如何使用特殊字符作爲分隔符在oracle中grep字符串？

Question

我想獲取從下面的圖案的第一特殊字符

SQL> select distinct cell_name from v$cell_state; 

CELL_NAME 
-------------------------------------------------------------------------------- 
10.160.0.39;10.160.0.40 
10.160.0.41;10.160.0.42 
10.160.0.43;10.160.0.44 
10.160.0.45;10.160.0.46 
10.160.0.47;10.160.0.48 
10.160.0.49;10.160.0.50 
10.160.0.51;10.160.0.52 

expected output: 
10.160.0.39 
10.160.0.41 
10.160.0.43 
10.160.0.45 
10.160.0.47 
10.160.0.49 
10.160.0.51 

SQL> select distinct cell_name from v$cell_state; 

CELL_NAME 
-------------------------------------------------------------------------------- 
10.160.0.39,10.160.0.40 
10.160.0.41,10.160.0.42 
10.160.0.43,10.160.0.44 
10.160.0.45,10.160.0.46 
10.160.0.47,10.160.0.48 
10.160.0.49,10.160.0.50 
10.160.0.51,10.160.0.52 

expected output: 
10.160.0.39 
10.160.0.41 
10.160.0.43 
10.160.0.45 
10.160.0.47 
10.160.0.49 
10.160.0.51 

SQL> select distinct cell_name from v$cell_state; 

CELL_NAME 
-------------------------------------------------------------------------------- 
190.160.14.3 
190.160.14.4 
190.160.14.5 
190.160.14.8 
190.160.14.9 

expected output: 
190.160.14.3 
190.160.14.4 
190.160.14.5 
190.160.14.8 
190.160.14.9 

我想編寫一個查詢來實現它獲取的IP在所有上述3個方案是，即使沒有特別的前結束該IP地址字符，那麼就應該取現有的IP輸出

Answer 1

[^;, ]任何包機比;，,和
*任意數量出現其他

select distinct 
     regexp_substr(cell_name,'[^;, ]*') 

from v$cell_state 
; 

演示

select regexp_substr('10.160.0.39;10.160.0.40' ,'[^;, ]*') 
     ,regexp_substr('10.160.0.39,10.160.0.40' ,'[^;, ]*') 
     ,regexp_substr('10.160.0.39'    ,'[^;, ]*') 

from dual 
;