我有
POW,POW,POWPRO,PRO,PRO,PROUTL,TNEUTL,TNEUTL,UTL,UTLTNE,UTL,UTLTNE
我想要
POW,POWPRO,PRO,PROUTL,TNEUTL,UTL,UTLTNE
我试过
select regexp_replace('POW,POW,POWPRO,PRO,PRO,PROUTL,TNEUTL,TNEUTL,UTL,UTLTNE,UTL,UTLTNE','([^,]+)(,\1)+','\1') from dual
我得到了输出
POWPROUTL,TNEUTL,UTLTNE,UTLTNE
但我希望输出
POW,POWPRO,PRO,PROUTL,TNEUTL,UTL,UTLTNE
请帮忙。
答案 0 :(得分:1)
两个只使用SQL的解决方案和第三个使用小/简单PL / SQL函数的解决方案,这使得最终的SQL查询非常简短。
Oracle安装程序:
CREATE TABLE data ( value ) AS
SELECT 'POW,POW,POWPRO,PRO,PRO,PROUTL,TNEUTL,TNEUTL,UTL,UTLTNE,UTL,UTLTNE' FROM DUAL;
CREATE TYPE stringlist AS TABLE OF VARCHAR2(4000);
/
查询1 :
SELECT LISTAGG( t.COLUMN_VALUE, ',' ) WITHIN GROUP ( ORDER BY t.COLUMN_VALUE ) AS list
FROM data d,
TABLE(
SET(
CAST(
MULTISET(
SELECT REGEXP_SUBSTR( d.value, '[^,]+', 1, LEVEL )
FROM DUAL
CONNECT BY LEVEL <= REGEXP_COUNT( d.value, '[^,]+' )
) AS stringlist
)
)
) t
GROUP BY d.value;
<强>输出强>:
LIST
---------------------------------------
POW,POWPRO,PRO,PROUTL,TNEUTL,UTL,UTLTNE
查询2 :
SELECT ( SELECT LISTAGG( COLUMN_VALUE, ',' ) WITHIN GROUP ( ORDER BY ROWNUM )
FROM TABLE( d.uniques ) ) AS list
FROM (
SELECT ( SELECT CAST(
COLLECT(
DISTINCT
REGEXP_SUBSTR( d.value, '[^,]+', 1, LEVEL )
)
AS stringlist
)
FROM DUAL
CONNECT BY LEVEL <= REGEXP_COUNT( d.value, '[^,]+' )
) uniques
FROM data d
) d;
<强>输出强>:
LIST
---------------------------------------
POW,POWPRO,PRO,PROUTL,TNEUTL,UTL,UTLTNE
Oracle安装程序:
小助手功能:
CREATE FUNCTION split_String(
i_str IN VARCHAR2,
i_delim IN VARCHAR2 DEFAULT ','
) RETURN stringlist DETERMINISTIC
AS
p_result stringlist := stringlist();
p_start NUMBER(5) := 1;
p_end NUMBER(5);
c_len CONSTANT NUMBER(5) := LENGTH( i_str );
c_ld CONSTANT NUMBER(5) := LENGTH( i_delim );
BEGIN
IF c_len > 0 THEN
p_end := INSTR( i_str, i_delim, p_start );
WHILE p_end > 0 LOOP
p_result.EXTEND;
p_result( p_result.COUNT ) := SUBSTR( i_str, p_start, p_end - p_start );
p_start := p_end + c_ld;
p_end := INSTR( i_str, i_delim, p_start );
END LOOP;
IF p_start <= c_len + 1 THEN
p_result.EXTEND;
p_result( p_result.COUNT ) := SUBSTR( i_str, p_start, c_len - p_start + 1 );
END IF;
END IF;
RETURN p_result;
END;
/
查询3 :
SELECT ( SELECT LISTAGG( COLUMN_VALUE, ',' ) WITHIN GROUP ( ORDER BY ROWNUM )
FROM TABLE( SET( split_String( d.value ) ) ) ) AS list
FROM data d;
或(如果您只想传递单个值):
SELECT LISTAGG( COLUMN_VALUE, ',' ) WITHIN GROUP ( ORDER BY ROWNUM ) AS list
FROM TABLE( SET( split_String(
'POW,POW,POWPRO,PRO,PRO,PROUTL,TNEUTL,TNEUTL,UTL,UTLTNE,UTL,UTLTNE'
) ) );
<强>输出强>:
LIST
---------------------------------------
POW,POWPRO,PRO,PROUTL,TNEUTL,UTL,UTLTNE
答案 1 :(得分:0)
下面提供的解决方案使用直接SQL(无PL / SQL)。它适用于任何可能的输入字符串,它可以删除重复的内容 - 它保持输入令牌的顺序,无论顺序如何。它还会删除连续的逗号(&#34;从输入字符串中删除空值&#34;),同时正确处理空输入。注意输入字符串的输出仅由逗号组成,并正确处理&#34;令牌&#34;分别由两个空格和一个空格组成。
查询运行相对缓慢;如果性能是一个问题,它可以重写为递归查询,使用&#34;传统&#34; substr
和instr
比正则表达式快一点。
with inputs (input_string) as (
select 'POW,POW,POWPRO,PRO,PRO,PROUTL,TNEUTL,TNEUTL,UTL,UTLTNE,UTL,UTLTNE' from dual
union all
select null from dual
union all
select 'ab,ab,st,ab,st, , , ,x,,,r' from dual
union all
select ',,,' from dual
),
tokens (input_string, rk, token) as (
select input_string, level,
regexp_substr(input_string, '([^,]+)', 1, level, null, 1)
from inputs
connect by level <= 1 + regexp_count(input_string, ',')
),
distinct_tokens (input_string, rk, token) as (
select input_string, min(rk) as rk, token
from tokens
group by input_string, token
)
select input_string, listagg(token, ',') within group (order by rk) output_string
from distinct_tokens
group by input_string
;
我创建的输入的结果:
INPUT_STRING OUTPUT_STRING
------------------------------------------------------------------ ----------------------------------------
,,, (null)
POW,POW,POWPRO,PRO,PRO,PROUTL,TNEUTL,TNEUTL,UTL,UTLTNE,UTL,UTLTNE POW,POWPRO,PRO,PROUTL,TNEUTL,UTL,UTLTNE
ab,ab,st,ab,st, , , ,x,,,r ab,st, , ,x,r
(null) (null)
4 rows selected.