Is this the correct windows for my mac?
I hope i'm not being really annoying with posting so many questions....
This site, http://support.apple.com/kb/HT5634?viewlocale=en_US&locale=en_US , says for my mac, the MacBook Pro 13-inch, Late 2011 should work for Windows 8 64bit, windows 7 64 bit and winodws 7 32 bit.
With that in mind, I chose what seems to be the newest, the windows 8 64 bit.
I found this on amazon and just wanted to make sure this is what i need. is this the full install cd and not just an upgrade? http://www.amazon.com/Windows-System-Builder-OEM-64-Bit/dp/B0094NY3R0/ref=sr_1_1 ?ie=UTF8&qid=1377372064&sr=8-1&keywords=windows+8+64+bit+full
i tried looking, but im not very tech savy, so i just got confused....
It is annoying to not just ask in the same thread you had already...
Title suggestion: Windows 8 OEM System Builder
That being said, 8.1 is due in October. Not sure if system builder will allow or not, so you might want to shop Microsoft for retail or wait for "buy now and get free upgrade"
8.1 runs smoother and nicer than 7 once you turn off Metro (which 8.1 makes easy).
Links to Amazon: anything from "ref" on are just extra and not needed. Search terms, user wish list location and ID.
Windows 8.1 Preview is free download but I think you need to be running 7 or 8 already, try:
http://www.windowsstore.com/
http://msdn.microsoft.com/windows/apps/bg182409
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its a shame to find Apple deletes posts that talks about the disappointing support offered by their partners...
is this the real forum for genuine mac supporters?
Since no one here works for Apple and we all volunteer our time, yes, this is a forum for genuine Mac supporters. However, it is hosted by Apple and we have to adhere to their rules. Your post was not a technical support question that anyone here could answer. It may have been deleted before you saw my reply, but I suggested you write a letter directly to Apple. It's extremely unlikely anyone who could address your issues would be reading this forum.
http://www.apple.com/contact/
Apple
1 Infinite Loop
Cupertino, CA 95014
Also, in the future, if you have questions about the functionality of the Apple Discussions themselves, like a question about a deleted post, you might post in the Feedback About Discussions forum, as this post is also not a technical question that any of us can answer and may also be inappropriate:
http://discussions.apple.com/forum.jspa?forumID=1076
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riley2 wrote:
Re: Is this the correct forum for iMovie HD 6.0.3 ?
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Peter -
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Have you checked System Preferences > Sound > Output?
Or are you interested in a phone modem adapter?
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Can't figure out the correct syntax for this select statement
Hello,
The following statement works great and gives the desired results:
prompt
prompt Using WITH t
prompt
with t as
select a.proj_id,
a.proj_start,
a.proj_end,
case when (
select min(a.proj_start)
from v b
where (a.proj_start = b.proj_end)
and (a.proj_id != b.proj_id)
is not null then 0 else 1
end as flag
from v a
order by a.proj_start
select proj_id,
proj_start,
proj_end,
flag,
-- the following select statement is what I am having a hard time
-- "duplicating" without using the WITH clause
select sum(t2.flag)
from t t2
where t2.proj_end <= t.proj_end
) s
from t;As an academic exercise I wanted to rewrite the above statement without using the WITH clause, I tried this (among dozens of other tries - I've hit a mental block and can't figure it out):
prompt
prompt without with
prompt
select c.proj_id,
c.proj_start,
c.proj_end,
c.flag,
-- This is what I've tried as the equivalent statement but, it is
-- syntactically incorrect. What's the correct syntax for what this
-- statement is intended ?
select sum(t2.flag)
from c t2
where t2.proj_end <= c.proj_end
) as proj_grp
from (
select a.proj_id,
a.proj_start,
a.proj_end,
case when (
select min(a.proj_start)
from v b
where (a.proj_start = b.proj_end)
and (a.proj_id != b.proj_id)
is not null then 0 else 1
end as flag
from v a
order by a.proj_start
) c;Thank you for helping, much appreciated.
John.
