T-SQL差距和排名查询的最佳方式是什么
这是输入表T-SQL差距和排名查询的最佳方式是什么,sql,sql-server,tsql,Sql,Sql Server,Tsql,这是输入表 IF OBJECT_ID('dbo.NumSeq', 'U') IS NOT NULL DROP TABLE dbo.NumSeq; CREATE TABLE dbo.NumSeq ( seqval INT NOT NULL CONSTRAINT PK_NumSeq PRIMARY KEY ); INSERT INTO dbo.NumSeq(seqval) VALUES (1), (5), (8), (9), (13); 下面是输入数据 SeqVal --
IF OBJECT_ID('dbo.NumSeq', 'U') IS NOT NULL
DROP TABLE dbo.NumSeq;
CREATE TABLE dbo.NumSeq
(
seqval INT NOT NULL CONSTRAINT PK_NumSeq PRIMARY KEY
);
INSERT INTO dbo.NumSeq(seqval)
VALUES (1), (5), (8), (9), (13);
SeqVal
------
1
5
8
9
13
SeqVal Rank
------------
1 1
2 1
3 1
4 1
5 2
6 2
7 2
8 3
9 4
10 5
11 5
12 5
13 5
默认情况下，如果在order By子句之后有一个more then use选项maxrecursion 0查询提示，则它限制为100个SEQVAL。您可以使用理货表和count…over来避免：
您可以使用理货台和计数…以避免：
Zohar Peled击败了我-这里的解决方案将涉及窗口函数和计数表。对于我的解决方案，我使用RangeAB-我自己的理货表函数：
CREATE FUNCTION dbo.rangeAB
(
@low bigint,
@high bigint,
@gap bigint,
@row1 bit
)
/****************************************************************************************
[Purpose]:
Creates up to 531,441,000,000 sequentia integers numbers beginning with @low and ending
with @high. Used to replace iterative methods such as loops, cursors and recursive CTEs
to solve SQL problems. Based on Itzik Ben-Gan's getnums function with some tweeks and
enhancements and added functionality. The logic for getting rn to begin at 0 or 1 is
based comes from Jeff Moden's fnTally function.
The name range because it's similar to clojure's range function. The name "rangeAB" as
used because "range" is a reserved SQL keyword.
[Author]: Alan Burstein
[Compatibility]:
SQL Server 2008+ and Azure SQL Database
[Syntax]:
SELECT r.RN, r.OP, r.N1, r.N2
FROM dbo.rangeAB(@low,@high,@gap,@row1) AS r;
[Parameters]:
@low = a bigint that represents the lowest value for n1.
@high = a bigint that represents the highest value for n1.
@gap = a bigint that represents how much n1 and n2 will increase each row; @gap also
represents the difference between n1 and n2.
@row1 = a bit that represents the first value of rn. When @row = 0 then rn begins
at 0, when @row = 1 then rn will begin at 1.
[Returns]:
Inline Table Valued Function returns:
rn = bigint; a row number that works just like T-SQL ROW_NUMBER() except that it can
start at 0 or 1 which is dictated by @row1.
op = bigint; returns the "opposite number that relates to rn. When rn begins with 0 and
ends with 10 then 10 is the opposite of 0, 9 the opposite of 1, etc. When rn begins
with 1 and ends with 5 then 1 is the opposite of 5, 2 the opposite of 4, etc...
n1 = bigint; a sequential number starting at the value of @low and incrimentingby the
value of @gap until it is less than or equal to the value of @high.
n2 = bigint; a sequential number starting at the value of @low+@gap and incrimenting
by the value of @gap.
[Dependencies]:
N/A
[Developer Notes]:
1. The lowest and highest possible numbers returned are whatever is allowable by a
bigint. The function, however, returns no more than 531,441,000,000 rows (8100^3).
2. @gap does not affect rn, rn will begin at @row1 and increase by 1 until the last row
unless its used in a query where a filter is applied to rn.
