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9.0

SQL Query - Identifying unique sender/receiver pairs

Asked by ardm in SQL Server 2005, SQL Server 2008

Tags: sql server 2005

I've been racking my brain trying to figure out a solution for a seemingly simple problem... it's a Friday so I guess I'm excused, if anyone can help I'd be grateful!

Given an SQL table which contains a list of messages, with sender/receiver columns identifying where the message is from (sentFrom) and where it went to (rcptTo)...

I'm trying to identify unique sender/receiver pairs from the above, regardless of whether someone is listed in sentFrom or rcptTo...

i.e. {A,B} and {B,A} should both be identified as "A<--->B"

So desired output given the table and sample values in the code block would be something like:
A<--->B
A<--->C
D<--->E
F<--->G
C<--->H

Formatting doesn't matter!

I also don't mind if "A<--->B" is identified as "B<--->A" instead (as long as they're not both there). The bottom line is I'm trying to establish that "A and B are having some sort of conversation, whether they are both talking or only one of them is"... etc for the rest of the pairs.

Thanks!

-Alex

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create table #test (
	sentFrom [varchar](20) null,
	rcptTo [varchar](20) null
)
 
insert #test (sentFrom, rcptTo) values ('A', 'B')
insert #test (sentFrom, rcptTo) values ('A', 'B')
insert #test (sentFrom, rcptTo) values ('A', 'B')
insert #test (sentFrom, rcptTo) values ('B', 'A')
insert #test (sentFrom, rcptTo) values ('A', 'C')
insert #test (sentFrom, rcptTo) values ('A', 'C')
insert #test (sentFrom, rcptTo) values ('C', 'A')
insert #test (sentFrom, rcptTo) values ('C', 'A')
insert #test (sentFrom, rcptTo) values ('D', 'E')
insert #test (sentFrom, rcptTo) values ('E', 'D')
insert #test (sentFrom, rcptTo) values ('F', 'G')
insert #test (sentFrom, rcptTo) values ('C', 'H')
[+][-]10/30/09 09:58 AM, ID: 25704847Expert Comment

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[+][-]10/30/09 10:09 AM, ID: 25704954Author Comment

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[+][-]10/30/09 10:53 AM, ID: 25705323Expert Comment

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[+][-]10/30/09 10:54 AM, ID: 25705327Assisted Solution

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[+][-]10/30/09 11:01 AM, ID: 25705389Accepted Solution

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About this solution

Zones: SQL Server 2005, SQL Server 2008
Tags: sql server 2005
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Solution Provided By: RajkumarGS
Participating Experts: 5
Solution Grade: A
 
[+][-]10/30/09 11:06 AM, ID: 25705440Assisted Solution

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[+][-]10/30/09 11:13 AM, ID: 25705487Expert Comment

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