Parent child Hierarchy based on non unique order number












0















I have a table containing system and order number column which is not unique.I would like to connect the system based on the order number



+----------------------+  
¦ system ¦OrderNo¦
¦--------------+-------+
¦ system1 ¦ 1 ¦
¦ system2 ¦ 2 ¦
¦ system3 ¦ 3 ¦
¦ system4 ¦ 4 ¦
¦ system1 ¦ 1 ¦
¦ system2 ¦ 2 ¦
¦ system5 ¦ 1 ¦
¦ system6 ¦ 2 ¦
¦ system7 ¦ 1 ¦
¦ system8 ¦ 2 ¦
+----------------------+


How do I generate a parent child relation based on the OrderNo Column. Where 1-2-3-4 are one set, 1-2,1-2,1-2 the other set



desired Output is as below



+----------------------+  
¦ Parent ¦Child ¦
¦--------------+-------+
¦ system1 ¦system2¦
¦ system2 ¦system3¦
¦ system3 ¦system4¦
¦ system4 ¦NULL ¦
¦ system1 ¦system2¦
¦ system2 ¦NULL ¦
¦ system5 ¦system6¦
¦ system6 ¦NULL ¦
¦ system7 ¦system8¦
¦ system8 ¦NULL ¦
+----------------------+









share|improve this question























  • Are those the only columns you have? If so, impossible with that data; you cannot guarantee the order of your data with the use of an ORDER BY clause, and thus you have no way to determine the relationship.

    – Larnu
    Nov 24 '18 at 21:46
















0















I have a table containing system and order number column which is not unique.I would like to connect the system based on the order number



+----------------------+  
¦ system ¦OrderNo¦
¦--------------+-------+
¦ system1 ¦ 1 ¦
¦ system2 ¦ 2 ¦
¦ system3 ¦ 3 ¦
¦ system4 ¦ 4 ¦
¦ system1 ¦ 1 ¦
¦ system2 ¦ 2 ¦
¦ system5 ¦ 1 ¦
¦ system6 ¦ 2 ¦
¦ system7 ¦ 1 ¦
¦ system8 ¦ 2 ¦
+----------------------+


How do I generate a parent child relation based on the OrderNo Column. Where 1-2-3-4 are one set, 1-2,1-2,1-2 the other set



desired Output is as below



+----------------------+  
¦ Parent ¦Child ¦
¦--------------+-------+
¦ system1 ¦system2¦
¦ system2 ¦system3¦
¦ system3 ¦system4¦
¦ system4 ¦NULL ¦
¦ system1 ¦system2¦
¦ system2 ¦NULL ¦
¦ system5 ¦system6¦
¦ system6 ¦NULL ¦
¦ system7 ¦system8¦
¦ system8 ¦NULL ¦
+----------------------+









share|improve this question























  • Are those the only columns you have? If so, impossible with that data; you cannot guarantee the order of your data with the use of an ORDER BY clause, and thus you have no way to determine the relationship.

    – Larnu
    Nov 24 '18 at 21:46














0












0








0








I have a table containing system and order number column which is not unique.I would like to connect the system based on the order number



+----------------------+  
¦ system ¦OrderNo¦
¦--------------+-------+
¦ system1 ¦ 1 ¦
¦ system2 ¦ 2 ¦
¦ system3 ¦ 3 ¦
¦ system4 ¦ 4 ¦
¦ system1 ¦ 1 ¦
¦ system2 ¦ 2 ¦
¦ system5 ¦ 1 ¦
¦ system6 ¦ 2 ¦
¦ system7 ¦ 1 ¦
¦ system8 ¦ 2 ¦
+----------------------+


How do I generate a parent child relation based on the OrderNo Column. Where 1-2-3-4 are one set, 1-2,1-2,1-2 the other set



desired Output is as below



+----------------------+  
¦ Parent ¦Child ¦
¦--------------+-------+
¦ system1 ¦system2¦
¦ system2 ¦system3¦
¦ system3 ¦system4¦
¦ system4 ¦NULL ¦
¦ system1 ¦system2¦
¦ system2 ¦NULL ¦
¦ system5 ¦system6¦
¦ system6 ¦NULL ¦
¦ system7 ¦system8¦
¦ system8 ¦NULL ¦
+----------------------+









share|improve this question














I have a table containing system and order number column which is not unique.I would like to connect the system based on the order number



