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Need to go down path to find if everything is settled (recursive possibly)

I have the following table.
GO
/****** Object: Table [dbo].[itTransactionPr ocess] Script Date:
05/01/2007 10:42:31 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFI ER ON
GO
CREATE TABLE [dbo].[itTransactionPr ocess](
[TransactionID] [int] IDENTITY(1,1) NOT NULL,
[LotNumber] [int] NOT NULL,
[CurrentProcessS tepID] [int] NOT NULL,
[NextProcessStep ID] [int] NULL,
[CategoryID] [int] NULL,
[ProductID] [int] NULL,
[ProductVariantI D] [int] NULL,
[ParentTransacti onID] [int] NULL,
[TransactionDate Entered] [datetime] NULL,
[TransactionDate Exit] [datetime] NULL,
[Settlement] [money] NULL,
[Completed] [int] NULL,
CONSTRAINT [PK_itTransactio nProcess] PRIMARY KEY CLUSTERED
(
[TransactionID] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORE COMPUTE = OFF, IGNORE_DUP_KEY
= OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCK S = ON) ON [PRIMARY]
) ON [PRIMARY]

Sample data is as follows
Basically what I need to do is return the lotid where all path have a
settlement date.

this is my current procedure

/****** Object: StoredProcedure [dbo].
[getPendingSettl ementDetails] Script Date: 05/01/2007 10:47:47
******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFI ER ON
GO
ALTER PROCEDURE [dbo].[getPendingSettl ementDetails]
AS

declare @LotNumbersTabl e table(LotNumber int)
insert into @LotNumbersTabl e EXEC GetPendingSettl ementsLotNumber s

Declare @ResultsTable table(LotNumber int, Company varchar(150),
Contact varchar(150), DateReceived datetime, DateComplete datetime,
SettlementLengt h int)
Declare @LotNumber int

Declare @DateRecieved datetime, @DateComplete datetime
Declare @NumberOfDaysFo rSettlement int
Declare @Company varchar(150)
Declare @Contact varchar(150)

select @LotNumber = min(LotNumber) from @LotNumbersTabl e
while @LotNumber is not null begin
Select @DateRecieved = min(Transaction DateEntered) from
itTransactionPr ocess where LotNumber = @LotNumber
Select @DateComplete = max(Transaction DateExit) from
itTransactionPr ocess where LotNumber = @LotNumber and Settlement is
not null
SET @NumberOfDaysFo rSettlement = DATEDIFF(DAY, @DateRecieved,
@DateComplete)
Select @Company = Company from SP_Active_Lot_D eliveries where LotID =
@LotNumber
Select @Contact = ContactName from SP_Active_Lot_D eliveries where
LotID = @LotNumber
INSERT INTO @ResultsTable (LotNumber, DateReceived, DateComplete,
SettlementLengt h, Company, Contact) Values
(@LotNumber, @DateRecieved, @DateComplete,
@NumberOfDaysFo rSettlement, @company, @contact)
select @LotNumber = min(LotNumber) from @LotNumbersTabl e where
LotNumber @LotNumber
end
Select * From @ResultsTable where SettlementLengt h is not null