3. @gap must be greater than 0 or the function will not return any rows.
4. Keep in mind that when @row1 is 0 then the highest row-number will be the number of
rows returned minus 1
5. If you only need is a sequential set beginning at 0 or 1 then, for best performance
use the RN column. Use N1 and/or N2 when you need to begin your sequence at any
number other than 0 or 1 or if you need a gap between your sequence of numbers.
6. Although @gap is a bigint it must be a positive integer or the function will
not return any rows.
7. The function will not return any rows when one of the following conditions are true:
* any of the input parameters are NULL
* @high is less than @low
* @gap is not greater than 0
To force the function to return all NULLs instead of not returning anything you can
add the following code to the end of the query:
UNION ALL
SELECT NULL, NULL, NULL, NULL
WHERE NOT (@high&@low&@gap&@row1 IS NOT NULL AND @high >= @low AND @gap > 0)
This code was excluded as it adds a ~5% performance penalty.
8. There is no performance penalty for sorting by rn ASC; there is a large performance
penalty for sorting in descending order WHEN @row1 = 1; WHEN @row1 = 0
If you need a descending sort the use op in place of rn then sort by rn ASC.
Best Practices:
--===== 1. Using RN (rownumber)
-- (1.1) The best way to get the numbers 1,2,3...@high (e.g. 1 to 5):
SELECT RN FROM dbo.rangeAB(1,5,1,1);
-- (1.2) The best way to get the numbers 0,1,2...@high-1 (e.g. 0 to 5):
SELECT RN FROM dbo.rangeAB(0,5,1,0);
--===== 2. Using OP for descending sorts without a performance penalty
-- (2.1) The best way to get the numbers 5,4,3...@high (e.g. 5 to 1):
SELECT op FROM dbo.rangeAB(1,5,1,1) ORDER BY rn ASC;
-- (2.2) The best way to get the numbers 0,1,2...@high-1 (e.g. 5 to 0):
SELECT op FROM dbo.rangeAB(1,6,1,0) ORDER BY rn ASC;
--===== 3. Using N1
-- (3.1) To begin with numbers other than 0 or 1 use N1 (e.g. -3 to 3):
SELECT N1 FROM dbo.rangeAB(-3,3,1,1);
-- (3.2) ROW_NUMBER() is built in. If you want a ROW_NUMBER() include RN:
SELECT RN, N1 FROM dbo.rangeAB(-3,3,1,1);
-- (3.3) If you wanted a ROW_NUMBER() that started at 0 you would do this:
SELECT RN, N1 FROM dbo.rangeAB(-3,3,1,0);
--===== 4. Using N2 and @gap
-- (4.1) To get 0,10,20,30...100, set @low to 0, @high to 100 and @gap to 10:
SELECT N1 FROM dbo.rangeAB(0,100,10,1);
-- (4.2) Note that N2=N1+@gap; this allows you to create a sequence of ranges.