+----------------------+  
¦ system ¦OrderNo¦
¦--------------+-------+
¦ system1 ¦ 1 ¦
¦ system2 ¦ 2 ¦
¦ system3 ¦ 3 ¦
¦ system4 ¦ 4 ¦
¦ system1 ¦ 1 ¦
¦ system2 ¦ 2 ¦
¦ system5 ¦ 1 ¦
¦ system6 ¦ 2 ¦
¦ system7 ¦ 1 ¦
¦ system8 ¦ 2 ¦
+----------------------+


How do I generate a parent child relation based on the OrderNo Column. Where 1-2-3-4 are one set, 1-2,1-2,1-2 the other set



desired Output is as below



+----------------------+  
¦ Parent ¦Child ¦
¦--------------+-------+
¦ system1 ¦system2¦
¦ system2 ¦system3¦
¦ system3 ¦system4¦
¦ system4 ¦NULL ¦
¦ system1 ¦system2¦
¦ system2 ¦NULL ¦
¦ system5 ¦system6¦
¦ system6 ¦NULL ¦
¦ system7 ¦system8¦
¦ system8 ¦NULL ¦
+----------------------+






sql sql-server sql-server-2008 hierarchy hierarchical-data






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asked Nov 24 '18 at 21:43









xoanonxoanon

1009




1009













  • Are those the only columns you have? If so, impossible with that data; you cannot guarantee the order of your data with the use of an ORDER BY clause, and thus you have no way to determine the relationship.

    – Larnu
    Nov 24 '18 at 21:46



















  • Are those the only columns you have? If so, impossible with that data; you cannot guarantee the order of your data with the use of an ORDER BY clause, and thus you have no way to determine the relationship.

    – Larnu
    Nov 24 '18 at 21:46

















Are those the only columns you have? If so, impossible with that data; you cannot guarantee the order of your data with the use of an ORDER BY clause, and thus you have no way to determine the relationship.

– Larnu
Nov 24 '18 at 21:46





Are those the only columns you have? If so, impossible with that data; you cannot guarantee the order of your data with the use of an ORDER BY clause, and thus you have no way to determine the relationship.

– Larnu
Nov 24 '18 at 21:46












1 Answer
1






active

oldest

votes


















0














As I mentioned in my comment, if these are the only columns you have, what you are after isn't achievable. Without some kind of a ascending unique key, you have no way of determining the relationship. If we do add a unique ascending key then we can achieve this.



If you were using SQL Server 2012+ (2008 is out of support, and very close to end of extended support, so upgrading should be a very high priority right now), then you could generate your islands using ROWS BETWEEN and then use LEAD:



CREATE TABLE dbo.SampleTable (ID int IDENTITY(1,1),
[system] varchar(8),
OrderNo int);
INSERT INTO dbo.SampleTable([System],OrderNo)
VALUES('system1',1),
('system2',2),
('system3',3),
('system4',4),
('system1',1),
('system2',2),
('system5',1),
('system6',2),
('system7',1),
('system8',2);
GO

WITH Groups AS(
SELECT ID,
[System],
COUNT(CASE WHEN OrderNo = 1 THEN 1 END) OVER (ORDER BY ID ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Grp
FROM dbo.SampleTable)
SELECT [System] AS Parent,
LEAD([System]) OVER (PARTITION BY Grp ORDER BY ID) AS Child
FROM Groups;
GO


Without access to LEAD and ROWS BETWEEN you'll have to be more creative; and the solution is going to be far slower:



WITH Groups AS(
SELECT ST.ID,
ST.[System],
G.Grp
FROM dbo.SampleTable ST
CROSS APPLY (SELECT COUNT(*) AS Grp
FROM dbo.SampleTable CA
WHERE CA.OrderNo =1
AND CA.ID <= ST.ID) G)
SELECT G1.[System] AS Parent,
G2.[System] AS Child
FROM Groups G1
LEFT JOIN Groups G2 ON G1.Grp = G2.Grp
AND G1.ID = G2.ID - 1;





share|improve this answer


























  • It works great on sql server 2008, but without LEAD the output is not the same

    – xoanon
    Nov 28 '18 at 12:40











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














As I mentioned in my comment, if these are the only columns you have, what you are after isn't achievable. Without some kind of a ascending unique key, you have no way of determining the relationship. If we do add a unique ascending key then we can achieve this.



