如何在oracle中对序列号进行分组？
我有一个用逗号分隔的字符串 例如200020012002200520062007和2010 我想把连续的数字分组 我的产出应该是2000-20032005-2007和2010年。在Oracle存储过程中有什么方法可以做到这一点吗？您必须使用，在这些年中循环，并执行以下操作：如何在oracle中对序列号进行分组？,oracle,Oracle,我有一个用逗号分隔的字符串 例如200020012002200520062007和2010 我想把连续的数字分组 我的产出应该是2000-20032005-2007和2010年。在Oracle存储过程中有什么方法可以做到这一点吗？您必须使用，在这些年中循环，并执行以下操作： CREATE OR REPLACE Function concatYears RETURN varchar; DECLARE oldYear number; concat varchar(300); newGr
CREATE OR REPLACE Function concatYears
RETURN varchar;
DECLARE
oldYear number;
concat varchar(300);
newGroup boolean;
cursor cur1 is
SELECT year
FROM dates
ORDER BY year ASC;
BEGIN
oldYear:=0;
newGroup:=true;
concat := '';
FOR row in c1
IF oldYear == 0 THEN
oldYear := row.year;
END IF;
IF newGroup == true THEN
concat := concat || CAST(oldYear AS varchar(4));
newGroup:= false;
ELSE
IF row.year > oldYear+1 THEN
concat:= concat || '-' || CAST(oldYear AS varchar(4)) || ' , ';
newGroup:=true;
END IF;
END IF;
oldYear:=row.year;
END LOOP;
RETURN concat;
END;
它可以通过使用分析函数使用SQL来完成 *更新1:*答案使用解析器功能更新，在以前的版本中丢失 *更新2:*添加了最终字符串组成
with p as ( -- Parameter string
select replace('2000,2001,2002,2005,2006,2007 and 2010',' and ',',') s from dual
),
ex as ( -- Parse string to sequence
select
to_number(
substr(
s,
decode( level, 1, 1, instr(s,',',1,level-1)+1 ),
decode( instr(s,',',1,level), 0, length(s)+1, instr(s,',',1,level) )
-
decode( level, 1, 1, instr(s,',',1,level-1)+1 )
)
) as y
from p
connect by instr(s,',',1,level-1) > 0
),
period_set as (
select -- Make final string for each interval start
y,
lag(y) over (order by y) prior_y,
max(y) over (partition by 1) max_y,
y || (case when is_end > 1 then null else '-' ||end_y end) as interval_string
from
( -- For each start find interval end
select
y,
is_start,
is_end,
lead(y) over (order by y) end_y
from
( -- Find if previous/next value differs more then by one.
-- If so, mark as start/end
select
y,
nvl(y - prev_y, 100) is_start,
nvl(next_y - y, 100) is_end
from
( -- Find previous/next value in sequence
select
y,
lag(y) over (order by y) prev_y,
lead(y) over (order by y) next_y
from ex
)
)
where
is_start > 1 or is_end > 1
)
where is_start > 1
)
select
replace(
substr(
sys_connect_by_path(
decode(y,max_y,'m', null) || interval_string,
','
),2
),
',m',
' and '
) result_str
from
period_set
where
connect_by_isleaf = 1
start with
prior_y is null
connect by
prior y = prior_y
这也可以在没有正则表达式的情况下完成假设您的数据是用逗号分隔的，并且输入可以用逗号分隔任何年份
DECLARE
v_str VARCHAR(100) := '&n';
v_instr NUMBER;
v_instr1 NUMBER;
v_g VARCHAR(50);
v_F VARCHAR(50);
v_OUT VARCHAR(50);
v_OUT1 VARCHAR(50);
v_TEMP VARCHAR(50);
v_TMP VARCHAR(50) := ' ';
v_cnt NUMBER :=0;
V_FLAG NUMBER :=1;
BEGIN
FOR i IN 1..Length(v_str)-Length(REPLACE(v_str,',',''))+1 LOOP
IF i = 1 THEN
v_g := SubStr(v_str,1,InStr(v_str,',',1,i)-1);
V_F := V_G;
ELSE
v_instr := InStr(v_str,',',1,i-1);
v_instr1 := InStr(v_str,',',1,i);
IF(v_cnt+1 <= Length(v_str)-Length(REPLACE(v_str,',',''))) then
v_g := SubStr(v_str,v_instr+1,v_instr1-v_instr-1);
IF V_FLAG = 0 THEN V_F := V_G; V_FLAG :=1; END IF;
ELSE
v_g := SubStr(v_str,v_instr+1);
IF V_FLAG = 0 THEN V_F := V_G; V_FLAG :=1; END IF;
END IF;
END IF;
v_cnt := v_cnt+1;
--IF(I>1) THEN
IF(V_TEMP+1 = V_G) THEN
IF(V_OUT IS not NULL) THEN
V_OUT := V_OUT||'-'||V_G;
ELSE
V_OUT := V_F||'-'||V_G;
END IF;
ELSE
V_OUT1 := SubStr(V_OUT,1,5)||SubStr(V_OUT,-4);
V_OUT := NULL;
v_out := v_g;
V_FLAG := 0;
END IF;
--END IF;
V_TEMP := To_Number(V_G);
--Dbms_Output.put_line(v_g);
IF(v_tmp <> v_out1) THEN
SELECT Decode(instr(v_OUT1,'-'),0,subStr(V_OUT1,1,4),v_out1) INTO v_out1 FROM dual;
Dbms_Output.put_line('Year span : '||v_OUT1);
v_tmp := v_out1;
END IF;
END LOOP;
SELECT Decode(Length(v_OUT),4,v_out,subStr(V_OUT,1,5)||SubStr(V_OUT,-4)) INTO v_out1 FROM dual;
Dbms_Output.put_line('Year span : '||v_out1);
END;
