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Posted by Mikhail Kovalev on 12/05/07 20:21
On 5 Des, 20:46, Norman Peelman <npeel...@cfl.rr.com> wrote:
> Mikhail Kovalev wrote:
> > On 5 Des, 18:42, Norman Peelman <npeel...@cfl.rr.com> wrote:
>
> <snipped>
>
>
>
> >> INSERT INTO nodes (node_address, node_count) VALUES ($node_address,
> >> $node_count) ON DUPLICATE KEY UPDATE node_count=node_count+1
>
> >> ...will INSERT new entries and UPDATE existing entries in one swoop.
>
> >> Norm
>
> > Ok, suppose I'm joining to structures which have been created
> > separately,
> > from before i have ('112/225/930', 3)
> > and i want to add ('112/225/930', 2), which also happens to be present
> > in the second table, only with a different count,
> > to make ('112/225/930', 5)
>
> > From there I want to make it the general case so that when updating an
> > entry with count 1 I am actually adding ('112/225/930', 1) to the
> > existing one, if it exists:
>
> > INSERT INTO nodes (node_address, node_count) VALUES (<new_address>,
> > <new_count>) ON DUPLICATE KEY UPDATE node_count = node_count +
> > <new_count>
>
> > Is this correct? (Do I have to use <> in VALUES, I'm following an
> > example which does it?)
>
> No replace everything '<..>' with your own variable name, that will do it.
>
> $new_count = 5
> ...ON DUPLICATE KEY UPDATE node_count = $new_count
>
> Not sure what you mean by 'second table'. How many do you have? Though
> this was one table...
>
> Norm
I have different databases for different types of sequences. Each
sequence is some tousands elements in length and is recorded by chunks
of small series each from 1 to 20 elements long. Sometimes I join
different databases together and analyze each one separately and then
in relation to the joint database.
I have been thinking, if a sequence is for instance 5.000 elements
long, chunks 1-20, and there are no duplicates (absolutely unlikely,
but still) I will end up having a database with around 105.000
entries...
My current solution is probably better because it saves so much
space(?)
An entry like this in the current solution:
112 (310)
225 (20)
930 (6)
700 (1)
7 (1)
812 (1)
Equals this in the new system I am about to implement:
('112', 310)
('112/225', 20)
('112/225/930', 6)
('112/225/930/700', 1)
('112/225/930/700/7', 1)
('112/225/930/812', 1)
I don't know how a flat MySQL database of 100.000-200.000 entries like
these will perform. I'm now also considering a filesystem as the
database, as C has suggested.
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