here is sample data
"TransactionID" ,"LotNumber","C urrentProcessSt epID","NextProc essStepID","Cat egoryID","Produ ctID","ProductV ariantID","Pare ntTransactionID ","TransactionD ateEntered","Tr ansactionDateEx it","Settlement ","Complete d"
"628","47","1", "2","5","",""," ","","2007-05-01
10:23:15.747000 000","",""
"629","47","1", "3","17","","", "","","2007-05-01
10:23:15.747000 000","0.25",""
"630","47","1", "4","34","","", "","","2007-05-01
10:23:15.747000 000","-0.15",""
"631","47","1", "3","38","","", "","","2007-05-01
10:23:15.747000 000","-0.15",""
"632","47","1", "4","33","","", "","","2007-05-01
10:23:15.747000 000","-0.35",""
"633","47","1", "3","15","","", "","","2007-05-01
10:23:15.747000 000","10",""
"634","47","2", "3","86","","", "628","2007-05-01
10:23:15.747000 000","2007-05-01 10:32:41.320000 000","-0.35",""
"635","47","3", "","17","",""," 629","2007-05-01
10:23:15.747000 000","","",""
"636","47","4", "","34","",""," 630","2007-05-01
10:23:15.747000 000","","",""
"637","47","3", "","38","",""," 631","2007-05-01
10:23:15.747000 000","","",""
"638","47","4", "","33","",""," 632","2007-05-01
10:23:15.747000 000","","",""
"639","47","3", "","15","",""," 633","2007-05-01
10:23:15.747000 000","","",""
"640","47","2", "3","85","","", "628","2007-05-01
10:24:47.983000 000","2007-05-01 10:32:41.320000 000","0.05",""
"641","47","2", "4","88","","", "628","2007-05-01
10:24:56.343000 000","2007-05-01 10:32:41.333000 000","0.8",""
"642","47","2", "4","9","",""," 628","2007-05-01
10:25:07.517000 000","2007-05-01 10:32:41.333000 000","-0.15",""
"643","47","2", "4","100","","" ,"628","2007-05-01
10:25:22.470000 000","2007-05-01 10:32:41.333000 000","-0.35",""
"644","47","2", "4","90","","", "628","2007-05-01
10:25:44.297000 000","2007-05-01 10:32:41.333000 000","-0.35",""
"645","47","2", "4","12","","", "628","2007-05-01
10:25:59.347000 000","2007-05-01 10:32:41.333000 000","-0.15",""
"646","47","2", "4","26","","", "628","2007-05-01
10:26:12.610000 000","2007-05-01 10:32:41.333000 000","-0.35",""
"647","47","2", "3","94","","", "628","2007-05-01
10:26:29.523000 000","2007-05-01 10:32:41.333000 000","-3",""
"648","47","2", "3","95","","", "628","2007-05-01
10:26:47.323000 000","2007-05-01 10:32:41.333000 000","-0.35",""
"649","47","2", "3","38","","", "628","2007-05-01
10:27:01.450000 000","2007-05-01 10:32:41.333000 000","-0.15",""
"650","47","2", "4","33","","", "628","2007-05-01
10:27:15.533000 000","2007-05-01 10:32:41.333000 000","-0.35",""
"651","47","2", "4","34","","", "628","2007-05-01
10:27:33.767000 000","2007-05-01 10:32:41.333000 000","-0.15",""
"652","47","2", "3","96","","", "628","2007-05-01
10:27:46.850000 000","2007-05-01 10:32:41.350000 000","-0.35",""
"653","47","2", "3","97","","", "628","2007-05-01
10:28:00.917000 000","2007-05-01 10:32:41.350000 000","0.05",""
"654","47","2", "4","36","","", "628","2007-05-01
10:28:10.813000 000","2007-05-01 10:32:41.350000 000","-15",""
"655","47","2", "4","37","","", "628","2007-05-01
10:28:25.347000 000","2007-05-01 10:32:41.350000 000","0.35",""
"656","47","2", "3","98","","", "628","2007-05-01
10:28:36.917000 000","2007-05-01 10:32:41.350000 000","-0.35",""
"694","47","2", "10","26","","" ,"628","2007-05-01
10:32:17.170000 000","2007-05-01 10:32:41.350000 000","",""
"695","47","2", "10","35","","" ,"628","2007-05-01
10:32:27.883000 000","2007-05-01 10:32:41.350000 000","45",""
"696","47","3", "","86","",""," 634","2007-05-01
10:32:41.320000 000","","",""
"697","47","3", "","85","",""," 640","2007-05-01
10:32:41.333000 000","","",""
"698","47","4", "","88","",""," 641","2007-05-01
10:32:41.333000 000","","",""
"699","47","4", "","9","","","6 42","2007-05-01
10:32:41.333000 000","","",""
"700","47","4", "","100","","", "643","2007-05-01
10:32:41.333000 000","","",""
"701","47","4", "","90","",""," 644","2007-05-01
10:32:41.333000 000","","",""
"702","47","4", "","12","",""," 645","2007-05-01
10:32:41.333000 000","","",""
"703","47","4", "","26","",""," 646","2007-05-01
10:32:41.333000 000","","",""
"704","47","3", "","94","",""," 647","2007-05-01
10:32:41.333000 000","","",""
"705","47","3", "","95","",""," 648","2007-05-01
10:32:41.333000 000","","",""
"706","47","3", "","38","",""," 649","2007-05-01
10:32:41.333000 000","","",""
"707","47","4", "","33","",""," 650","2007-05-01
10:32:41.333000 000","","",""
"708","47","4", "","34","",""," 651","2007-05-01
10:32:41.333000 000","","",""
"709","47","3", "","96","",""," 652","2007-05-01
10:32:41.350000 000","","",""
"710","47","3", "","97","",""," 653","2007-05-01
10:32:41.350000 000","","",""
"711","47","4", "","36","",""," 654","2007-05-01
10:32:41.350000 000","","",""
"712","47","4", "","37","",""," 655","2007-05-01
10:32:41.350000 000","","",""
"713","47","3", "","98","",""," 656","2007-05-01
10:32:41.350000 000","","",""
"714","47","10" ,"","26","","", "694","2007-05-01
10:32:41.350000 000","","",""
"715","47","10" ,"","35","","", "695","2007-05-01
10:32:41.350000 000","","",""
If you follow transaction id 714 up through the parent transaction ids
it doesn't not have a settlement cost yet lot 47 shows up as settled.

