Question

我有一个用逗号隔开多年的字符串。

例如2000,2001,2002,2005,2006,2007 and 2010。

我想将连续数字分组。

我的输出应为2000-2003,2005-2007 and 2010。有没有办法在Oracle存储过程中执行此操作？

Answer 1

免责声明 - 我不建议“按原样”使用此解决方案，但它可以提供创意，编写它很有趣

我假设您在表中有一个包含csv字符串的列。

如果您正在使用oracle 11gR2，那么您可以使用递归CTE- Here is a sqlfiddle demo

with t as 
(
  select replace(replace(v, ' and ', ','), ' ','') v
  from strings
  ),
rcte(text, token, res) as
(
  select v, regexp_substr(v, '^\d*[^,]'), regexp_substr(v, '^\d*[^,]') ||'-'
  from t
  union all
  select regexp_replace(text, '^\d*,', ''), 
         regexp_substr(text, '^\d*[^,]'),
         case when regexp_substr(text, '^\d*[^,]') = token then
                   res
              when regexp_substr(text, '^\d*[^,]') = token+1 then  
                   regexp_replace(res, '-\d*$', '-'||(token+1))
              else rtrim(res, '-') || ',' || regexp_substr(text, '^\d*[^,]') || '-' 
         end
  from rcte
  where text <> token
  )
select rtrim(res, '-') from rcte
where text = regexp_substr(rtrim(res, '-'), '\d*$');

（这可以在没有正则表达式的情况下完成）

Answer 2

可以使用分析函数通过SQL完成。

* 更新1：* 使用解析器功能更新了答案，错过了以前的版本。

* 更新2：* 添加了最终字符串组合

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

可以找到SQL小提琴here。

Answer 3

假设您的数据将以逗号分隔并输入，您可以提及以逗号分隔的任何年份。

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;

Answer 4

使用数学

select min(seq) as range_from , max(seq) range_to , count(*) as cnt 
from   sometable 
group by ceil(seq/3) * 3

此部分ceil(seq/3) * 3将数字四舍五入为最接近的三的倍数。如果你想要5的范围，只需使用ceil(seq/5) * 5。干杯！

Answer 5

一个很棒的问题！ 请检查我的逻辑......

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

Answer 6

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

Answer 7

您必须使用cursor来循环使用这些年，并执行以下操作：

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;

如何在oracle中对序列号进行分组？

7 个答案: