Viewing 15 posts - 196 through 210 (of 331 total)
Why dont we break it down on an average:
1200rent
400groceries
300gas car
100ins car
200eletric
100gas
300health ins
300car payment
2900total monthly cost
34800total year cost
44544with tax cost tossed in
So on an Average to live in america you need...
July 15, 2011 at 9:53 am
Revenant (7-15-2011)
bopeavy (7/15/2011)
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SQLRNNR (7-15-2011)
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motivation
observation
Conclusion
Success
July 15, 2011 at 9:38 am
SQLRNNR (7-15-2011)
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motivation
observation
July 15, 2011 at 9:13 am
JohnDBA
The easiest thing for you to do is correct the design failure something like this:
alter table #MySample add addessCode varchar(18)
Then update it with the proper values, after that make...
July 15, 2011 at 9:11 am
JohnDBA
froget the address table and look at the mysample table. You have 2 records that have that value. Once again how do you know that the record belong to ZZ...There...
July 15, 2011 at 8:49 am
So how can I determine that based on the info you gave me? There is nothing in the #MySample table letting me know that row belongs to ZZ group.
July 15, 2011 at 8:02 am
JohnDBA
I must join ID with address code where code start with 'A B ' + 9 digits.
There are 3 spaces between A and B, and there is 1 space after...
July 15, 2011 at 7:48 am
J Thaddeus Klopcic
$50k - $75k plus nice benefits isn't enough? Sorry -- our systems can't spin gold from straw.
$50K...
July 15, 2011 at 7:23 am
Brandie Tarvin (7/15/2011)
Daniel Bowlin (7/15/2011)
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perfect
freaky
absurd
July 15, 2011 at 7:16 am
John
did you try my suggestion no reverse at all:
SELECT * FROM #MyAddress a join #mysample s on
Right(replace(a.AddressCode,' ',''),9) = LEFT(s.ID,9)
where replace(a.AddressCode,' ','') like ('AB%')
July 15, 2011 at 7:12 am
Jeff
Careful, now... The code above will only allow each person to live in one place... ever.
So true Jeff thats why I stated not sure what he...
July 15, 2011 at 6:10 am
JohnDBA
I think this is what you wanted:
SELECT * FROM #MyAddress a join #mysample s on
Right(replace(a.AddressCode,' ',''),9) = LEFT(s.ID,9)
where replace(a.AddressCode,' ','') like ('AB%')
July 15, 2011 at 6:02 am
JohnDBA
I must join ID with address code where code start with 'A B ' + 9 digits.
There are 3 spaces between A and B, and there is 1 space after...
July 14, 2011 at 8:38 pm
Viewing 15 posts - 196 through 210 (of 331 total)