I'd like to find the first "gap" in a counter column in an SQL table. For example, if there are values 1,2,4 and 5 I'd like to find out 3.
I can of course get the values in order and go through it manually, but I'd like to know if there would be a way to do it in SQL.
In addition, it should be quite standard SQL, working with different DBMSes.
In MySQL and 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
In 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
In 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 (works everywhere, least efficient):
SELECT MIN(id) + 1
FROM mytable mo
WHERE NOT EXISTS
(
SELECT NULL
FROM mytable mi
WHERE mi.id = mo.id + 1
)
Systems supporting sliding window functions:
SELECT -- TOP 1
-- Uncomment above for SQL Server 2012+
previd + 1
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
URL as well, though it may be QR coded I think. –
Lesson [1, 2, 11, 12], then this would find only 3. What I'd love it to find is 3-10 instead - basically the beginning and the end of every gap. I understand that I might have to write my own python script that leverages SQL (in my case MySql), but it would be nice if SQL could get me closer to what I want (I have a table with 2 million rows that has gaps, so I will need to slice it into smaller pieces and run some SQL on it). I suppose I could run one query to find the start of a gap, then another to find the end of a gap, and them "merge sort" the two sequences. –
Fervid NULL, not 0, if the table is empty. This is true for all databases. –
Lesson where clause and also why the naming choice of mo and mi? I'm guessing o is for original but why mi? –
Mohun o and i stand for "outer" and "inner", this might not be the obvious choice but it works for me. The query basically says "look at all the records in id order and for each of them, see if the record with the next consecutive id exists. This means: for id = 1 check if id = 2 exists, for id = 2 check if id = 3 exists and so on. As soon as you find a record for which the next consecutive id does not exist, return its id plus one". For op's data that would mean "id = 2 is the first record for which id = 3 does not exist, so let's return 2 + 1 = 3". –
Lesson SELECT [TOP 1] previd + 1 .... (missing the +1) to be consistent with the other solutions, right? –
Tameshatamez Your answers all work fine if you have a first value id = 1, otherwise this gap will not be detected. For instance if your table id values are 3,4,5, your queries will return 6.
I did something like this
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)
The first thing that came into my head. Not sure if it's a good idea to go this way at all, but should work. Suppose the table is t and the column is 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
Edit: This one may be a tick faster (and shorter!):
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
LEFT OUTER JOING t2 would require you to have t2 table, which is just an alias. –
Dickey There isn't really an extremely standard SQL way to do this, but with some form of limiting clause you can do
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)
or
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)
or
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)
This works in SQL Server - can't test it in other systems but it seems standard...
SELECT MIN(t1.ID)+1 FROM mytable t1 WHERE NOT EXISTS (SELECT ID FROM mytable WHERE ID = (t1.ID + 1))
You could also add a starting point to the where clause...
SELECT MIN(t1.ID)+1 FROM mytable t1 WHERE NOT EXISTS (SELECT ID FROM mytable WHERE ID = (t1.ID + 1)) AND ID > 2000
So if you had 2000, 2001, 2002, and 2005 where 2003 and 2004 didn't exist, it would return 2003.
The following solution:
Numbers the ordered rows sequentially in the "with" clause and then reuses the result twice with an inner join on the row number, but offset by 1 so as to compare the row before with the row after, looking for IDs with a gap greater than 1. More than asked for but more widely applicable.
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
The inner query produces:
gap_start gap_end
3 3
9 11
The outer query produces:
gap_start
3
Inner join to a view or sequence that has a all possible values.
No table? Make a table. I always keep a dummy table around just for this.
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
Then,
select a.id from artificial_range a
where not exists ( select * from your_table b
where b.counter = a.id) ;
This one accounts for everything mentioned so far. It includes 0 as a starting point, which it will default to if no values exist as well. I also added the appropriate locations for the other parts of a multi-value key. This has only been tested on SQL Server.
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
For PostgreSQL
An example that makes use of recursive query.
This might be useful if you want to find a gap in a specific range (it will work even if the table is empty, whereas the other examples will not)
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
My guess:
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;
I wrote up a quick way of doing it. Not sure this is the most efficient, but gets the job done. Note that it does not tell you the gap, but tells you the id before and after the gap (keep in mind the gap could be multiple values, so for example 1,2,4,7,11 etc)
I'm using sqlite as an example
If this is your table structure
create table sequential(id int not null, name varchar(10) null);
and these are your rows
id|name
1|one
2|two
4|four
5|five
9|nine
The query is
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
Here is an alternative to show the range of all possible gap values in portable and more compact way :
Assume your table schema looks like this :
> SELECT id FROM your_table;
+-----+
| id |
+-----+
| 90 |
| 103 |
| 104 |
| 118 |
| 119 |
| 120 |
| 121 |
| 161 |
| 162 |
| 163 |
| 185 |
+-----+
To fetch the ranges of all possible gap values, you have the following query :
lowerbound column being smaller than upperbound column, then use GROUP BY and MIN(m2.id) to reduce number of useless records.lowerbound is exactly upperbound - 1(YOUR_MIN_ID_VALUE, 89) and (186, YOUR_MAX_ID_VALUE) at both ends, that implicitly means any number in both of the ranges hasn't been used in your_table so far.> 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 |
+-------------------+-------------------+
select min([ColumnName]) from [TableName]
where [ColumnName]-1 not in (select [ColumnName] from [TableName])
and [ColumnName] <> (select min([ColumnName]) from [TableName])
Here is standard a SQL solution that runs on all database servers with no change:
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);
See in action for;
It works for empty tables or with negatives values as well. Just tested in SQL Server 2012
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
If You use Firebird 3 this is most elegant and simple:
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
-- 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
Found most of approaches run very, very slow in mysql. Here is my solution for mysql < 8.0. Tested on 1M records with a gap near the end ~ 1sec to finish. Not sure if it fits other SQL flavours.
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
If your counter is starting from 1 and you want to generate first number of sequence (1) when empty, here is the corrected piece of code from first answer valid for Oracle:
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
)
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]
If the numbers in the column are positive integers (starting from 1) then here is how to solve it easily. (assuming ID is your column name)
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
SELECT ID+1 FROM table WHERE ID+1 NOT IN (SELECT ID FROM table) ORDER BY 1;
try this one:
create table #numbers ( id int,num int );
insert into #numbers values (1,1),(2,2), (3,4),(4,5),(5,5),(6,6),(7,10), (8,15);
select
d1.id start_gap_id,
d2.id end_gap_id,
d1.num start_gap_num,
d2.num end_gap_num,
d2.num - d1.num - 1 gap from #numbers d1
join #numbers d2 on d1.id + 1 = d2.id
where (d2.num - d1.num - 1) > 0
drop table #numbers
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LAG(id, 1, null)function withOVER (ORDER BY id)clause. – Headman