Solved

SQL Select query

Posted on 2014-04-30
3
145 Views
Last Modified: 2014-04-30
I have a table such as the table below.   The BegMTypw identifies where a particular feature begins, EndMType identifies where a particular feature ends.  CasingID is the foreignkey to the Feature.  
ID      Coordinate      BegMType      EndMType      CasingID
1      0                      Casing                                             1
2      5                  
3      7                  
4      8                  
5      9                  
6      10                                               Casing                       1
7      20                  
8      30                  
9      35                  
10      50                       Casing                                             1
11      51                  
12      62                  
13      100                                               Casing                       1


My output should look like

BegM    EndM     CasingID
0            10            1
50           100          1

The coordinates will always be entered in order.  Any ideas on how I can get the output above?  I can't do min and max because the CasingID is not unique, this would yeild

BegM   EndM   CasingID
0             100        1

This result would be incorrect, I need the two records for beg and end.
0
Comment
Question by:yanci1179
3 Comments
 
LVL 5

Accepted Solution

by:
dannygonzalez09 earned 500 total points
ID: 40032386
Something like this?
SELECT * INTO #Test
FROM
(
SELECT Id = 1,CoOrdinate = 0, BegMType = 'Casing',EndMtype = '',CasingId = 1
UNION
SELECT Id = 2,CoOrdinate = 5, BegMType = '',EndMtype = '',CasingId = ''
UNION
SELECT Id = 3,CoOrdinate = 7, BegMType = '',EndMtype = '',CasingId = ''
UNION
SELECT Id = 4,CoOrdinate = 10, BegMType = '',EndMtype = 'Casing',CasingId = 1
UNION
SELECT Id = 5,CoOrdinate = 50, BegMType = 'Casing',EndMtype = '',CasingId = 1
UNION
SELECT Id = 6,CoOrdinate = 100, BegMType = '' ,EndMtype = 'Casing',CasingId = 1

)x

With BegCTE AS
(
SELECT CoOrdinate,CasingId,RANK() OVER (PARTITION BY BegMType ORDER BY Id) BegRnk FROM #Test
WHERE LEN(BegMType) > 1
)
,EndCTE AS
(
SELECT CoOrdinate,CasingId,RANK() OVER (PARTITION BY EndMType ORDER BY Id) EndRnk FROM #Test
WHERE LEN(EndMType) > 1
)

SELECT B.CoOrdinate,E.CoOrdinate,B.CasingId FROM BegCTE B
LEFT JOIN EndCTE E
ON B.BegRnk = E.EndRnk and B.CasingId = E.CasingId

Open in new window

0
 
LVL 40

Expert Comment

by:Kyle Abrahams
ID: 40032387
you could look at the rank() function.

http://technet.microsoft.com/en-us/library/ms176102.aspx

it's a CTE with a join

something like


;with cte as
(select *, rank() over (Partition By CasingID order by coordinate) myRank
  from table
  where isnull(BegMType, '') != ''  or isnull(EndMType, '') != ''
)  


select  B.Coordinate, E.Coordinate, B.CasingId
from cte B
join cte E on B.CasingID = E.CasingID and E.MyRank = B.MyRank +1 and isnull(B.BegMType, '') != ''
0
 

Author Closing Comment

by:yanci1179
ID: 40033216
Thanks!!
0

Featured Post

PRTG Network Monitor: Intuitive Network Monitoring

Network Monitoring is essential to ensure that computer systems and network devices are running. Use PRTG to monitor LANs, servers, websites, applications and devices, bandwidth, virtual environments, remote systems, IoT, and many more. PRTG is easy to set up & use.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Nowadays, some of developer are too much worried about data. Who is using data, who is updating it etc. etc. Because, data is more costlier in term of money and information. So security of data is focusing concern in days. Lets' understand the Au…
Let's review the features of new SQL Server 2012 (Denali CTP3). It listed as below: PERCENT_RANK(): PERCENT_RANK() function will returns the percentage value of rank of the values among its group. PERCENT_RANK() function value always in be…
Via a live example, show how to extract insert data into a SQL Server database table using the Import/Export option and Bulk Insert.
Viewers will learn how to use the SELECT statement in SQL and will be exposed to the many uses the SELECT statement has.

910 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

23 Experts available now in Live!

Get 1:1 Help Now