT_2
------------------------ ------------------------ ------------------------
1241 - ASGEF - AFZHE     ASGEF - AFZHE            ASGEF - AFZHE
1242 - ASGEF1            ASGEF1                   ASGEF1
1243 - UJHAHAHG - AGHBNG UJHAHAHG - AGHBNG        UJHAHAHG - AGHBNG

SQL>