-- For example, to get (0,10),(10,20),(20,30).... (90,100):
SELECT N1, N2 FROM dbo.rangeAB(0,90,10,1);
-- (4.3) Remember that a rownumber is included and it can begin at 0 or 1:
SELECT RN, N1, N2 FROM dbo.rangeAB(0,90,10,1);
[Examples]:
--===== 1. Generating Sample data (using rangeAB to create "dummy rows")
-- The query below will generate 10,000 ids and random numbers between 50,000 and 500,000
SELECT
someId = r.rn,
someNumer = ABS(CHECKSUM(NEWID())%450000)+50001
FROM rangeAB(1,10000,1,1) r;
--===== 2. Create a series of dates; rn is 0 to include the first date in the series
DECLARE @startdate DATE = '20180101', @enddate DATE = '20180131';
SELECT r.rn, calDate = DATEADD(dd, r.rn, @startdate)
FROM dbo.rangeAB(1, DATEDIFF(dd,@startdate,@enddate),1,0) r;
GO
--===== 3. Splitting (tokenizing) a string with fixed sized items
-- given a delimited string of identifiers that are always 7 characters long
DECLARE @string VARCHAR(1000) = 'A601225,B435223,G008081,R678567';
SELECT
itemNumber = r.rn, -- item's ordinal position
itemIndex = r.n1, -- item's position in the string (it's CHARINDEX value)
item = SUBSTRING(@string, r.n1, 7) -- item (token)
FROM dbo.rangeAB(1, LEN(@string), 8,1) r;
GO
--===== 4. Splitting (tokenizing) a string with random delimiters
DECLARE @string VARCHAR(1000) = 'ABC123,999F,XX,9994443335';
SELECT
itemNumber = ROW_NUMBER() OVER (ORDER BY r.rn), -- item's ordinal position
itemIndex = r.n1+1, -- item's position in the string (it's CHARINDEX value)
item = SUBSTRING
(
@string,
r.n1+1,
ISNULL(NULLIF(CHARINDEX(',',@string,r.n1+1),0)-r.n1-1, 8000)
) -- item (token)
FROM dbo.rangeAB(0,DATALENGTH(@string),1,1) r
WHERE SUBSTRING(@string,r.n1,1) = ',' OR r.n1 = 0;
-- logic borrowed from: http://www.sqlservercentral.com/articles/Tally+Table/72993/
--===== 5. Grouping by a weekly intervals
-- 5.1. how to create a series of start/end dates between @startDate & @endDate
DECLARE @startDate DATE = '1/1/2015', @endDate DATE = '2/1/2015';
SELECT
WeekNbr = r.RN,
WeekStart = DATEADD(DAY,r.N1,@StartDate),
WeekEnd = DATEADD(DAY,r.N2-1,@StartDate)
FROM dbo.rangeAB(0,datediff(DAY,@StartDate,@EndDate),7,1) r;
GO
-- 5.2. LEFT JOIN to the weekly interval table
BEGIN
DECLARE @startDate datetime = '1/1/2015', @endDate datetime = '2/1/2015';
-- sample data
DECLARE @loans TABLE (loID INT, lockDate DATE);
INSERT @loans SELECT r.rn, DATEADD(dd, ABS(CHECKSUM(NEWID())%32), @startDate)
FROM dbo.rangeAB(1,50,1,1) r;
-- solution
SELECT
WeekNbr = r.RN,
WeekStart = dt.WeekStart,
WeekEnd = dt.WeekEnd,
total = COUNT(l.lockDate)
FROM dbo.rangeAB(0,datediff(DAY,@StartDate,@EndDate),7,1) r
CROSS APPLY (VALUES (
CAST(DATEADD(DAY,r.N1,@StartDate) AS DATE),
CAST(DATEADD(DAY,r.N2-1,@StartDate) AS DATE))) dt(WeekStart,WeekEnd)
LEFT JOIN @loans l ON l.lockDate BETWEEN dt.WeekStart AND dt.WeekEnd
GROUP BY r.RN, dt.WeekStart, dt.WeekEnd ;
END;
--===== 6. Identify the first vowel and last vowel in a along with their positions
DECLARE @string VARCHAR(200) = 'This string has vowels';
SELECT TOP(1) position = r.rn, letter = SUBSTRING(@string,r.rn,1)
FROM dbo.rangeAB(1,LEN(@string),1,1) r
WHERE SUBSTRING(@string,r.rn,1) LIKE '%[aeiou]%'
ORDER BY r.rn;
