SQL 需要从列中返回范围

Posted 2023-04-12

【中文标题】SQL 需要从列中返回范围

【英文标题】：SQL need to return ranges from a column

【发布时间】：2011-11-28 18:42:02

【问题描述】：

我有一个表，其中有一个名为 ID 的整数列，其中的值可能有间隙（例如 1、2、3、4、7、8、10、14、15、16、20）

我想找到一个查询，在上面的示例中，结果如下：

1-4 7-8 10 14-16 20

= 更新 =

感谢下面的代码（它似乎在 SQL Server 中运行良好），我觉得我非常接近让它在 MS-Access 中运行，这是目标。虽然在我的陈述中我仍然遇到语法错误，但我无法弄清楚......

选择值从 (

SELECT islands.PORTID, CSTR(islands.PORTID ) as val 
FROM MYTABLE islands
WHERE NOT EXISTS (SELECT * FROM MYTABLE t2 WHERE t2.PORTID = islands.PORTID - 1)
AND NOT EXISTS (SELECT * FROM MYTABLE t2 WHERE t2.PORTID = islands.PORTID + 1)

UNION 


SELECT 
    rngStart.PORTID
    ,CSTR(rngStart.PORTID ) + '-' 
        + CSTR(MIN(rngEnd.PORTID)) as val
FROM MYTABLE rngStart 
INNER JOIN MYTABLE checkNext ON checkNext.PORTID = rngStart.PORTID + 1
INNER JOIN
(
    SELECT PORTID 
    FROM MYTABLE tblRangeEnd 
    WHERE NOT EXISTS (SELECT * FROM MYTABLE t2 WHERE t2.PORTID = tblRangeEnd.PORTID + 1)
) rngEnd on rngEnd.PORTID > rngStart.PORTID 
WHERE NOT EXISTS (SELECT * FROM MYTABLE t2 WHERE t2.PORTID = rngStart.PORTID - 1)
GROUP BY rngStart.PORTID

) 作为 tbl 按 PORTID ASC 排序

【问题讨论】：

Sql 服务器？ PostgreSQL？甲骨文？语法可能会有所不同，因此了解正确的语法会很有用。

最终我需要能够支持 SQL Server、Oracle 和 Access。不幸的是，MS-Access 是最重要的。

@user1069916：请参阅我的编辑以了解如何在 Access 中进行这项工作。

SQL Query for Grouping the results based on sequence的可能重复

【参考方案1】：

SELECT val FROM
(
    -- Get the islands
    SELECT islands.ID, CAST(islands.ID as varchar(10)) as val 
    FROM @t1 islands
    WHERE NOT EXISTS (SELECT * FROM @t1 t2 WHERE t2.ID = islands.ID - 1)
    AND NOT EXISTS (SELECT * FROM @t1 t2 WHERE t2.ID = islands.ID + 1)

    UNION 
    
    -- Get the ranges
    SELECT 
        rngStart.ID
        ,CAST(rngStart.ID as varchar(10)) + '-' 
            + CAST(MIN(rngEnd.ID) as varchar(10)) as val
    FROM @t1 rngStart 
    INNER JOIN @t1 checkNext ON checkNext.ID = rngStart.ID + 1
    INNER JOIN
    (
        SELECT ID 
        FROM @t1 tblRangeEnd 
        WHERE NOT EXISTS (SELECT * FROM @t1 t2 WHERE t2.ID = tblRangeEnd.ID + 1)
    ) rngEnd on rngEnd.ID > rngStart.ID 
    WHERE NOT EXISTS (SELECT * FROM @t1 t2 WHERE t2.ID = rngStart.ID - 1)
    GROUP BY rngStart.ID
) as tbl
ORDER BY ID ASC

我使用了一个名为 @t1 的表变量，但只需将其替换为您的表名即可。

Here 正在行动中。

编辑

要在 Access 中进行这项工作，您必须稍微更改联接。试试这个：

SELECT val FROM
(
    SELECT islands.PORTID, CSTR(islands.PORTID) as val 
    FROM MYTABLE islands
    WHERE NOT EXISTS 
        (SELECT * FROM MYTABLE t2 WHERE t2.PORTID = islands.PORTID - 1)
    AND NOT EXISTS 
        (SELECT * FROM MYTABLE t2 WHERE t2.PORTID = islands.PORTID + 1)

    UNION 

    SELECT 
         rngStart.PORTID
        ,CSTR(rngStart.PORTID) + '-' + CSTR(MIN(endPORTID)) as val
    FROM MYTABLE rngStart 
    INNER JOIN 
    (
        SELECT checkNext.PORTID as nextPORTID, rngEnd.PORTID as endPORTID
        FROM MYTABLE checkNext
        INNER JOIN
        (
            SELECT rngEnd.PORTID
            FROM MYTABLE rngEnd 
            WHERE NOT EXISTS 
                (SELECT * FROM MYTABLE t2 WHERE t2.PORTID = rngEnd.PORTID + 1)
        ) AS rngEnd on rngEnd.PORTID > checkNext.PORTID - 1
    ) AS checkNext ON checkNext.nextPORTID = rngStart.PORTID + 1
    WHERE NOT EXISTS 
        (SELECT * FROM MYTABLE t2 WHERE t2.PORTID = rngStart.PORTID - 1)
    GROUP BY rngStart.PORTID
) as tbl
ORDER BY PORTID ASC

【讨论】：

这简直太棒了。非常感谢。

【参考方案2】：

create table #sequence (id int not null primary KEY)

insert into #sequence(id)
    select 1
    union all select 2
    union all select 3
    union all select 4
    union all select 7
    union all select 8
    union all select 10
    union all select 14
    union all select 15
    union all select 16
    union all select 20


--Find Contig ranges
select l.id as start,
    (
        select min(a.id) as id
        from #sequence as a
            left outer join #sequence as b on a.id = b.id - 1
        where b.id is null
            and a.id >= l.id
    ) AS fend
from #sequence as l
    left outer join #sequence as r on r.id = l.id - 1
where r.id is null;

--Find missing values in sequence
select l.id + 1 as start, min(fr.id) - 1 as stop
from #sequence as l
    left outer join #sequence as r on l.id = r.id - 1
    left outer join #sequence as fr on l.id < fr.id
where r.id is null and fr.id is not null
group by l.id, r.id;

drop table #sequence

这将为您提供范围和范围之间的差距，以便您了解自己拥有什么或需要什么。提供的样本数据

【参考方案3】：

如果您访问函数row_number()，这将起作用。

with C as
(
  select ID,
         ID - row_number() over(order by ID) as grp
  from YourTable
)
select min(ID) as MinID,
       max(ID) as MaxID
from C
group by grp

或使用子查询代替公用表表达式。

select min(ID) as MinID,
       max(ID) as MaxID
from (select ID,
             ID - row_number() over(order by ID) as grp
      from YourTable) as C
group by grp

结果：

MinID       MaxID
----------- -----------
1           4
7           8
10          10
14          16
20          20

在 SQL Server 上试用https://data.stackexchange.com/***/q/119411/

【参考方案4】：

再试一次：

SELECT
    MIN(i), MAX(i)
FROM (
    select 
        i - (SELECT COUNT(*) FROM tbl t WHERE t.i < tbl.i) g,
        i 
    from tbl
) t
GROUP BY g

肯定会很慢，但我没有看到其他方法来对访问中的行进行编号。