我们希望在每个“，”分隔符后收集4个字符
以下是使用regexp_replace的方法：

select regexp_replace(
    '123456,852369,7852159,1596357',
    '([^,]{4})[^,]*(,|$)',
    '\1\2'
)
from dual

正则表达式分解：

([^,]{4})    4 characters others than "," (capture that group as \1)
[^,]*        0 to n characters other than "," (no capture)
(,|$)        either character "," or the end of string (capture this as \2)

该函数用capture 1（我们需要的4个字符）和capture 2（分隔符，如果有的话）替换每个匹配。

Demo：

| 结果|
| - -|
| 1234，8523，7852，1596个单位|

一个选项可能是拆分字符串，提取4个字符并将其聚合回来：

SQL> with test (col) as
  2    (select '123456,852369,7852159,1596357' from dual)
  3  select listagg(regexp_substr(col, '[^,]{4}', 1, level), ',')
  4         within group (order by level) result
  5  from test
  6  connect by level <= regexp_count(col, ',') + 1;

RESULT
--------------------------------------------------------------------------------
1234,8523,7852,1596

SQL>

使用REGEX_REPLACE：

select regexp_replace(the_string, '(^|,)([^,]{4})[^,]*', '\1\2')
from mytable;

这看起来
1.字符串的开头或逗号
1.然后是四个不是逗号的字符
1.然后是任意数量的非逗号结尾字符
只保留了
1.开头或逗号
1.后面的四个字符
演示：https://dbfiddle.uk/efUFvKyO