-- To avoid a sort in the execution plan we'll use op instead of rn
SELECT TOP(1) position = r.op, letter = SUBSTRING(@string,r.op,1)
FROM dbo.rangeAB(1,LEN(@string),1,1) r
WHERE SUBSTRING(@string,r.rn,1) LIKE '%[aeiou]%'
ORDER BY r.rn;
---------------------------------------------------------------------------------------
[Revision History]:
Rev 00 - 20140518 - Initial Development - Alan Burstein
Rev 01 - 20151029 - Added 65 rows to make L1=465; 465^3=100.5M. Updated comment section
- Alan Burstein
Rev 02 - 20180613 - Complete re-design including opposite number column (op)
Rev 03 - 20180920 - Added additional CROSS JOIN to L2 for 530B rows max - Alan Burstein
****************************************************************************************/
RETURNS TABLE WITH SCHEMABINDING AS RETURN
WITH L1(N) AS
(
SELECT 1
FROM (VALUES
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0)) T(N) -- 90 values
),
L2(N) AS (SELECT 1 FROM L1 a CROSS JOIN L1 b CROSS JOIN L1 c),
iTally AS (SELECT rn = ROW_NUMBER() OVER (ORDER BY (SELECT 1)) FROM L2 a CROSS JOIN L2 b)
SELECT
r.RN,
r.OP,
r.N1,
r.N2
FROM
(
SELECT
RN = 0,
OP = (@high-@low)/@gap,
N1 = @low,
N2 = @gap+@low
WHERE @row1 = 0
UNION ALL -- COALESCE required in the TOP statement below for error handling purposes
SELECT TOP (ABS((COALESCE(@high,0)-COALESCE(@low,0))/COALESCE(@gap,0)+COALESCE(@row1,1)))
RN = i.rn,
OP = (@high-@low)/@gap+(2*@row1)-i.rn,
N1 = (i.rn-@row1)*@gap+@low,
N2 = (i.rn-(@row1-1))*@gap+@low
FROM iTally AS i
ORDER BY rn
) AS r
WHERE @high&@low&@gap&@row1 IS NOT NULL AND @high >= @low AND @gap > 0;
-- Sample data I used
IF OBJECT_ID('dbo.NumSeq', 'U') IS NOT NULL DROP TABLE dbo.NumSeq;
CREATE TABLE dbo.NumSeq(seqval INT NOT NULL CONSTRAINT PK_NumSeq PRIMARY KEY);
INSERT INTO dbo.NumSeq(seqval) VALUES (1), (5), (8), (9), (13);
-- Solution
SELECT seqval = r.N1, rnk = SUM(CASE WHEN n.seqval IS NULL THEN 0 ELSE 1 END) OVER (ORDER BY r.RN)
FROM
(
SELECT MIN(n.seqval), MAX(n.seqval)
FROM dbo.NumSeq AS n
) AS rng(mn,mx)
CROSS APPLY dbo.rangeAB(rng.mn,rng.mx,1,1) AS r
LEFT JOIN dbo.NumSeq AS n ON r.N1 = n.seqval;
Zohar Peled击败了我-这里的解决方案将涉及窗口函数和计数表。对于我的解决方案，我使用RangeAB-我自己的理货表函数：
CREATE FUNCTION dbo.rangeAB
(
@low bigint,
@high bigint,
@gap bigint,
@row1 bit
)
/****************************************************************************************
[Purpose]:
Creates up to 531,441,000,000 sequentia integers numbers beginning with @low and ending
with @high. Used to replace iterative methods such as loops, cursors and recursive CTEs
to solve SQL problems. Based on Itzik Ben-Gan's getnums function with some tweeks and
enhancements and added functionality. The logic for getting rn to begin at 0 or 1 is
based comes from Jeff Moden's fnTally function.
The name range because it's similar to clojure's range function. The name "rangeAB" as
used because "range" is a reserved SQL keyword.
[Author]: Alan Burstein
[Compatibility]:
SQL Server 2008+ and Azure SQL Database
[Syntax]:
SELECT r.RN, r.OP, r.N1, r.N2
FROM dbo.rangeAB(@low,@high,@gap,@row1) AS r;
[Parameters]:
@low = a bigint that represents the lowest value for n1.
@high = a bigint that represents the highest value for n1.
