You might save some space if you compress() before storing. Depending on
file content I'm seeing 0-50% savings?
select length(load_file('c:/temp/SomeFile.pdf')) as old_size
,length(compress(load_file('c:/temp/SomeFile.pdf'))) as new_size
Ed
-----Original Message-----
From: Michael Stassen [mailto:[EMAIL PROTECTED]
Sent: Tuesday, June 22, 2004 11:01 PM
To: Justin Crone
Cc: [EMAIL PROTECTED]
Subject: Re: Some BLOB help please.
Justin Crone wrote:
<snip>
>
> So as I said, I am rather pleased with the performance and the ease at
> which I was able to get this up and running. However.... The problem
is
> I do have limits, and one of those being disk space. Those 10,000
files
> are taking up 21 GB of space in the database. However the actual space
> required by the files is around 5GB on the file system. The average
file
> size is about 1.9MB, so it would seem that each row inserted into the
> database is conforming to that 1.9MB average, giving me this 21GB
table.
Could you explain that again? If average file size is 1.9Mb, then 21 Gb
for
10,780 files is about right. On the other hand, if the total is 5 Gb,
then
5Gb/10,780 yields about .47 Mb average per file. So which is it? Do
your
files average 1.9Mb, in which case we must wonder how you stored them in
only 5Gb, or do your files average .47 Mb, in which case we must figure
out
why they are roughly 4 times as big when stored as BLOBs? (The manual
<http://dev.mysql.com/doc/mysql/en/Storage_requirements.html> says BLOBs
take length + 2bytes to store .)
> I would like to know if there is something that I can change to get
these
> numbers in line with each other, or if this is just the way of things.
> Current projections for the total documents needed to complete the
rotation
> of these files is 720,000 documents. Which if the 1.9MB average keeps,
that
> puts me in the neighborhood of 1.4TB of storage.
Even at .47Mb per file, that's about 330 Gb of storage required just for
the
720,000 files.
Michael
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