If you were using SQL Server 2012+ (2008 is out of support, and very close to end of extended support, so upgrading should be a very high priority right now), then you could generate your islands using ROWS BETWEEN and then use LEAD:



CREATE TABLE dbo.SampleTable (ID int IDENTITY(1,1),
[system] varchar(8),
OrderNo int);
INSERT INTO dbo.SampleTable([System],OrderNo)
VALUES('system1',1),
('system2',2),
('system3',3),
('system4',4),
('system1',1),
('system2',2),
('system5',1),
('system6',2),
('system7',1),
('system8',2);
GO

WITH Groups AS(
SELECT ID,
[System],
COUNT(CASE WHEN OrderNo = 1 THEN 1 END) OVER (ORDER BY ID ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Grp
FROM dbo.SampleTable)
SELECT [System] AS Parent,
LEAD([System]) OVER (PARTITION BY Grp ORDER BY ID) AS Child
FROM Groups;
GO


Without access to LEAD and ROWS BETWEEN you'll have to be more creative; and the solution is going to be far slower:



WITH Groups AS(
SELECT ST.ID,
ST.[System],
G.Grp
FROM dbo.SampleTable ST
CROSS APPLY (SELECT COUNT(*) AS Grp
FROM dbo.SampleTable CA
WHERE CA.OrderNo =1
AND CA.ID <= ST.ID) G)
SELECT G1.[System] AS Parent,
G2.[System] AS Child
FROM Groups G1
LEFT JOIN Groups G2 ON G1.Grp = G2.Grp
AND G1.ID = G2.ID - 1;





share|improve this answer


























  • It works great on sql server 2008, but without LEAD the output is not the same

    – xoanon
    Nov 28 '18 at 12:40
















0














As I mentioned in my comment, if these are the only columns you have, what you are after isn't achievable. Without some kind of a ascending unique key, you have no way of determining the relationship. If we do add a unique ascending key then we can achieve this.



If you were using SQL Server 2012+ (2008 is out of support, and very close to end of extended support, so upgrading should be a very high priority right now), then you could generate your islands using ROWS BETWEEN and then use LEAD:



CREATE TABLE dbo.SampleTable (ID int IDENTITY(1,1),
[system] varchar(8),
OrderNo int);
INSERT INTO dbo.SampleTable([System],OrderNo)
VALUES('system1',1),
('system2',2),
('system3',3),
('system4',4),
('system1',1),
('system2',2),
('system5',1),
('system6',2),
('system7',1),
('system8',2);
GO

WITH Groups AS(
SELECT ID,
[System],
COUNT(CASE WHEN OrderNo = 1 THEN 1 END) OVER (ORDER BY ID ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Grp
FROM dbo.SampleTable)
SELECT [System] AS Parent,
LEAD([System]) OVER (PARTITION BY Grp ORDER BY ID) AS Child
FROM Groups;
GO


Without access to LEAD and ROWS BETWEEN you'll have to be more creative; and the solution is going to be far slower:



WITH Groups AS(
SELECT ST.ID,
ST.[System],
G.Grp
FROM dbo.SampleTable ST
CROSS APPLY (SELECT COUNT(*) AS Grp
FROM dbo.SampleTable CA
WHERE CA.OrderNo =1
AND CA.ID <= ST.ID) G)
SELECT G1.[System] AS Parent,
G2.[System] AS Child
FROM Groups G1
LEFT JOIN Groups G2 ON G1.Grp = G2.Grp
AND G1.ID = G2.ID - 1;





share|improve this answer


























  • It works great on sql server 2008, but without LEAD the output is not the same

    – xoanon
    Nov 28 '18 at 12:40














0












0








0







As I mentioned in my comment, if these are the only columns you have, what you are after isn't achievable. Without some kind of a ascending unique key, you have no way of determining the relationship. If we do add a unique ascending key then we can achieve this.