@gap = a bigint that represents how much n1 and n2 will increase each row; @gap also
represents the difference between n1 and n2.
@row1 = a bit that represents the first value of rn. When @row = 0 then rn begins
at 0, when @row = 1 then rn will begin at 1.
[Returns]:
Inline Table Valued Function returns:
rn = bigint; a row number that works just like T-SQL ROW_NUMBER() except that it can
start at 0 or 1 which is dictated by @row1.
op = bigint; returns the "opposite number that relates to rn. When rn begins with 0 and
ends with 10 then 10 is the opposite of 0, 9 the opposite of 1, etc. When rn begins
with 1 and ends with 5 then 1 is the opposite of 5, 2 the opposite of 4, etc...
n1 = bigint; a sequential number starting at the value of @low and incrimentingby the
value of @gap until it is less than or equal to the value of @high.
n2 = bigint; a sequential number starting at the value of @low+@gap and incrimenting
by the value of @gap.
[Dependencies]:
N/A
[Developer Notes]:
1. The lowest and highest possible numbers returned are whatever is allowable by a
bigint. The function, however, returns no more than 531,441,000,000 rows (8100^3).
2. @gap does not affect rn, rn will begin at @row1 and increase by 1 until the last row
unless its used in a query where a filter is applied to rn.
3. @gap must be greater than 0 or the function will not return any rows.
4. Keep in mind that when @row1 is 0 then the highest row-number will be the number of
rows returned minus 1
5. If you only need is a sequential set beginning at 0 or 1 then, for best performance
use the RN column. Use N1 and/or N2 when you need to begin your sequence at any
number other than 0 or 1 or if you need a gap between your sequence of numbers.
6. Although @gap is a bigint it must be a positive integer or the function will
not return any rows.
7. The function will not return any rows when one of the following conditions are true:
* any of the input parameters are NULL
* @high is less than @low
* @gap is not greater than 0
To force the function to return all NULLs instead of not returning anything you can
add the following code to the end of the query:
UNION ALL
SELECT NULL, NULL, NULL, NULL
WHERE NOT (@high&@low&@gap&@row1 IS NOT NULL AND @high >= @low AND @gap > 0)
This code was excluded as it adds a ~5% performance penalty.
8. There is no performance penalty for sorting by rn ASC; there is a large performance
penalty for sorting in descending order WHEN @row1 = 1; WHEN @row1 = 0
If you need a descending sort the use op in place of rn then sort by rn ASC.
Best Practices:
--===== 1. Using RN (rownumber)
-- (1.1) The best way to get the numbers 1,2,3...@high (e.g. 1 to 5):
SELECT RN FROM dbo.rangeAB(1,5,1,1);
-- (1.2) The best way to get the numbers 0,1,2...@high-1 (e.g. 0 to 5):
SELECT RN FROM dbo.rangeAB(0,5,1,0);
--===== 2. Using OP for descending sorts without a performance penalty
-- (2.1) The best way to get the numbers 5,4,3...@high (e.g. 5 to 1):
SELECT op FROM dbo.rangeAB(1,5,1,1) ORDER BY rn ASC;
-- (2.2) The best way to get the numbers 0,1,2...@high-1 (e.g. 5 to 0):
SELECT op FROM dbo.rangeAB(1,6,1,0) ORDER BY rn ASC;
--===== 3. Using N1
-- (3.1) To begin with numbers other than 0 or 1 use N1 (e.g. -3 to 3):
SELECT N1 FROM dbo.rangeAB(-3,3,1,1);
-- (3.2) ROW_NUMBER() is built in. If you want a ROW_NUMBER() include RN:
SELECT RN, N1 FROM dbo.rangeAB(-3,3,1,1);
-- (3.3) If you wanted a ROW_NUMBER() that started at 0 you would do this:
SELECT RN, N1 FROM dbo.rangeAB(-3,3,1,0);
--===== 4. Using N2 and @gap
-- (4.1) To get 0,10,20,30...100, set @low to 0, @high to 100 and @gap to 10:
SELECT N1 FROM dbo.rangeAB(0,100,10,1);
-- (4.2) Note that N2=N1+@gap; this allows you to create a sequence of ranges.
