sql - 如何在使用 SQL 的运行计数器中找到“差距”？

Question

我想在 SQL 表的计数器列中找到第一个“间隙”。例如，如果有值 1、2、4 和 5，我想找出 3。

我当然可以按顺序获取值并手动完成它，但我想知道是否有办法在 SQL 中完成它。

此外，它应该是非常标准的 SQL，可以与不同的 DBMS 一起使用。

score 214 · Accepted Answer

在MySQL和PostgreSQL：

SELECT  id + 1
FROM    mytable mo
WHERE   NOT EXISTS
        (
        SELECT  NULL
        FROM    mytable mi 
        WHERE   mi.id = mo.id + 1
        )
ORDER BY
        id
LIMIT 1

在SQL Server：

SELECT  TOP 1
        id + 1
FROM    mytable mo
WHERE   NOT EXISTS
        (
        SELECT  NULL
        FROM    mytable mi 
        WHERE   mi.id = mo.id + 1
        )
ORDER BY
        id

在Oracle：

SELECT  *
FROM    (
        SELECT  id + 1 AS gap
        FROM    mytable mo
        WHERE   NOT EXISTS
                (
                SELECT  NULL
                FROM    mytable mi 
                WHERE   mi.id = mo.id + 1
                )
        ORDER BY
                id
        )
WHERE   rownum = 1

ANSI（无处不在，效率最低）：

SELECT  MIN(id) + 1
FROM    mytable mo
WHERE   NOT EXISTS
        (
        SELECT  NULL
        FROM    mytable mi 
        WHERE   mi.id = mo.id + 1
        )

支持滑动窗口功能的系统：

SELECT  -- TOP 1
        -- Uncomment above for SQL Server 2012+
        previd
FROM    (
        SELECT  id,
                LAG(id) OVER (ORDER BY id) previd
        FROM    mytable
        ) q
WHERE   previd <> id - 1
ORDER BY
        id
-- LIMIT 1
-- Uncomment above for PostgreSQL

score 14 · Accepted Answer

如果您的第一个值 id = 1，则您的答案都可以正常工作，否则将无法检测到此差距。例如，如果您的表 id 值为 3、4、5，您的查询将返回 6。

我做了这样的事情

SELECT MIN(ID+1) FROM (
    SELECT 0 AS ID UNION ALL 
    SELECT  
        MIN(ID + 1)
    FROM    
        TableX) AS T1
WHERE
    ID+1 NOT IN (SELECT ID FROM TableX)

score 10 · Accepted Answer

实际上并没有一种非常标准的 SQL 方法来执行此操作，但是您可以使用某种形式的限制子句

SELECT `table`.`num` + 1
FROM `table`
LEFT JOIN `table` AS `alt`
ON `alt`.`num` = `table`.`num` + 1
WHERE `alt`.`num` IS NULL
LIMIT 1

（MySQL、PostgreSQL）

或者

SELECT TOP 1 `num` + 1
FROM `table`
LEFT JOIN `table` AS `alt`
ON `alt`.`num` = `table`.`num` + 1
WHERE `alt`.`num` IS NULL

（SQL 服务器）

或者

SELECT `num` + 1
FROM `table`
LEFT JOIN `table` AS `alt`
ON `alt`.`num` = `table`.`num` + 1
WHERE `alt`.`num` IS NULL
AND ROWNUM = 1

（甲骨文）

score 9 · Accepted Answer

第一个出现在我脑海中的东西。不确定这样做是否是个好主意，但应该可以。假设表是t，列是c：

SELECT 
    t1.c + 1 AS gap 
FROM t as t1 
LEFT OUTER JOIN t as t2 ON (t1.c + 1 = t2.c)
WHERE t2.c IS NULL 
ORDER BY gap ASC 
LIMIT 1

编辑：这个可能会更快（更短！）：

SELECT 
    min(t1.c) + 1 AS gap 
FROM t as t1 
LEFT OUTER JOIN t as t2 ON (t1.c + 1 = t2.c) 
WHERE t2.c IS NULL

score 6 · Accepted Answer

这适用于 SQL Server - 无法在其他系统中测试它，但它似乎是标准的......

SELECT MIN(t1.ID)+1 FROM mytable t1 WHERE NOT EXISTS (SELECT ID FROM mytable WHERE ID = (t1.ID + 1))

您还可以在 where 子句中添加一个起点...