Thanks for you help.

May 1 '07 #1
6 1717
The procedure I entered above is incorrect.

here is the correct one.

ALTER Procedure [dbo].[GetPendingSettl ementsLotNumber s]
(
@CurrentParentT ransactionID int = 0
)
as
/*------------------------------------------------------------
Lists the contents of a table designed to represent a multi-
branch tree. The result set takes the form:
ItemID TreeLevel Label
Tree traversal is done non-recursively (to avoid SQL Server's
limit of 32 nested procedure calls)
------------------------------------------------------------*/
--table to hold the result set: a tree structure arranged by level
declare @IndentedTree table(Transacti onID int, ParentTransacti onID
int, TreeLevel int, LotNumber int, SettlementPrice decimal(32,9))
--table to track where we are in the tree
--this represents a stack turned upside down (with the most resent
item on
--the bottom)
declare @UnvisitedNodes table(StackID int identity(1,1), TransactionID
int, ParentTransacti onID int, TreeLevel int, LotNumber int,
SettlementPrice decimal(32,9))
declare @LastTreeLevel int
set nocount on
--initialize the unvisited nodes list
set @LastTreeLevel = 0
INSERT INTO @UnvisitedNodes (TransactionID, ParentTransacti onID,
TreeLevel, LotNumber, SettlementPrice )
--SELECT TransactionID, @LastTreeLevel, Label
--FROM tree
--WHERE ParentTransacti onID = @CurrentParentT ransactionID
--ORDER BY Label desc
Select TransactionID, ParentTransacti onID, @LastTreeLevel,
LotNumber, Settlement
From itTransactionPr ocess
Where ParentTransacti onID = @CurrentParentT ransactionID and
Settlement is null
ORDER BY ParentTransacti onID desc
--Select * from @UnvisitedNodes
--loop through levels of the tree structure
while ((SELECT count(*) FROM @UnvisitedNodes ) <0)
begin
--add the top item to the result set
INSERT INTO @IndentedTree
SELECT TransactionID, ParentTransacti onID, TreeLevel, LotNumber,
SettlementPrice
FROM @UnvisitedNodes
WHERE StackID = (SELECT max(StackID) FROM @UnvisitedNodes )
--Select * from @IndentedTree
Delete From @IndentedTree where TransactionID in (Select
ParentTransacti onID from itTransactionPr ocess)

--get the top item's ID
set @CurrentParentT ransactionID = (
SELECT TransactionID
FROM @UnvisitedNodes
WHERE StackID = (SELECT max(StackID) FROM @UnvisitedNodes ))

--get the top item's indentation level
set @LastTreeLevel = (
SELECT TreeLevel
FROM @UnvisitedNodes
WHERE StackID = (SELECT max(StackID) FROM @UnvisitedNodes ))
--delete that item from the list

DELETE FROM @UnvisitedNodes WHERE TransactionID =
@CurrentParentT ransactionID

--add the children of the current item to the top of the list of
unvisited
--nodes
INSERT INTO @UnvisitedNodes (TransactionID, ParentTransacti onID,
TreeLevel, LotNumber)
--SELECT TransactionID, @LastTreeLevel + 1, LotNumber
--FROM tree
--WHERE ParentTransacti onID = @CurrentParentT ransactionID
--ORDER BY LotNumber desc
Select TransactionID, ParentTransacti onID, @LastTreeLevel + 1,
LotNumber
From itTransactionPr ocess
Where ParentTransacti onID = @CurrentParentT ransactionID and
Settlement is null
ORDER BY ParentTransacti onID desc
end
--return the result set
Select LotId From ItLots where LotID not in
(SELECT LotNumber
FROM @IndentedTree) and LotID in (Select LotNumber from
itTransactionPr ocess)

May 1 '07 #2
Please disreguard post.