-- For example, to get (0,10),(10,20),(20,30).... (90,100):
SELECT N1, N2 FROM dbo.rangeAB(0,90,10,1);
-- (4.3) Remember that a rownumber is included and it can begin at 0 or 1:
SELECT RN, N1, N2 FROM dbo.rangeAB(0,90,10,1);
[Examples]:
--===== 1. Generating Sample data (using rangeAB to create "dummy rows")
-- The query below will generate 10,000 ids and random numbers between 50,000 and 500,000
SELECT
someId = r.rn,
someNumer = ABS(CHECKSUM(NEWID())%450000)+50001
FROM rangeAB(1,10000,1,1) r;
--===== 2. Create a series of dates; rn is 0 to include the first date in the series
DECLARE @startdate DATE = '20180101', @enddate DATE = '20180131';
SELECT r.rn, calDate = DATEADD(dd, r.rn, @startdate)
FROM dbo.rangeAB(1, DATEDIFF(dd,@startdate,@enddate),1,0) r;
GO
--===== 3. Splitting (tokenizing) a string with fixed sized items
-- given a delimited string of identifiers that are always 7 characters long
DECLARE @string VARCHAR(1000) = 'A601225,B435223,G008081,R678567';
SELECT
itemNumber = r.rn, -- item's ordinal position
itemIndex = r.n1, -- item's position in the string (it's CHARINDEX value)
item = SUBSTRING(@string, r.n1, 7) -- item (token)
FROM dbo.rangeAB(1, LEN(@string), 8,1) r;
GO
--===== 4. Splitting (tokenizing) a string with random delimiters
DECLARE @string VARCHAR(1000) = 'ABC123,999F,XX,9994443335';
SELECT
itemNumber = ROW_NUMBER() OVER (ORDER BY r.rn), -- item's ordinal position
itemIndex = r.n1+1, -- item's position in the string (it's CHARINDEX value)
item = SUBSTRING
(
@string,
r.n1+1,
ISNULL(NULLIF(CHARINDEX(',',@string,r.n1+1),0)-r.n1-1, 8000)
) -- item (token)
FROM dbo.rangeAB(0,DATALENGTH(@string),1,1) r
WHERE SUBSTRING(@string,r.n1,1) = ',' OR r.n1 = 0;
-- logic borrowed from: http://www.sqlservercentral.com/articles/Tally+Table/72993/
--===== 5. Grouping by a weekly intervals
-- 5.1. how to create a series of start/end dates between @startDate & @endDate
DECLARE @startDate DATE = '1/1/2015', @endDate DATE = '2/1/2015';
SELECT
WeekNbr = r.RN,
WeekStart = DATEADD(DAY,r.N1,@StartDate),
WeekEnd = DATEADD(DAY,r.N2-1,@StartDate)
FROM dbo.rangeAB(0,datediff(DAY,@StartDate,@EndDate),7,1) r;
GO
-- 5.2. LEFT JOIN to the weekly interval table
BEGIN
DECLARE @startDate datetime = '1/1/2015', @endDate datetime = '2/1/2015';
-- sample data
DECLARE @loans TABLE (loID INT, lockDate DATE);