SELECT MIN(t1.ID)+1 FROM mytable t1 WHERE NOT EXISTS (SELECT ID FROM mytable WHERE ID = (t1.ID + 1)) AND ID > 2000

因此，如果您有 2000、2001、2002 和 2005，而 2003 和 2004 不存在，它将返回 2003。

score 4 · Accepted Answer

以下解决方案：

提供测试数据；
产生其他间隙的内部查询；和
它适用于 SQL Server 2012。

在“ with ”子句中按顺序对有序行进行编号，然后在行号上通过内连接重复使用结果两次，但偏移1，以便比较前行和后行，寻找间隙大于的ID 1. 超出要求，但适用范围更广。

create table #ID ( id integer );

insert into #ID values (1),(2),    (4),(5),(6),(7),(8),    (12),(13),(14),(15);

with Source as (
    select
         row_number()over ( order by A.id ) as seq
        ,A.id                               as id
    from #ID as A WITH(NOLOCK)
)
Select top 1 gap_start from (
    Select 
         (J.id+1) as gap_start
        ,(K.id-1) as gap_end
    from       Source as J
    inner join Source as K
    on (J.seq+1) = K.seq
    where (J.id - (K.id-1)) <> 0
) as G

内部查询产生：

gap_start   gap_end

3           3

9           11

外部查询产生：

gap_start

3

score 2 · Accepted Answer

内连接到具有所有可能值的视图或序列。

没有桌子？做一张桌子。我总是为此保留一张假桌子。

create table artificial_range( 
  id int not null primary key auto_increment, 
  name varchar( 20 ) null ) ;

-- or whatever your database requires for an auto increment column

insert into artificial_range( name ) values ( null )
-- create one row.

insert into artificial_range( name ) select name from artificial_range;
-- you now have two rows

insert into artificial_range( name ) select name from artificial_range;
-- you now have four rows

insert into artificial_range( name ) select name from artificial_range;
-- you now have eight rows

--etc.

insert into artificial_range( name ) select name from artificial_range;
-- you now have 1024 rows, with ids 1-1024

然后，

 select a.id from artificial_range a
 where not exists ( select * from your_table b
 where b.counter = a.id) ;

score 2 · Accepted Answer

这个解释了到目前为止提到的所有内容。它包括 0 作为起点，如果不存在任何值，它将默认为起点。我还为多值键的其他部分添加了适当的位置。这仅在 SQL Server 上进行了测试。

select
    MIN(ID)
from (
    select
        0 ID
    union all
    select
        [YourIdColumn]+1
    from
        [YourTable]
    where
        --Filter the rest of your key--
    ) foo
left join
    [YourTable]
    on [YourIdColumn]=ID
    and --Filter the rest of your key--
where
    [YourIdColumn] is null

score 2 · Accepted Answer

为了PostgreSQL

使用递归查询的示例。

如果您想在特定范围内查找间隙，这可能很有用（即使表为空，它也可以工作，而其他示例则不会）

WITH    
    RECURSIVE a(id) AS (VALUES (1) UNION ALL SELECT id + 1 FROM a WHERE id < 100), -- range 1..100  
    b AS (SELECT id FROM my_table) -- your table ID list    
SELECT a.id -- find numbers from the range that do not exist in main table
FROM a
LEFT JOIN b ON b.id = a.id
WHERE b.id IS NULL
-- LIMIT 1 -- uncomment if only the first value is needed

score 1 · Accepted Answer

我猜：

SELECT MIN(p1.field) + 1 as gap
FROM table1 AS p1  
INNER JOIN table1 as p3 ON (p1.field = p3.field + 2)
LEFT OUTER JOIN table1 AS p2 ON (p1.field = p2.field + 1)
WHERE p2.field is null;

score 1 · Accepted Answer

我写了一个快速的方法。不确定这是最有效的，但可以完成工作。请注意，它不会告诉您间隙，而是告诉您间隙前后的 id（请记住，间隙可能是多个值，例如 1、2、4、7、11 等）

我以 sqlite 为例

如果这是您的表结构

create table sequential(id int not null, name varchar(10) null);