select min(seq) as range_from , max(seq) range_to , count(*) as cnt
from sometable
group by ceil(seq/3) * 3
with test as
(
select '2000,2002,2003,2004,2006,2007' str from dual
)
,test1 as (
select
split1,
lead(split1, 1, null) over (order by split1 asc) lead_no,
level1
from
(
select to_number(regexp_substr (str, '[^,]+', 1, rownum)) split1, level as level1
from test b
connect by level <= length (regexp_replace (str, '[^,]+')) + 1
)x
)
--select * from test1
,test2 (split1, lead_no, level1, op, op1) as(
select
split1,
lead_no,
level1,
(case when split1+1=lead_no then to_char(split1) else NULL end),
(case when split1+1=lead_no then NULL else to_char(split1) end)
from test1
where level1=1
union all
select
a.split1,
a.lead_no,
b.level1+1,
(case when a.split1+1=a.lead_no and to_char(b.op) is not null then to_char(b.op)
when a.split1+1=a.lead_no and to_char(b.op) is null then to_char(a.split1)
else null end),
(case when (a.split1+1<>a.lead_no and to_char(b.op)<>to_char(a.split1)) OR
(a.lead_no is null and to_char(b.op) is not null) then to_char(b.op) ||'-'||to_char(a.split1)
when a.lead_no is null then to_char(a.split1)
else null end)
from test1 a inner join test2 b on a.level1 = b.level1+1
)
select op1 from test2
where op1 is not null
您需要以表格或逗号分隔字符串的形式显示结果？输出字符串中是真实的2000-2003需求，还是错误的，必须是2000-2002？@techdo:逗号分隔字符串。请参考以下各种场景的示例：I/P 2000、2002、2006、2007、2009-O/P应为2000、2002、2006-2007和2009 I/P 2000、2002，2006年、2007年、2009年、2010年、2011年-O/P应为2000年、2002年、2006年-2007年和2009年-2011年I/P-2000年、2007年、2011年-O/P应为2000年、2007年和2011年-您何时提交和提交？在最后一组之前？我真的不知道PL/SQL，所以如果有一些错误，很抱歉。。。就靠打字吧！请你解释一下否决票，帮我改进我的答案。。。还有我的知识！作为参考，我犯了一个错误。我在康考特治疗中一年一年地换了一行。谢谢你的回复。我执行了这个脚本。第一次工作。下一次就不行了。显示06:36:08.423 DBMS ORA-01722:无效error@bhuvana，胡猜，值之间可能有一个空格，而不仅仅是，。我更新了我的答案我试过这个密码。但当我输入'200020022003200420062007'时，它缺少2000。我希望输出为20002002-20042006-2007您能解释一下逻辑吗？我有点理解。你能告诉我你在哪里发现理解困难吗？我试过这个代码。但当我输入'200020022003200420062007'时，它缺少2000。我预计产量为20002002-20042006-2007
with test as
(
select '2000,2002,2003,2004,2006,2007' str from dual
)
,test1 as (
select
split1,
lead(split1, 1, null) over (order by split1 asc) lead_no,
level1
from
(
select to_number(regexp_substr (str, '[^,]+', 1, rownum)) split1, level as level1
from test b
connect by level <= length (regexp_replace (str, '[^,]+')) + 1
)x
)
--select * from test1
,test2 (split1, lead_no, level1, op, op1) as(
select
split1,
lead_no,
level1,
(case when split1+1=lead_no then to_char(split1) else NULL end),
(case when split1+1=lead_no then NULL else to_char(split1) end)
from test1
where level1=1
union all
select
a.split1,
a.lead_no,
b.level1+1,
(case when a.split1+1=a.lead_no and to_char(b.op) is not null then to_char(b.op)
when a.split1+1=a.lead_no and to_char(b.op) is null then to_char(a.split1)
else null end),
(case when (a.split1+1<>a.lead_no and to_char(b.op)<>to_char(a.split1)) OR
(a.lead_no is null and to_char(b.op) is not null) then to_char(b.op) ||'-'||to_char(a.split1)
when a.lead_no is null then to_char(a.split1)
else null end)
from test1 a inner join test2 b on a.level1 = b.level1+1
)
select op1 from test2
where op1 is not null
18:42:15 SYSTEM@dwal> l
1 with p as (
2 select replace('2000,2001,2002,2005,2006,2007 and 2010',' and ',',') s, '[0-9]{4}' r from dual
3 ), ex as (
4 select regexp_substr(s,r, 1, level) as y
5 from p
6 connect by level <= regexp_count(s, r)
7 ), grp as (
8 select connect_by_root(y) s, y
9 from ( select e1.y y, e2.y p from ex e1, ex e2 where e1.y - 1 = e2.y(+) )
10 connect by prior y = p
11 start with p is null
12 ), agg as (
13 select listagg(s||decode(max(y), s, null, '-'||max(y)), ',') within group (order by s) str
14 from grp group by s
15 )
16* select regexp_replace(str, ',', ' and ', 1, regexp_count(str, ',')) result from agg
18:42:16 SYSTEM@dwal> /
RESULT
------------------------------
2000-2002,2005-2007 and 2010
Elapsed: 00:00:00.02