Anybody can help me understand if the Ipads becomes ready to be used in all world regions with 3G and or 4G infrastructure? I mean, are all of them released or when I buy one in the US to be used in Argentina I need to ask for a released one, less cheaper
anybody can help me understand if the Ipads becomes ready to be used in all world regions with 3G and or 4G infrastructure? I mean, are all of them released or when I buy one in the US to be used in Argentina I need to ask for a released one, with add costs?
There are two versions of the current iPad, the WiFi only (which will work anywhere in the world, but only connect to WiFi networks) and the 3G/4G model. The latter will connect to 3G networks worldwide, as I understand it, but the only 4G networks it can connect to are in the US and Canada.
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Hi all, I finally solved the issue. Here I wish to share with you all. First, I delete my hotmail account from mail account. Secondly I reboot my ipad2 with holding sleep and home button at once untill the apple logo is appeared. Then I add back the hotmail account. Done. Hope it helps. Thank you.
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Hi talero,
Contact AppleCare or bring it into an Apple Store or AASP. -
Hi i don't no if anyone can help me, i have a ipad 1 and have been trying to upgrade it to the latest version and cant find anywhere. on my phone it is in the settings to upgrade, iv been online and looked but cant find any link. can someone send me a link or direct me the right way many thanks.
Upgrading the iOS "over-the-air" was introduced with iOS 5. If you are currently running an iOS version < 5.0 you can only update through iTunes. Connect your iPad to your PC or Mac. Open iTunes and click on "Check for updates"
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You will need to restore your iPhone, take a look at this link, http://support.apple.com/kb/HT1212
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Contact Apple, or if you bought the computer secondhand, the previous owner.
(118258) -
Hello, I can't open the FluidTunes. Anybody can help me?
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You didn't give any information other than you have have some kind of iMac. You should probably contact the maker of the software http://majicjungle.com/about.html and click on the email for support.
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This is happening to me too. I'm looking for anyway to fix this its getting very annoying not being able to update apps.
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Get a lens cleaner disc. One with brushes and try that. CDs and DVDs use different lenses, so since CDs load and play, I strongly suspect a dirty lens.
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How can I cancel my subscripction before 30 days are passed? I don't find any telephone number or addresse where I can do the cancel. Anybody can help me? thanks
Cancel http://helpx.adobe.com/x-productkb/policy-pricing/return-cancel-or-change-order.html
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Lightroom 5.2 won't open after "successfully installed" - can anybody please help me understand why?
Hi everyone
I've bought and downloaded Adobe Lightroom 5 for a new iMac 27" intel core 5 3,4 GHz. The machine is only three days old and no more updates are to be found when searching for it. However; after the message that "lightroom 5 has successfully been installed" I don't get to open the program. So I never get to the point where I can register the serial number / registration key for purchased product.
I've followed the instructions that comes with Lightroom 5, but no luck. Also tried to uninstall and reinstall.
No secret that I'm total incompetent when it comes to understanding of computers and software....
Hope someone can help me understand what I'm doing wrong, and sugest some good solutions :-)
ThxI think you may need to uninstall and then start over. There is no installer on the Mac for Lightroom so select it in Applications and send it to the trash. I suggest you download again from a different source. You can use the link below to download the free trial and then validate the software after installation with your product serial number. Ensure you choose your language version and MAC platform from the dropdown list at the top of the page.
After you click download it starts by downloading or updating (replace) the Adobe Download Assistant. You then browse the products and select Photoshop Lightroom 5 to start the installation.
https://www.adobe.com/cfusion/tdrc/index.cfm?product=photoshop_lightroom&promoid&promoid=D TEML -
When I bought my Mac they gave me a free hp printer. My mac is a 10.7 and the printer is for 10.6 and below.... so why did they give me this printer? Can I use it somehow or should I ask for a different one that is compatible with my computer?
Can I use it somehow or should I ask for a different one that is compatible with my computer?
No, that's not necessary.
You Mac most likely sat in inventory before it was sold so the latest drivers were not installed.
If you tried using the printer and couldn'tget it to work, go here > http://www8.hp.com/us/en/support-drivers.html
If this is a USB connected printer, make sure it's connected to your Mac, then at that link click: Detect Your Product
Download and install drivers for v10.7 / Lion. -
Hello, my email is connected to Lotus Notes. When I updated my iPad to iOS 6, it REQUIRED that I have a password for the iPad or else my email wouldn't work or synchronize with Lotus Notes (on my work laptop). This is really frustrating, because I cannot disable the password under General<Passcode. Thanks!
This would almost certainly be a setting that is being enforced by your company. You'll need to talk to them about this.
Regards. -
my iphone 4's backlight turns off ocassionally.,some say update it and it will be ok..,anybody can help me?
yes the phone is already on..,autolock is set to 1 min..,what i mean is that whenever im using my iphone the backlight turns off and i got a black sceen but i can see the icons on my iphone if i put it on a bright light..,even if i press the wake up botton the backlight just lit at little and it turns off again,,,and after about 30 mins on sleep mode and press the wake up button it will function as a normal again,few minutes after backlights turns off again..thanks!