INSERT @loans SELECT r.rn, DATEADD(dd, ABS(CHECKSUM(NEWID())%32), @startDate)
FROM dbo.rangeAB(1,50,1,1) r;
-- solution
SELECT
WeekNbr = r.RN,
WeekStart = dt.WeekStart,
WeekEnd = dt.WeekEnd,
total = COUNT(l.lockDate)
FROM dbo.rangeAB(0,datediff(DAY,@StartDate,@EndDate),7,1) r
CROSS APPLY (VALUES (
CAST(DATEADD(DAY,r.N1,@StartDate) AS DATE),
CAST(DATEADD(DAY,r.N2-1,@StartDate) AS DATE))) dt(WeekStart,WeekEnd)
LEFT JOIN @loans l ON l.lockDate BETWEEN dt.WeekStart AND dt.WeekEnd
GROUP BY r.RN, dt.WeekStart, dt.WeekEnd ;
END;
--===== 6. Identify the first vowel and last vowel in a along with their positions
DECLARE @string VARCHAR(200) = 'This string has vowels';
SELECT TOP(1) position = r.rn, letter = SUBSTRING(@string,r.rn,1)
FROM dbo.rangeAB(1,LEN(@string),1,1) r
WHERE SUBSTRING(@string,r.rn,1) LIKE '%[aeiou]%'
ORDER BY r.rn;
-- To avoid a sort in the execution plan we'll use op instead of rn
SELECT TOP(1) position = r.op, letter = SUBSTRING(@string,r.op,1)
FROM dbo.rangeAB(1,LEN(@string),1,1) r
WHERE SUBSTRING(@string,r.rn,1) LIKE '%[aeiou]%'
ORDER BY r.rn;
---------------------------------------------------------------------------------------
[Revision History]:
Rev 00 - 20140518 - Initial Development - Alan Burstein
Rev 01 - 20151029 - Added 65 rows to make L1=465; 465^3=100.5M. Updated comment section
- Alan Burstein
Rev 02 - 20180613 - Complete re-design including opposite number column (op)
Rev 03 - 20180920 - Added additional CROSS JOIN to L2 for 530B rows max - Alan Burstein
****************************************************************************************/
RETURNS TABLE WITH SCHEMABINDING AS RETURN
WITH L1(N) AS
(
SELECT 1
FROM (VALUES
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0)) T(N) -- 90 values
),
L2(N) AS (SELECT 1 FROM L1 a CROSS JOIN L1 b CROSS JOIN L1 c),
iTally AS (SELECT rn = ROW_NUMBER() OVER (ORDER BY (SELECT 1)) FROM L2 a CROSS JOIN L2 b)
SELECT
r.RN,
r.OP,
r.N1,
r.N2
FROM
(
SELECT
RN = 0,
OP = (@high-@low)/@gap,
N1 = @low,
N2 = @gap+@low
WHERE @row1 = 0
UNION ALL -- COALESCE required in the TOP statement below for error handling purposes
SELECT TOP (ABS((COALESCE(@high,0)-COALESCE(@low,0))/COALESCE(@gap,0)+COALESCE(@row1,1)))
RN = i.rn,
OP = (@high-@low)/@gap+(2*@row1)-i.rn,
N1 = (i.rn-@row1)*@gap+@low,
N2 = (i.rn-(@row1-1))*@gap+@low
FROM iTally AS i
ORDER BY rn
) AS r
