Re: how to efficiently query for the next in MySQL Community Edition 5.1.34?

[Peter Brawley]

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, 
> 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 holding the next integer that T has for S (U having no 
> row for the last integer of each string).  I could index T on (S,I) and 
> write this query as
>
> select t1.*, t2.I as J from T as t1, T as t2
> where t1.S=t2.S and t1.I < t2.I
> and not exists (select * from T as t12 where t12.S=t1.S and t1.I < t12.I 
> and t12.I < t2.I)
>
> but the query planner says this is quite expensive to run: it will 
> enumerate all of T as t1, do a nested enumeration of all t2's entries for 
> S=t1.S, and inside that do a further nested enumeration of t12's entries 
> for S=t1.S --- costing about 10,000 times the size of T.  There has to be 
> a better way!
>
> Thanks,
> Mike Spreitzer
>
>   
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