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Anybody can help with this SQL?
The table is simple, only 2 columns:
create table personpay(
id integer primary key,
pay number(8,2) not null);
So the original talbe looks like this:
ID PAY
1 800
2 400
3 1200
4 500
5 600
6 1900
The requirement is to use one single query(no pl/sql) to show something lile the following, in other words, for each ID, the pay is the sum of all before itself and itself. So the query result looks like this:
ID PAY
1 800
2 1200
3 2400
4 2900
5 3500
6 5400
Again, just use one sql. No pl/sql. Anybody can help with this? I really appreciate that.
thanks,Eh, people are so "analytically minded" that can't even notice a simple join?
Counting Ordered Rows
Let’s start with a basic counting problem. Suppose we are given a list of integers, for example:
x
2
3
4
6
9
and want to enumerate all of them sequentially like this:
x #
2 1
3 2
4 3
6 4
9 5
Enumerating rows in the increasing order is the same as counting how many rows precede a given row.
SQL enjoys success unparalleled by any rival query language. Not the last reason for such popularity might be credited to its proximity to English . Let examine the informal idea
Enumerating rows in increasing order is counting how many rows precede a given row.
carefully. Perhaps the most important is that we referred to the rows in the source table twice: first, to a given row, second, to a preceding row. Therefore, we need to join our number list with itself (fig 1.1).
Cartesian Product
Surprisingly, not many basic SQL tutorials, which are so abundant on the web today, mention Cartesian product. Cartesian product is a join operator with no join condition
select A.*, B.* from A, B
Figure 1.1: Cartesian product of the set A = {2,3,4,6,9} by itself. Counting all the elements x that are no greater than y produces the sequence number of y in the set A.
Carrying over this idea into formal SQL query is straightforward. As it is our first query in this book, let’s do it step by step. The Cartesian product itself is
select t.x x, tt.x y
from T t, T tt
Next, the triangle area below the main diagonal is
select t.x x, tt.x y
from T t, T tt
where tt.x <= t.x
Finally, we need only one column – t.x – which we group the previous result by and count
select t.x, count(*) seqNum
from T t, T tt
where tt.x <= t.x
group by t.x
What if we modify the problem slightly and ask for a list of pairs where each number is coupled with its predecessor?
x predecessor
2
3 2
4 3
6 4
9 6
Let me provide a typical mathematician’s answer, first -- it is remarkable in a certain way. Given that we already know how to number list elements successively, it might be tempted to reduce the current problem to the previous one:
Enumerate all the numbers in the increasing order and match each sequence number seq# with predecessor seq#-1. Next!
This attitude is, undoubtedly, the most economical way of thinking, although not necessarily producing the most efficient SQL. Therefore, let’s revisit our original approach, as illustrated on fig 1.2.
Figure 1.2: Cartesian product of the set A = {2,3,4,6,9} by itself. The predecessor of y is the maximal number in a set of x that are less than y. There is no predecessor for y = 2.
This translates into the following SQL query
select t.x, max(tt.x) predecessor
from T t, T tt
where tt.x < t.x
group by t.x
Both solutions are expressed in standard SQL leveraging join and grouping with aggregation. Alternatively, instead of joining and grouping why don’t we calculate the count or max just in place as a correlated scalar subquery:
select t.x,
(select count(*) from T tt where tt.x <= t.x) seq#
from T t
group by t.x
The subquery always returns a single value; this is why it is called scalar. The tt.x <= t.x predicate connects it to the outer query; this is why it is called correlated. Arguably, leveraging correlated scalar subqueries is one the most intuitive techniques to write SQL queries.
How about counting rows that are not necessarily distinct? This is where our method breaks. It is challenging to distinguish duplicate rows by purely logical means, so that various less “pure” counting methods were devised. They all, however, require extending the SQL syntactically, which was the beginning of slipping along the ever increasing language complexity slope.
Here is how analytic SQL extension counts rows
select x, rank() over(order by x) seq# from T; -- first problem
select x, lag() over(order by x) seq# from T; -- second problem
Many people suggest that it’s not only more efficient, but more intuitive. The idea that “analytics rocks” can be challenged in many ways. The syntactic clarity has its cost: SQL programmer has to remember (or, at least, lookup) the list of analytic functions. The performance argument is not evident, since non-analytical queries are simpler construction from optimizer perspective. A shorter list of physical execution operators implies fewer query transformation rules, and less dramatic combinatorial explosion of the optimizer search space.
It might even be argued that the syntax could be better. The partition by and order by clauses have similar functionality to the group by and order by clauses in the main query block. Yet one name was reused, and the other had been chosen to have a new name. Unlike other scalar expressions, which can be placed anywhere in SQL query where scalar values are accepted, the analytics clause lives in the scope of the select clause only. I have never been able to suppress an impression that analytic extension could be designed in more natural way.
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