WHERE @high&@low&@gap&@row1 IS NOT NULL AND @high >= @low AND @gap > 0;
-- Sample data I used
IF OBJECT_ID('dbo.NumSeq', 'U') IS NOT NULL DROP TABLE dbo.NumSeq;
CREATE TABLE dbo.NumSeq(seqval INT NOT NULL CONSTRAINT PK_NumSeq PRIMARY KEY);
INSERT INTO dbo.NumSeq(seqval) VALUES (1), (5), (8), (9), (13);
-- Solution
SELECT seqval = r.N1, rnk = SUM(CASE WHEN n.seqval IS NULL THEN 0 ELSE 1 END) OVER (ORDER BY r.RN)
FROM
(
SELECT MIN(n.seqval), MAX(n.seqval)
FROM dbo.NumSeq AS n
) AS rng(mn,mx)
CROSS APPLY dbo.rangeAB(rng.mn,rng.mx,1,1) AS r
LEFT JOIN dbo.NumSeq AS n ON r.N1 = n.seqval;
这里的Betters解决方案：No recursive Ctest这不会给我期望的输出，我希望输出只有13行。这里的Betters解决方案：No recursive Ctest这不会给我期望的输出，我希望输出只有13行。为什么你说下面是输入数据，然后插入了完全不同的值？还有什么逻辑是10在输出中没有与9配对并且具有不同的秩，但是5与6配对并且具有相同的秩？我同意Martin Smith的观点-我认为9到12都应该具有秩4，13应该具有秩5-但是，我可能误解了逻辑，因为您没有解释我更新了输入数据。谢谢。这种逻辑就像为从1到Max的所有序列分配密集秩，并且在遇到Seqval时增加秩。那么为什么秩在10处增加呢？这里没有seqval。为什么你说下面是输入数据，然后插入了完全不同的值？还有什么逻辑是10在输出中没有与9配对并且具有不同的秩，但是5与6配对并且具有相同的秩？我同意Martin Smith的观点-我认为9到12都应该具有秩4，13应该具有秩5-但是，我可能误解了逻辑，因为您没有解释我更新了输入数据。谢谢。这种逻辑就像为从1到Max的所有序列分配密集秩，并且在遇到Seqval时增加秩。那么为什么秩在10处增加呢？这里没有seqval，回答得很好+1@AlanBurstein坦斯克！回答得好+1@AlanBurstein坦斯克！谢谢你，艾伦，你的回答给了我最需要的。这个解决方案无法扩展。你能帮忙吗我使用的样本数据插入到dbo.NumSeqid中，seqval值为1,1,1,5,1,8,1,9,1,13,2,1,2,5；Select*From NumSeq-解决方案选择rng.id，seqval=r.N1，rnk=SUMCASE当n.seqval为NULL时，则0否则1按rng.id结束分区，按r.RN从Select id开始排序，MINn.seqval，MAXn.seqval从dbo.NumSeq开始作为n按id分组作为rngid，mn，mx交叉应用dbo.rangeabrung.mn，rng.mx，1,1作为r左连接dbo.NumSeq作为r.N1上的n=n.seqval顺序为1,2；让它工作起来解决方案选择rng.id，seqval=r.N1，rnk=SUMCASE当n.seqval为NULL时，则0否则1按r.RN结束分区，按r.RN从SELECT id开始排序，MINn.seqval，MAXn.seqval从dbo.NumSeq开始排序，n按id分组，按rngid，mn，mx交叉应用dbo.rangeabrung.mn，rng.mx，1作为r左连接dbo.NumSeq作为n在r.N1=n.seqval和n.id=rng.id开始排序，2.谢谢你，艾伦，你的回答给了我最需要的。这个解决方案无法扩展。你能帮忙吗我使用的样本数据插入到dbo.NumSeqid中，seqval值为1,1,1,5,1,8,1,9,1,13,2,1,2,5；Select*From NumSeq-解决方案选择rng.id，seqval=r.N1，rnk=SUMCASE当n.seqval为NULL时，则0否则1按rng.id结束分区，按r.RN从Select id开始排序，MINn.seqval，MAXn.seqval从dbo.NumSeq开始作为n按id分组作为rngid，mn，mx交叉应用dbo.rangeabrung.mn，rng.mx，1,1作为r左连接dbo.NumSeq作为r.N1上的n=n.seqval顺序为1,2；让它工作起来解决方案选择rng.id，seqval=r.N1，rnk=SUMCASE当n.seqval为NULL时，则0否则1按r.RN结束分区，按r.RN从SELECT id开始排序，MINn.seqval，MAXn.seqval从dbo.NumSeq开始排序，n按id分组，按rngid，mn，mx交叉应用dbo.rangeabrung.mn，rng.mx，1作为r左连接dbo.NumSeq作为n在r.N1=n.seqval和n.id=rng.id开始排序，2.
seqval rnk
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