PS: The DDL for the table v used by the above statements is:
drop table v;
create table v (
proj_id number,
proj_start date,
proj_end date
insert into v values
( 1, to_date('01-JAN-2005', 'dd-mon-yyyy'),
to_date('02-JAN-2005', 'dd-mon-yyyy'));
insert into v values
( 2, to_date('02-JAN-2005', 'dd-mon-yyyy'),
to_date('03-JAN-2005', 'dd-mon-yyyy'));
insert into v values
( 3, to_date('03-JAN-2005', 'dd-mon-yyyy'),
to_date('04-JAN-2005', 'dd-mon-yyyy'));
insert into v values
( 4, to_date('04-JAN-2005', 'dd-mon-yyyy'),
to_date('05-JAN-2005', 'dd-mon-yyyy'));
insert into v values
( 5, to_date('06-JAN-2005', 'dd-mon-yyyy'),
to_date('07-JAN-2005', 'dd-mon-yyyy'));
insert into v values
( 6, to_date('16-JAN-2005', 'dd-mon-yyyy'),
to_date('17-JAN-2005', 'dd-mon-yyyy'));
insert into v values
( 7, to_date('17-JAN-2005', 'dd-mon-yyyy'),
to_date('18-JAN-2005', 'dd-mon-yyyy'));
insert into v values
( 8, to_date('18-JAN-2005', 'dd-mon-yyyy'),
to_date('19-JAN-2005', 'dd-mon-yyyy'));
insert into v values
( 9, to_date('19-JAN-2005', 'dd-mon-yyyy'),
to_date('20-JAN-2005', 'dd-mon-yyyy'));
insert into v values
(10, to_date('21-JAN-2005', 'dd-mon-yyyy'),
to_date('22-JAN-2005', 'dd-mon-yyyy'));
insert into v values
(11, to_date('26-JAN-2005', 'dd-mon-yyyy'),
to_date('27-JAN-2005', 'dd-mon-yyyy'));
insert into v values
(12, to_date('27-JAN-2005', 'dd-mon-yyyy'),
to_date('28-JAN-2005', 'dd-mon-yyyy'));
insert into v values
(13, to_date('28-JAN-2005', 'dd-mon-yyyy'),
to_date('29-JAN-2005', 'dd-mon-yyyy'));
insert into v values
(14, to_date('29-JAN-2005', 'dd-mon-yyyy'),
to_date('30-JAN-2005', 'dd-mon-yyyy'));Hi, John,
Not that you asked, but as you proabably know, analytic functions are much better at doing this kind of thing.
You may be amazed (as I continually am) by how simple and efficient these queries can be.
For example:
WITH got_grp AS
SELECT proj_id, proj_start, proj_end
, proj_end - SUM (proj_end - proj_start) OVER (ORDER BY proj_start) AS grp
FROM v
SELECT ROW_NUMBER () OVER (ORDER BY grp) AS proj_grp
, MIN (proj_start) AS proj_start
, MAX (proj_end) AS proj_end
FROM got_grp
GROUP BY grp
ORDER BY proj_start
;Produces the results you want:
PROJ_GRP PROJ_START PROJ_END
1 01-Jan-2005 05-Jan-2005
2 06-Jan-2005 07-Jan-2005
3 16-Jan-2005 20-Jan-2005
4 21-Jan-2005 22-Jan-2005
5 26-Jan-2005 30-Jan-2005This is problem is an example of Neighbor-Defined Groups . You want to GROUP BY something that has 5 distinct values, to get the 5 rows above, but there's nothing in the table itself that tells you to which group each row belongs. The groups are not defined by any column in hte table, but by relationships between rows. In this case, a row is in the same group as its neighbor (the row immediatly before or after it when sorted by proj_start or proj_end) if proj_end of the earlier row is the same as proj_start of the later row. That is, there is nothing about 03-Jan-2005 that says the row with proj_id=2 is in the first group, or even that it is in the same group with its neighbor, the row with proj_id=3. Only the relation between those rows, the fact that the earlier row has end_date=03-Jan-2005 and the later row has start_date=03-Jan-2003, that says these neighbors belong to the same group.
You're figuring out when a new group starts, and then counting how many groups have already started to see to which group each row belongs. That's a prefectly natural procedural way of approaching the problem. But SQL is not a procedural language, and sometimes another approach is much more efficient. In this case, as in many others, a Constant Difference defines the groups. The difference between proj_end (or proj_start, it doesn't matter in this case) and the total duratiojn of the rows up to that date determines a group. The actual value of that difference means nothing to you or anybody else, so I used ROW_NUMBER in the query above to map those distinct values into consecutive integers 1, 2, 3, ... which are a much simpler way to identify the groups.
Note that the query above only requires one pass through the table, and only requires one sub-query. It does not need a WITH clause; you could easily make got_grp an in-line view.
If you used analytic functions (LEAD or LAG) to compute flag, and then to compute proj_grp (COUNT or SUM), you would need two sub-queries, one for each analytic function, but you would still only need one pass through the table. Also, those sub-queries could be in-line views; yiou would not need to use a WITH clause. -
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Thank you. -
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