Question

我有一些像这样的表

row chequeNo
 1     15
 2     19
 3     20
 4     35
 5     16

我需要得到像这样的结果

row  from   to    
 1    15    16     
 2    19    20    
 3    35    35

所以我需要chequeNo组，其中值将是连续的而没有任何中断。 chequeNo是唯一的专栏。另外，它应该使用一个sql select查询，因为我没有权限创建除select查询之外的任何sql结构。

有可能吗？

非常感谢任何帮助

Answer 1

你可以在这里使用Aketi Jyuuzou的技术Tabibitosan：

SQL> create table mytable (id,chequeno)
  2  as
  3  select 1, 15 from dual union all
  4  select 2, 19 from dual union all
  5  select 3, 20 from dual union all
  6  select 4, 35 from dual union all
  7  select 5, 16 from dual
  8  /

Table created.

SQL> with tabibitosan as
  2  ( select chequeno
  3         , chequeno - row_number() over (order by chequeno) grp
  4      from mytable
  5  )
  6  select row_number() over (order by grp) "row"
  7       , min(chequeno) "from"
  8       , max(chequeno) "to"
  9    from tabibitosan
 10   group by grp
 11  /

       row       from         to
---------- ---------- ----------
         1         15         16
         2         19         20
         3         35         35

3 rows selected.

的问候，
罗布。

Answer 2

这应该适用于Oracle 10（仅使用Oracle 11测试）

select group_nr + 1,
       min(chequeno) as start_value,
       max(chequeno) as end_value
from (
  select chequeno,
         sum(group_change_flag) over (order by rn) as group_nr
  from (
    select row_number() over (order by chequeno) as rn,
           chequeno, 
           case 
             when chequeno - lag(chequeno,1,chequeno) over (order by chequeno) <= 1 then 0 
             else 1
           end as group_change_flag
    from foo
  ) t1
) t2
group by group_nr
order by group_nr

（它应该适用于任何支持标准SQL窗口函数的DBMS，例如PostgreSQL，DB2，SQL Server 2012）

Answer 3

这是一种“普通的香草”方法：

SELECT T1.chequeNo, T2.chequeNo
FROM Table1 AS T1 INNER JOIN Table1 AS T2 ON T2.chequeNo >= T1.chequeNo
WHERE
NOT EXISTS (SELECT T0.chequeNo FROM Table1 T0 WHERE T0.chequeNo IN ((T1.chequeNo-1), (T2.chequeNo+1)))
AND (SELECT COUNT(*) FROM Table1 T0 WHERE T0.chequeNo BETWEEN T1.chequeNo AND T2.chequeNo)=(T2.chequeNo - T1.chequeNo + 1)
ORDER BY 1,2

如果对于大型数据集效率太低，请告诉我。

Answer 4

CREATE TABLE YOUR_TABLE (
    chequeNo NUMBER PRIMARY KEY
);

INSERT INTO YOUR_TABLE VALUES (15);
INSERT INTO YOUR_TABLE VALUES (19);
INSERT INTO YOUR_TABLE VALUES (20);
INSERT INTO YOUR_TABLE VALUES (35);
INSERT INTO YOUR_TABLE VALUES (16);

SELECT T1.chequeNo "from", T2.chequeNo "to"
FROM
    (
        SELECT chequeNo, ROW_NUMBER() OVER (ORDER BY chequeNo) RN
        FROM (
            SELECT chequeNo, LAG(chequeNo) OVER (ORDER BY chequeNo) PREV
            FROM YOUR_TABLE
        )
        WHERE PREV IS NULL OR chequeNo > PREV + 1
    ) T1
    JOIN
    (
        SELECT chequeNo, ROW_NUMBER() OVER (ORDER BY chequeNo) RN
        FROM (
            SELECT chequeNo, LEAD(chequeNo) OVER (ORDER BY chequeNo) NEXT
            FROM YOUR_TABLE
        )
        WHERE NEXT IS NULL OR chequeNo < NEXT - 1
    ) T2
    USING (RN);

结果：

from                   to                     
---------------------- ---------------------- 
15                     16                     
19                     20                     
35                     35

如果我们稍微调整一下......

INSERT INTO YOUR_TABLE VALUES (17);
INSERT INTO YOUR_TABLE VALUES (18);

......我们得到：

from                   to                     
---------------------- ---------------------- 
15                     20                     
35                     35

按顺序分组值

4 个答案: