I can help with a buffering problem in a video-based eLearning..

Hello,
I hope I can give a little back here for once. : ) I have a video-based eLearning that is hosted on a LMS that has been giving the users with low bandwidth a lot of problems. I was doing some testing with Chrome Dev Tools today, and I think I find some info that could be of use to some people. The eLearning is hosted on Moodle.
Here is what I found :
Contrary to what I believed, the Chrome browser (and it is probably the same with other browsers) will not download all of the files, or at least not all of the video files in a SCORM packet without the user first interacting with that video in some way. As you see in the graphic, I first arrived at the intro video shortly before 2.5 minutes, and only then did the browser begin downloading this video.
At about three minutes, see red line, I used the forward button to jump to the next video, and because the browser was not yet finished downloading the intro video, it was downloading two simultaneously. This is when I started experiencing a buffering issue.
A few seconds later, I used the slider in the progress bar to move to the 4th video, see blue bar, which "set off" the third video as well. At this point my browser was trying to download four videos at once, which caused major buffering issues.
This means, however, that jumping around an eLearning with several videos in it, even using the TOC or forward and backward buttons will cause all of the videos you "touch" to begin downloading, and if you have low bandwidth, they can be downloading simultaneously for a long time.
Possible solution --> Allow navigation only with with TOC and make it so the user can only go to the next slide/video after he has finished viewing the current one, ensuring,that only one video will be downloaded at once. (Needs to be tested).
Anyway, I hope this has provided some insight to those experiencing problems similar to this!

Under the advanced tab in the button inspector you can set it to activate your subtitle stream.

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Hi
I've a big problem with adobe acrobat reader XI pro and I hope you can help me.
The problem is; when I past copied text from some pdf books (not all of them) it past symbols only! wherever I past it! and even if I coped that text from another pdf reader (adobe pdf reader, internet browsers, ...etc.).
This problem started to happen since yesterday when I installed adobe acrobat reader XI pro to try it before I buy it, and before that when I was using the free adobe pdf reader I was totally able to copy any text from any pdf and past it anywhere with nothing wrong.
What can I do?
thank you a lot.

There is no product called Adobe Acrobat Reader Pro. There is
- Adobe Acrobat Pro ($$)
- Adobe Reader (free)
Which do you have? And are you a programmer?

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Try the suggestions here to see if they resolve your problem:
http://support.apple.com/kb/ts1559
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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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http://www.bitstorm.org/extensions/tweak/

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Hello Anubha,
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Hi, rynodino -
As Tom has suggested, plugins need to be installed in the Plug-Ins folder for the browser.
Typically each browser will have its own Plug-Ins folder, usually located in the same folder containing the browser itself. In order for each of several browsers to access the same plugins, the plugins must be replicated and placed in each of the Plug-Ins folders for each of the browsers. The easy way to do that is to hold down the Option key while dragging a plugin to the new location - a copy (not an alias) of the plugin will be generated at the location it is dragged to.
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Please help with Nokia N95 problem.

If anyone can help with this I’d be much appreciated, I have a N95 on contract with T-Mobile. When I try to play videos from myspace, or you tube etc… I get a message telling me the following.
“Hello, Either you either have JavaScript turned off or an old version of macromedia's flash player click here to get the latest flash player”.
I then get the option of clicking on a link within the previous sentence statement that directs me to acrobat downloads home page where I get the following message.
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daperk,
You should post this to board "Smartphones, Nseries and Eseries Devices".

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
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4 2900
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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.

Who can help with download of Acrobat XI Pro to Mac OS X 10.5.8?

Who can help with download of Acrobat XI Pro to Mac OS X 10.5.8.

Downloadable installation files available:
Suites and Programs: CC 2014 | CC | CS6 | CS5.5 | CS5 | CS4, CS4 Web Standard | CS3
Acrobat: XI, X | 9,8 | 9 standard
Premiere Elements: 13 | 12 | 11, 10 | 9, 8, 7 win | 8 mac | 7 mac
Photoshop Elements: 13 |12 | 11, 10 | 9,8,7 win | 8 mac | 7 mac
Lightroom: 5.7.1| 5 | 4 | 3
Captivate: 8 | 7 | 6 | 5.5, 5
Contribute: CS5 | CS4, CS3
Download and installation help for Adobe links
Download and installation help for Prodesigntools links are listed on most linked pages. They are critical; especially steps 1, 2 and 3. If you click a link that does not have those steps listed, open a second window using the Lightroom 3 link to see those 'Important Instructions'.