If you were using SQL Server 2012+ (2008 is out of support, and very close to end of extended support, so upgrading should be a very high priority right now), then you could generate your islands using ROWS BETWEEN and then use LEAD:



CREATE TABLE dbo.SampleTable (ID int IDENTITY(1,1),
[system] varchar(8),
OrderNo int);
INSERT INTO dbo.SampleTable([System],OrderNo)
VALUES('system1',1),
('system2',2),
('system3',3),
('system4',4),
('system1',1),
('system2',2),
('system5',1),
('system6',2),
('system7',1),
('system8',2);
GO

WITH Groups AS(
SELECT ID,
[System],
COUNT(CASE WHEN OrderNo = 1 THEN 1 END) OVER (ORDER BY ID ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Grp
FROM dbo.SampleTable)
SELECT [System] AS Parent,
LEAD([System]) OVER (PARTITION BY Grp ORDER BY ID) AS Child
FROM Groups;
GO


Without access to LEAD and ROWS BETWEEN you'll have to be more creative; and the solution is going to be far slower:



WITH Groups AS(
SELECT ST.ID,
ST.[System],
G.Grp
FROM dbo.SampleTable ST
CROSS APPLY (SELECT COUNT(*) AS Grp
FROM dbo.SampleTable CA
WHERE CA.OrderNo =1
AND CA.ID <= ST.ID) G)
SELECT G1.[System] AS Parent,
G2.[System] AS Child
FROM Groups G1
LEFT JOIN Groups G2 ON G1.Grp = G2.Grp
AND G1.ID = G2.ID - 1;





share|improve this answer















As I mentioned in my comment, if these are the only columns you have, what you are after isn't achievable. Without some kind of a ascending unique key, you have no way of determining the relationship. If we do add a unique ascending key then we can achieve this.



If you were using SQL Server 2012+ (2008 is out of support, and very close to end of extended support, so upgrading should be a very high priority right now), then you could generate your islands using ROWS BETWEEN and then use LEAD:



CREATE TABLE dbo.SampleTable (ID int IDENTITY(1,1),
[system] varchar(8),
OrderNo int);
INSERT INTO dbo.SampleTable([System],OrderNo)
VALUES('system1',1),
('system2',2),
('system3',3),
('system4',4),
('system1',1),
('system2',2),
('system5',1),
('system6',2),
('system7',1),
('system8',2);
GO

WITH Groups AS(
SELECT ID,
[System],
COUNT(CASE WHEN OrderNo = 1 THEN 1 END) OVER (ORDER BY ID ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Grp
FROM dbo.SampleTable)
SELECT [System] AS Parent,
LEAD([System]) OVER (PARTITION BY Grp ORDER BY ID) AS Child
FROM Groups;
GO


Without access to LEAD and ROWS BETWEEN you'll have to be more creative; and the solution is going to be far slower:



WITH Groups AS(
SELECT ST.ID,
ST.[System],
G.Grp
FROM dbo.SampleTable ST
CROSS APPLY (SELECT COUNT(*) AS Grp
FROM dbo.SampleTable CA
WHERE CA.OrderNo =1
AND CA.ID <= ST.ID) G)
SELECT G1.[System] AS Parent,
G2.[System] AS Child
FROM Groups G1
LEFT JOIN Groups G2 ON G1.Grp = G2.Grp
AND G1.ID = G2.ID - 1;






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 24 '18 at 22:10

























answered Nov 24 '18 at 22:03









LarnuLarnu

20.2k51732




20.2k51732













  • It works great on sql server 2008, but without LEAD the output is not the same

    – xoanon
    Nov 28 '18 at 12:40



















  • It works great on sql server 2008, but without LEAD the output is not the same

    – xoanon
    Nov 28 '18 at 12:40

















It works great on sql server 2008, but without LEAD the output is not the same

– xoanon
Nov 28 '18 at 12:40





It works great on sql server 2008, but without LEAD the output is not the same

– xoanon
Nov 28 '18 at 12:40




















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