Everything work but when I moved the database over to a new server the
default values didn't stick to so the parent trasactionid was never
initiallized to 0.

Thanks,

May 1 '07 #3
Designing Solutions WD (mi************ **@gmail.com) writes:
Please disreguard post.

Everything work but when I moved the database over to a new server the
default values didn't stick to so the parent trasactionid was never
initiallized to 0.
Which version of SQL Server are you? If you are on SQL 2005, you should
look into a new features know as Common Table Expressions, or CTEs for
short. A special form is recursive CTE which permits you to handle a
hierarchy in a single statement. Look up the topic "WITH common table
expression" in Books Online for further details.
--
Erland Sommarskog, SQL Server MVP, es****@sommarsk og.se

Books Online for SQL Server 2005 at
http://www.microsoft.com/technet/pro...ads/books.mspx
Books Online for SQL Server 2000 at
http://www.microsoft.com/sql/prodinf...ons/books.mspx
May 1 '07 #4
> Lists the contents of a table designed to represent a multi-branch tree. The result set takes the form: ItemID TreeLevel Label Tree traversal is done non-recursively (to avoid SQL Server's limit of 32 nested procedure calls) <<

Have you gotten a copy of TREES & HIERARCHIES IN SQL? There are
several ways to model trees without *any* procedural traversal code at
all.

You also have more NULL-able columns int his one table than the entire
payroll data base of a major automobile.

You used an IDENTITY as a key; is the real key (lot_nbr,
process_step) ?

Many of your data element names are just plain wrong. Think how silly
"category_i d" is for an attribute; it is either "<some kind
of>_category" or "<some kind of>_id", but not both an identifier
(unique to one and only one instance of an entity) and a category
(value appears in many entities). Ditto monsters like
"current_proces s_step_id" as opposed to a mere "process_st ep" or
"process_id " instead. Get a copy of ISO-11179 rules for data element
names.

Learn what data and meta data are so you will not mix them in the
table.

May 2 '07 #5


"--CELKO--" <jc*******@eart hlink.netwrote in message
news:11******** **************@ o5g2000hsb.goog legroups.com...
>> Lists the contents of a table designed to represent a multi-branch
tree. The result set takes the form: ItemID TreeLevel Label Tree
traversal is done non-recursively (to avoid SQL Server's limit of 32
nested procedure calls) <<

Have you gotten a copy of TREES & HIERARCHIES IN SQL? There are
several ways to model trees without *any* procedural traversal code at
all.

You also have more NULL-able columns int his one table than the entire
payroll data base of a major automobile.
Now I'm curious, how many cars have a payroll database. :-)


--
Greg Moore
SQL Server DBA Consulting Remote and Onsite available!
Email: sql (at) greenms.com http://www.greenms.com/sqlserver.html
May 2 '07 #6
On May 1, 11:00 pm, "Greg D. Moore \(Strider\)"
<mooregr_delete t...@greenms.co mwrote:
"--CELKO--" <jcelko...@eart hlink.netwrote in message

news:11******** **************@ o5g2000hsb.goog legroups.com...
> Lists the contents of a table designed to represent a multi-branch
tree. The result set takes the form: ItemID TreeLevel Label Tree
traversal is done non-recursively (to avoid SQL Server's limit of 32
nested procedure calls) <<
Have you gotten a copy of TREES & HIERARCHIES IN SQL? There are
several ways to model trees without *any* procedural traversal code at
all.
You also have more NULL-able columns int his one table than the entire
payroll data base of a major automobile.

Now I'm curious, how many cars have a payroll database. :-)
You gotta be careful... they'll try to sneak one in with their option
packages...

May 2 '07 #7

This thread has been closed and replies have been disabled. Please start a new discussion.

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