Can anyone help with Address Book problem ?

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running os x 10.6.4 for a few months now since my husband upgraded it...???
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Pop in your Snow Leopard DVD then open the Optional Installs folder. Double-click on the Optional Installs package. You can selectively reinstall some applications by checking the ones you want. Check Address Book then click on the Install button.
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Here's a situation I need some help with:
I live in a semi-detached house. My neighbour has broadband, I don't. The phone company (BT) has said it would cost too much to install another line to enable me to have broadband, and would involve digging up the road and the garden, which they're not prepared to do.
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Now, I consider myself Mac-savvy, but when it coms to wireless stuff I confess I know very little! So if someone could explain in simple terms what I would need to buy and approx costs I be very grateful!
I have a G5, and I believe my neighbour is a PC user.
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Many thanks,
Andy

Hi Andy, before going too far down the 'sharing' route I would rattle some cages chez BT. If your neighbour has broadband then you and they are within a hair's breadth (line length speaking) of the same distance to the exchange. If they have enabled BB next door, why can't they enable BB for you without the hassle? I live on a mobile home site in the back of beyond in Devon and when I upgraded from dial-up to BB, and then to 1 meg there was no problem. There was a two week hiatus while BT did their stuff 'on the line' but it all went without a hitch. Er...I take it you do have an exclusive landline?
However, if the situation re the line is as per, then the simplest way (to obviate any wifi interference issues) would be for you or neighbour to route an ethernet cable through walls/window frames from your Mac to a router on their side of the wall. (Approx £30?) This would be connected to their phone line. Their PC would also plug into the router. (Which they may already have if they're on BB, though they may have a nasty USB BB modem &nbsp:&nbsp: No good for what you want.
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If done via either of the above, then no, your neighbour doesn't have to have their PC on all the time, if ever.
You don't have to sign up with an ISP, it would have to be a 'gentleman's agreement' that you bung him £x pm for your share. He is already paying his ISP.
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I hope the above is some help but I would have a word with the neighbour first and sound them out, after rattling BT that is...
Come back here if you want more help.
Good luck!
Adrian

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I need help with the following book problem...could someone write this code for me?? Thanks!!
First design and implement a class called PairOfDice, composed of two six-sided Die objects. Using the PairOfDice class, design and implement a class to play a game called Pig. In this game, the user competes against the computer. On each turn, the current player rolls a pair of dice and accumulates points. The goal is to reach 100 points before your opponent does. If, on any turn, the player rolls a 1, all points accumulated for that round are forfeited and control of the dice moves to the other player. If the player rolls two 1s in one turn, the player loses all points accumulated thus far in the game and loses control of the dice. The player may voluntarily turn over the dice after each roll. Therefore the player must decide to either roll again (be a pig) and risk losing points, or relinquish control of the dice, possibly allowing the other player to win. Implement the computer player such that it always relinquishes the dice after accumulating 20 or more points in any given round.
I realize this is a long code, so it would be greatly appreciated to anyone who writes this one for me, or can at least give me any parts of it. I honestly have no clue how to do this because our professor is a dumbass and just expects us to know how to do this...he doesn't teach us this stuff...don't ask. Anyways, thanks for taking the time to read this and I hope someone helps out!! Thank you in advance to anyone who does!!!

Nasty comments? It's not a matter of not liking you, it's a matter of responding to the
"I'ts everyone else's fault but mine" attitude of your post.
If you are genuine, the program is very easy to write, so, your Professor is correct when he said you should
be able to do it. I'm still very much in the beginner category at java, and it took me only 20 mins to write
(+5 mins to design it on paper). 2 classes in one file < 100 lines. Most of the regulars here would do it in half the time / half the lines.
All of the clues are in the description, break them down into their various if / else conditions. Write it down on paper.
Have a go. Try to achieve something. Post what you've tried, no-one will laugh at you, and you will be
pleasanty surprised at the level of help you will get when you've "obviously" made some effort.
Again, have a go. If you don't like problem-solving, then you don't like programming. So, you gotta ask
yourself - "what am I doin' here?"

Sreaming at the top of my lungs for HELP with Java Script problem!

Hello,
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Hi,
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You can safely ignore it:
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I can help with a buffering problem in a video-based eLearning..

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