我知道你用的是10g，所以这不管用。但是为了完整性，LISTAGG()“正确”地处理NULL值。不过，要做到这一点，你必须更新到11g2：

-- Some sample data, roughly equivalent to yours
with t as (
  select 'foo' as x from dual union all
  select null       from dual union all
  select 'bar'      from dual
)
-- Use the listagg aggregate function to join all values
select listagg(x, ';') within group (order by rownum)
from t;

或者更简洁一点，如果您想列出表中的列：

-- I use SYS.ORA_MINING_VARCHAR2_NT as a TABLE TYPE. Use your own, if you prefer
select listagg(column_value, ';') within group (order by rownum)
from table(ORA_MINING_VARCHAR2_NT('foo', null, 'bar'));

或在实际的table上：

select listagg(column_value, ';') 
       within group (order by rownum)
from Table1
cross join table(ORA_MINING_VARCHAR2_NT(Table1.a, Table1.b, Table1.c))
group by Table1.id;

现在我不确定这是否比您的原始示例更好(更具可读性)：-)

select trim(';' from REGEXP_REPLACE (a || ';' || b || ';' || c , ';+' , ';')) abc 
from Table1

AFAIK，没有简明的方法可以做到这一点。

在过去，我曾求助于

SELECT a
||     DECODE(b
       ,      NULL, NULL
       ,      ';' || b)
||     DECODE(c
       ,      NULL, NULL
       ,      ';' || c)
||     DECODE(d
       ,      NULL, NULL
       ,      ';' || d)
...
FROM   table1

但这并不比你的例子更好。

Not all fields will have data sequentially. This is for fields that are blank and without spaces:

1. Evaluate if the field is blank, if so concatenate an empty field so it is ignored and if blank, concatenate the field with a space on the end.
2. Then [TRIM] spaces (leading and trailing), then [REPLACE] spaces with your delimiter.

You can get creative with this until Oracle gives a function to do this.

Code snippet:

SELECT 
    f1,f2,f3,f4
  , REPLACE (TRIM(
     CASE WHEN f1 IS NULL THEN '' ELSE f1||' ' END||
     CASE WHEN f2 IS NULL THEN '' ELSE f2||' ' END||
     CASE WHEN f3 IS NULL THEN '' ELSE f3||' ' END||
     CASE WHEN f4 IS NULL THEN '' ELSE f4||' ' END),' ','|') concat_result
from 
    (SELECT 
         'I' f1
        ,'want' f2
        ,'' f3
        ,'concat' f4
 FROM dual) c'