这些是你的行

id|name
1|one
2|two
4|four
5|five
9|nine

查询是

select a.* from sequential a left join sequential b on a.id = b.id + 1 where b.id is null and a.id <> (select min(id) from sequential)
union
select a.* from sequential a left join sequential b on a.id = b.id - 1 where b.id is null and a.id <> (select max(id) from sequential);

https://gist.github.com/wkimeria/7787ffe84d1c54216f1b320996b17b7e

score 0 · Accepted Answer

select min([ColumnName]) from [TableName]
where [ColumnName]-1 not in (select [ColumnName] from [TableName])
and [ColumnName] <> (select min([ColumnName]) from [TableName])

score 0 · Accepted Answer

这是一个标准的 SQL 解决方案，可以在所有数据库服务器上运行而无需更改：

select min(counter + 1) FIRST_GAP
    from my_table a
    where not exists (select 'x' from my_table b where b.counter = a.counter + 1)
        and a.counter <> (select max(c.counter) from my_table c);

见行动；

通过Oracle 的 livesql进行 PL/SQL ，
MySQL 通过sqlfiddle，
PostgreSQL 通过sqlfiddle
MS Sql 通过sqlfiddle

score 0 · Accepted Answer

它也适用于空表或负值。刚刚在 SQL Server 2012 中测试过

 select min(n) from (
select  case when lead(i,1,0) over(order by i)>i+1 then i+1 else null end n from MyTable) w

score 0 · Accepted Answer

如果您使用 Firebird 3，这是最优雅和简单的：

select RowID
  from (
    select `ID_Column`, Row_Number() over(order by `ID_Column`) as RowID
      from `Your_Table`
        order by `ID_Column`)
    where `ID_Column` <> RowID
    rows 1

score 0 · Accepted Answer

            -- PUT THE TABLE NAME AND COLUMN NAME BELOW
            -- IN MY EXAMPLE, THE TABLE NAME IS = SHOW_GAPS AND COLUMN NAME IS = ID

            -- PUT THESE TWO VALUES AND EXECUTE THE QUERY

            DECLARE @TABLE_NAME VARCHAR(100) = 'SHOW_GAPS'
            DECLARE @COLUMN_NAME VARCHAR(100) = 'ID'


            DECLARE @SQL VARCHAR(MAX)
            SET @SQL = 
            'SELECT  TOP 1
                    '+@COLUMN_NAME+' + 1
            FROM    '+@TABLE_NAME+' mo
            WHERE   NOT EXISTS
                    (
                    SELECT  NULL
                    FROM    '+@TABLE_NAME+' mi 
                    WHERE   mi.'+@COLUMN_NAME+' = mo.'+@COLUMN_NAME+' + 1
                    )
            ORDER BY
                    '+@COLUMN_NAME

            -- SELECT @SQL

            DECLARE @MISSING_ID TABLE (ID INT)

            INSERT INTO @MISSING_ID
            EXEC (@SQL)

            --select * from @MISSING_ID

            declare @var_for_cursor int
            DECLARE @LOW INT
            DECLARE @HIGH INT
            DECLARE @FINAL_RANGE TABLE (LOWER_MISSING_RANGE INT, HIGHER_MISSING_RANGE INT)
            DECLARE IdentityGapCursor CURSOR FOR   
            select * from @MISSING_ID
            ORDER BY 1;  

            open IdentityGapCursor

            fetch next from IdentityGapCursor
            into @var_for_cursor

            WHILE @@FETCH_STATUS = 0  
            BEGIN
            SET @SQL = '
            DECLARE @LOW INT
            SELECT @LOW = MAX('+@COLUMN_NAME+') + 1 FROM '+@TABLE_NAME
                    +' WHERE '+@COLUMN_NAME+' < ' + cast( @var_for_cursor as VARCHAR(MAX))

            SET @SQL = @sql + '
            DECLARE @HIGH INT
            SELECT @HIGH = MIN('+@COLUMN_NAME+') - 1 FROM '+@TABLE_NAME
                    +' WHERE '+@COLUMN_NAME+' > ' + cast( @var_for_cursor as VARCHAR(MAX))

            SET @SQL = @sql + 'SELECT @LOW,@HIGH'

            INSERT INTO @FINAL_RANGE
             EXEC( @SQL)
            fetch next from IdentityGapCursor
            into @var_for_cursor
            END

            CLOSE IdentityGapCursor;  
            DEALLOCATE IdentityGapCursor;  

