Sql 两种varchar的比较

Sql 两种varchar的比较,sql,oracle,Sql,Oracle,我需要您的帮助来找到解决方案: char1 : "1,2,3" char2 : "1,2,3,4,5" 并返回以下结果: char3 : "4,5" 我怎样才能做到这一点呢?像这样的事情 SQL> with test (char1, char2) as 2 (select '1,2,3', '1,2,3,4,5' from dual), 3 ch1 as 4 (select regexp_substr(char1, '[^,]+', 1, level) va

我需要您的帮助来找到解决方案:

char1 : "1,2,3"
char2 : "1,2,3,4,5" 
并返回以下结果:

char3 : "4,5"
我怎样才能做到这一点呢?

像这样的事情

SQL> with test (char1, char2) as
  2    (select '1,2,3', '1,2,3,4,5' from dual),
  3  ch1 as
  4    (select regexp_substr(char1, '[^,]+', 1, level) val
  5     from test
  6     connect by level <= regexp_count(char1, ',') + 1
  7    ),
  8  ch2 as
  9    (select regexp_substr(char2, '[^,]+', 1, level) val
 10     from test
 11     connect by level <= regexp_count(char2, ',') + 1
 12    )
 13  --
 14  select listagg(x.val, ',') within group (order by x.val) result
 15  from
 16  ( (select val from ch1
 17     minus
 18     select val from ch2
 19    )
 20    union
 21    (select val from ch2
 22     minus
 23     select val from ch1
 24    )
 25  ) x;

RESULT
-----------------------------------------------------------------------
4,5

SQL>
还有一个变体:

with t as 
( select 
 '1,2,3' as char1, 
 '1,2,3,4,5' as char2 
from dual )
select listagg(coalesce(a.parsed_char,b.parsed_char),',') within group (order 
by 1) as res
from (  select regexp_substr(char1, '[^,]+',1,level) as parsed_char 
    from t
    connect by regexp_substr(char1, '[^,]+',1,level) is not null ) a
full join 
 (  select regexp_substr(char2, '[^,]+',1,level) as parsed_char 
    from t
    connect by regexp_substr(char2, '[^,]+',1,level) is not null ) b
on (a.parsed_char = b.parsed_char)  
where a.parsed_char is null or b.parsed_char is null;

你的价值观是有序的吗?是否可以多次使用相同的值?这些值总是数字吗?char1总是char2的子字符串吗?如果是这样,它是否总是char2的第一部分?双引号是打字错误还是有什么意思?到目前为止,您尝试了什么?事实上,第一个字符串是动态影响的,它可以包含不同于我为示例修复的值!好吧,那么-你为什么不发这样一个例子呢?我原以为会是一样的,但我可能错了。每个字符串的值已经存储在两个变量中,比如var1和var2,因此处理将在var1和var2上进行。对不起,谢谢你的帮助