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Views: 569 | Replies: 5
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Join Date: Sep 2007
Posts: 13
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Newbie Poster
Hi
can any one help me in following case?
Table A
ID | Name
1 , Apple
2 , Bat
3 , Cat
4 , Dog
Table B
ID | Name
1 , A
2 , B
3 , C
4 , D
5 , E
6 , F
7 , G
How will i get the below result??
Results
ID | Name
1 , Apple
2 , Bat
3 , Cat
4 , Dog
5 , E
6 , F
7 , G
How can I achieve this.
Thanks
can any one help me in following case?
Table A
ID | Name
1 , Apple
2 , Bat
3 , Cat
4 , Dog
Table B
ID | Name
1 , A
2 , B
3 , C
4 , D
5 , E
6 , F
7 , G
How will i get the below result??
Results
ID | Name
1 , Apple
2 , Bat
3 , Cat
4 , Dog
5 , E
6 , F
7 , G
How can I achieve this.
Thanks
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Join Date: Jul 2005
Location: Dallas, TX
Posts: 481
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Rep Power: 4
Solved Threads: 19
Posting Pro in Training
ok, i am assuming that you want a name for all id's. if a has an entry use that, otherwise use the name from b. also assuming that B has an entry for every id. if those assumptions are correct, then this should work
select b.id, isnull(a.name,b.name) as [Name] from b left join a on b.id = a.id
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Join Date: Feb 2005
Location: Braintree, UK
Posts: 1,165
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Solved Threads: 59
Or maybe:
select b.id, isnull(a.name,b.name) as [Name] from b left join a on b.[Name] = LEFT(a.[Name],1)
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Join Date: Jul 2005
Location: Dallas, TX
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Posting Pro in Training
that would add the assumption that the names in b are the first letter of the names in a
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Join Date: Feb 2005
Location: Braintree, UK
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Hmm not really, it replaces the assumption that a.id is a foreign key to b.id.
The OP does not specify. If there is a relation between the id's (which there should be in my opinion, and if the data provided represents all possibilities, then this is indeed the case) your solution is the more efficient.
However if the id's are not related then that only leaves the initial letters, in which case my alternate solution will do.
The OP does not specify. If there is a relation between the id's (which there should be in my opinion, and if the data provided represents all possibilities, then this is indeed the case) your solution is the more efficient.
However if the id's are not related then that only leaves the initial letters, in which case my alternate solution will do.
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Join Date: Sep 2007
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Originally Posted by campkev 
ok, i am assuming that you want a name for all id's. if a has an entry use that, otherwise use the name from b. also assuming that B has an entry for every id. if those assumptions are correct, then this should work
select b.id, isnull(a.name,b.name) as [Name] from b left join a on b.id = a.id
Thanks to reply my query. It worked.
Last edited by bhakti.thakkar : Jan 27th, 2008 at 11:36 pm.
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