            SELECT ROW_NUMBER() OVER(ORDER BY LOWER_MISSING_RANGE) AS 'Gap Number',* FROM @FINAL_RANGE

score 0 · Accepted Answer

发现大多数方法在mysql. 这是我的解决方案mysql < 8.0。在 1M 条记录上进行测试，在接近尾声时有一个间隙 ~ 1 秒完成。不确定它是否适合其他 SQL 风格。

SELECT cardNumber - 1
FROM
    (SELECT @row_number := 0) as t,
    (
        SELECT (@row_number:=@row_number+1), cardNumber, cardNumber-@row_number AS diff
        FROM cards
        ORDER BY cardNumber
    ) as x
WHERE diff >= 1
LIMIT 0,1

我假设序列从“1”开始。

score 0 · Accepted Answer

如果您的计数器从 1 开始，并且您想在为空时生成序列的第一个数字 (1)，则以下是对 Oracle 有效的第一个答案的更正代码：

SELECT
  NVL(MIN(id + 1),1) AS gap
FROM
  mytable mo  
WHERE 1=1
  AND NOT EXISTS
      (
       SELECT  NULL
       FROM    mytable mi 
       WHERE   mi.id = mo.id + 1
      )
  AND EXISTS
     (
       SELECT  NULL
       FROM    mytable mi 
       WHERE   mi.id = 1
     )

score 0 · Accepted Answer

DECLARE @Table AS TABLE(
[Value] int
)

INSERT INTO @Table ([Value])
VALUES
 (1),(2),(4),(5),(6),(10),(20),(21),(22),(50),(51),(52),(53),(54),(55)
 --Gaps
 --Start    End     Size
 --3        3       1
 --7        9       3
 --11       19      9
 --23       49      27


SELECT [startTable].[Value]+1 [Start]
     ,[EndTable].[Value]-1 [End]
     ,([EndTable].[Value]-1) - ([startTable].[Value]) Size 
 FROM 
    (
SELECT [Value]
    ,ROW_NUMBER() OVER(PARTITION BY 1 ORDER BY [Value]) Record
FROM @Table
)AS startTable
JOIN 
(
SELECT [Value]
,ROW_NUMBER() OVER(PARTITION BY 1 ORDER BY [Value]) Record
FROM @Table
)AS EndTable
ON [EndTable].Record = [startTable].Record+1
WHERE [startTable].[Value]+1 <>[EndTable].[Value]

score 0 · Accepted Answer

如果列中的数字是正整数（从 1 开始），那么这里是如何轻松解决它。（假设 ID 是您的列名）

    SELECT TEMP.ID 
    FROM (SELECT ROW_NUMBER() OVER () AS NUM FROM 'TABLE-NAME') AS TEMP 
    WHERE ID NOT IN (SELECT ID FROM 'TABLE-NAME')
    ORDER BY 1 ASC LIMIT 1

score 0 · Accepted Answer

这是一种以便携和更紧凑的方式显示所有可能间隙值范围的替代方法：

假设您的表架构如下所示：

> SELECT id FROM your_table;
+-----+
| id  |
+-----+
|  90 |
| 103 |
| 104 |
| 118 |
| 119 |
| 120 |
| 121 |
| 161 |
| 162 |
| 163 |
| 185 |
+-----+

要获取所有可能的间隙值的范围，您可以使用以下查询：

子查询列出了成对的id，每对id的lowerboundcolumn小于upperboundcolumn，然后使用GROUP BYandMIN(m2.id)来减少无用记录的数量。
外部查询进一步删除了lowerbound确切位置的记录upperbound - 1
我的查询没有（明确地）输出 2 条记录(YOUR_MIN_ID_VALUE, 89)，并且在两端，这隐含地意味着到目前为止(186, YOUR_MAX_ID_VALUE)尚未使用两个范围中的任何数字。your_table

> SELECT  m3.lowerbound + 1, m3.upperbound - 1 FROM
  (
    SELECT m1.id as lowerbound, MIN(m2.id) as upperbound FROM
    your_table m1 INNER JOIN your_table
    AS m2 ON m1.id < m2.id GROUP BY m1.id
  )
  m3 WHERE m3.lowerbound < m3.upperbound - 1;

+-------------------+-------------------+
| m3.lowerbound + 1 | m3.upperbound - 1 |
+-------------------+-------------------+
|                91 |               102 |
|               105 |               117 |
|               122 |               160 |
|               164 |               184 |
+-------------------+-------------------+

sql - 如何在使用 SQL 的运行计数器中找到“差距”？

21 回答 21

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