regexp\u substr（列1，“[^”，]+”，1，1）
它只会返回 NON-NULL 数组中的值。可能有 NULL 数组中的值，因此下面的查询将返回数组中元素的确切位置，并考虑空值：

with data as 
(
  select '["abc","def","xyz","klm","nop","qrs"]' arr from dual union all
  select '["abc",,"xyz","klm","nop","qrs"]' arr from dual union all
  select '["abc","def",,"lkj",]' arr from dual
)
select 
    regexp_substr(arr, '(.*?)(,|$)', 1, 1, NULL, 1) col1,
    regexp_substr(arr, '(.*?)(,|$)', 1, 2, NULL, 1) col2,
    regexp_substr(arr, '(.*?)(,|$)', 1, 3, NULL, 1) col3,
    regexp_substr(arr, '(.*?)(,|$)', 1, 4, NULL, 1) col4,
    regexp_substr(arr, '(.*?)(,|$)', 1, 5, NULL, 1) col5,
    regexp_substr(arr, '(.*?)(,|$)', 1, 6, NULL, 1) col6
from 
(
 select replace(replace(replace(arr, '['), '"'), ']') arr from data
);

COL1 COL2 COL3 COL4 COL5 COL6
---- ---- ---- ---- ---- ----
abc  def  xyz  klm  nop  qrs 
abc       xyz  klm  nop  qrs 
abc  def       lkj