oracle中使用connect展開帶有分隔符字符串時帶有空值時的處理

使用版本爲11g及以上

對於aaa,bbb,ddd,ggg這樣的字符串，可以使用connect語句將其拆分

with tab1 as (
select 'aaa,bbb,ddd,ggg' str from dual
)
select t1.str,
       regexp_substr(t1.str, '[^,]+', 1, level) res,
       level
  from tab1 t1
connect by level <= regexp_count(t1.str, ',') + 1
;

STR	RES	LEVEL
aaa,bbb,ddd,ggg	aaa	1
aaa,bbb,ddd,ggg	bbb	2
aaa,bbb,ddd,ggg	ddd	3
aaa,bbb,ddd,ggg	ggg	4

網上經常可以找到這樣的例子，這寫的也沒有什麼問題。但是當所分隔的值中存在null時，就有問題了。

with tab1 as (
select 'aaa,bbb,,ddd,,,ggg' str from dual
)
select t1.str,
       regexp_substr(t1.str, '[^,]+', 1, level) res,
       level
  from tab1 t1
connect by level <= regexp_count(t1.str, ',') + 1
;

STR	RES	LEVEL
aaa,bbb,ddd,ggg	aaa	1
aaa,bbb,ddd,ggg	bbb	2
aaa,bbb,ddd,ggg	ddd	3
aaa,bbb,ddd,ggg	ggg	4
aaa,bbb,ddd,ggg		5
aaa,bbb,ddd,ggg		6
aaa,bbb,ddd,ggg		7

而我們真正想要的結果應該是這樣的

STR	RES	LV
aaa,bbb,ddd,ggg	aaa	1
aaa,bbb,ddd,ggg	bbb	2
aaa,bbb,ddd,ggg		3
aaa,bbb,ddd,ggg	ddd	4
aaa,bbb,ddd,ggg		5
aaa,bbb,ddd,ggg		6
aaa,bbb,ddd,ggg	ggg	7

這裏本應該使用更加巧妙的正則表達式來解決的，但是我一時沒想到，就先用下面的方法來湊合了

with tab1 as (
select 'aaa,bbb,,ddd,,,ggg' str from dual
)
, tab2 as (
select t1.str,
       regexp_substr(t1.str, '[^,]*', 1, level) res, 
       lag(regexp_substr(t1.str, '[^,]*', 1, level)) over(order by level) lg,
       max(regexp_substr(t1.str, '[^,]*', 1, level)) 
           over(order by level rows between current row and unbounded following) mx,
       level lv
  from tab1 t1
connect by level <= regexp_count(t1.str, ',') * 2 + 1
)
select t1.str,
       t1.res,
       row_number() over(order by t1.lv) lv
  from tab2 t1
 where t1.mx is not null
    and (t1.res is not null
    or (t1.res is null and t1.lg is null)
    )
;