从多个方括号中获取文本

Question

对于Oracle SQL

i具有以下标记格式：{ABCDE，12345} {47，CE}

SELECT 
[column a] 
[column b] 
[column c] 
FROM 
[table name] 
WHERE 
[*content of column d*] IN {ABCDE,12345} 
[*content of column e*] IN {47,CE}

所以我需要查询以查看标记中的内容，并在以下情况下选择所有具有相同内容的行：-D列=第一组括号-E列=第二组中的内容

Answer 1

一个选项可能是这个（请参阅代码中的注释）：

SQL> with test (col) as
  2    (select '{ABCDE,12345}{47,CE}' from dual),
  3  temp as
  4    -- remove leading and trailing brackets
  5    -- replace "middle" brackets with semi-colon (a new separator)
  6    (select replace(substr(col, 2, length(col) - 2), '}{', ';') val
  7     from test
  8    )
  9  select regexp_substr(val, '[^;]+', 1, level) result
 10  from temp
 11  connect by level <= regexp_count(val, ';') + 1;

RESULT
--------------------------------------------------------------------
ABCDE,12345
47,CE

SQL>

---

啊，是的-单独的列：那么，substr可能有帮助：

SQL> with test (col) as
  2    (select '{ABCDE,12345}{47,CE}' from dual)
  3  select substr(col, 2, instr(col, '}{') - 2) val1,
  4         substr(substr(col, instr(col, '}{') + 2),
  5                1,
  6                length(substr(col, instr(col, '}{') + 2)) - 1
  7               ) val2
  8  from test;

VAL1        VAL2
----------- -----
ABCDE,12345 47,CE

SQL>