ve another factor of 100. How do
I save (even a portion of) that?
Thanks,
Mike Spreitzer
Peter Brawley
06/20/09 06:56 PM
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mspre...@us.ibm.com, Lotus Notes: Mike Spreitzer/Watson/IBM
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*Peter Brawley *
06/20/09 03:59 PM
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Re: how to efficiently query for the next in MySQL Community Edition
5.1.34?
>I do not follow why you suggested a join to get the associated S,
>that can be done in the original query (and I did NOT say a given
>integer I is ass
ke Spreitzer
*Peter Brawley *
06/20/09 12:39 PM
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5.1.34?
Mike,
>Y
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5.1.34?
Mike,
>Yes, for each (S, I) pair the goal is to efficiently find the next
largest
>integer associated with S in T. For the highest integer I associated
with
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Re: how to efficiently query for the next in MySQL Community Edition
5.1.34?
Mike
J holding the next integer that T has for S
You mean for each i, the next value of i with that s?
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Re: how to efficiently query for the next in MySQL Community Edition
5.1.34?
Mike
>J holding the next integer that T has for S
You mean for each i, the next value of i with that s?
>(U having no row for the last inte
Huh???
On Saturday, June 20, 2009, Peter Brawley wrote:
> Mike
>
>>J holding the next integer that T has for S
>
> You mean for each i, the next value of i with that s?
>
>>(U having no row for the last integer of each string).
>
> I do not understand that at all.
>
> PB
>
>
> Mike Spreitzer wrot
Mike
>J holding the next integer that T has for S
You mean for each i, the next value of i with that s?
>(U having no row for the last integer of each string).
I do not understand that at all.
PB
Mike Spreitzer wrote:
Suppose I have a table T with two column, S holding strings (say,
VARCHA
Suppose I have a table T with two column, S holding strings (say,
VARCHAR(200)) and I holding integers. No row appears twice. A given
string appears many times, on average about 100 times. Suppose I have
millions of rows. I want to make a table U holding those same columns
plus one more, J
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