Question

我想在SQL表的计数器列中找到第一个“缺口”。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中执行此操作。

此外，它应该是非常标准的SQL，使用不同的DBMS。

Answer 1

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

Answer 2

如果你的第一个值为id = 1，你的答案一切正常，否则将无法检测到这个差距。例如，如果您的表ID值为3,4,5，则查询将返回6.

我做了类似的事

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)

Answer 3

首先出现在我脑海中。不确定这样做是否一个好主意，但应该有效。假设该表为t且列为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

Answer 4

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年，2001年，2002年和2005年，2003年和2004年不存在，它将返回2003年。

Answer 5

实际上没有极标准的SQL方法，但是你可以使用某种形式的限制条款

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

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

（SQL Server）

或

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

（Oracle）的

Answer 6

内部联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。

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) ;

Answer 7

PostgreSQL

使用递归查询的示例。

如果您想在特定范围内找到差距，这可能很有用（即使表是空的，它也会起作用，而其他的例子则不会）

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

Answer 8

我的猜测：

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;

Answer 9

这个记录到目前为止所提到的一切。它包含0作为起始点，如果不存在任何值，它将默认为起始点。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

Answer 10

我写了一个快速的方法。不确定这是最有效的，但是可以完成工作。请注意，它不会告诉您间隙，而是告诉您间隙之前和之后的ID（请注意，间隙可以是多个值，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

Answer 11

这是一种以可移植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，您有以下查询：

子查询列出了成对的id，每个id的lowerbound列都小于upperbound列，然后使用GROUP BY和MIN(m2.id)来减少无用记录的数量.
外部查询进一步删除了 lowerbound 正好是 upperbound - 1 的记录
我的查询没有（显式地）在两端输出 2 条记录 (YOUR_MIN_ID_VALUE, 89) 和 (186, YOUR_MAX_ID_VALUE)，这隐式意味着 {{1} 中没有使用这两个范围中的任何数字} 到目前为止。

your_table

Answer 12

如果列中的数字是正整数（从1开始），那么这是如何轻松求解。（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

Answer 13

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]

Answer 14

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），那么以下是对Oracle有效的第一个答案中的经过纠正的代码段：

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
     )

Answer 15

以下解决方案：

提供测试数据；
内部查询会产生其他差距；和
它可用于SQL Server 2012。

在“ with ”子句中对有序行进行顺序编号，然后将结果重复使用两次，并在行号上进行内部联接，但偏移量为1，以便将之前的行与该行进行比较之后，寻找间隔大于1的ID。要求更多，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

Answer 16

发现大多数方法在mysql中运行非常慢，非常慢。这是我对mysql < 8.0的解决方案。在1M记录上进行了测试，间隔接近尾声〜1秒才能完成。不确定是否适合其他SQL版本。

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。

Answer 17

            -- 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

Answer 18

如果你使用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

Answer 19

它适用于空表或负值。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

Answer 20

这是标准的SQL解决方案，可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);

请参阅以下内容;

PL / SQL via Oracle's livesql，
MySQL通过sqlfiddle，
PostgreSQL通过sqlfiddle
MS Sql via sqlfiddle

Answer 21

import UIKit

class LoadingScreen: UIViewController {

    var time : Float = 0.0
    var timer: NSTimer?

    @IBOutlet weak var progressView: UIProgressView!

    override func viewDidLoad() {
        super.viewDidLoad()

        // Do stuff

        timer = NSTimer.scheduledTimerWithTimeInterval(0.1, target: self, selector:Selector("setProgress"), userInfo: nil, repeats: true)

    } //close viewDidLoad

    func setProgress() {
        time += 0.1
        progressView.progress = time / 3
        if time >= 3 {
            timer!.invalidate()
        }
    }
}

如何在运行SQL的计数器中找到“差距”？

21 个答案: