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August 12

Excel queries and pivot tables

I frequently write queries to our database by way of the MS query function in Excel. When I've done my thing and am ready to bring the data back, I'm given three main options: to dump the result into a table, into a PivotTable, or into a PivotChart. I usually use one of the first two, depending on what I want to do next. At the bottom of the dialog box is the option to import the data into the Data Model version of a pivot table, but it's greyed out, which is the problem. Searching online finds people with a similar situation, but they are in relation to creating a Pivot Table from an existing table, while I am coming at the problem from a different source. Any thoughts as to why the option is greyed out for me and how I can get around this? Perhaps alternatively, is there any way to convert a "normal" pivot table into a Data Model one after said PT is in existence? Searching online finds this unanswered question on Mr Excel from two years ago. Matt Deres (talk) 15:23, 12 August 2019 (UTC)[reply]

jQuery interpreting widths wrongly

I've encountered a strange problem at work. The code uses jQuery to interpret the widths of elements and set child elements styles accordingly.

The problem is, jQuery interprets the widths wrongly. No matter what the actual widths are, the jQuery function width() always returns 100. This happens even if I only call the function from within $(document).ready() and $(window).load().

What is strange here is that the jQuery code works perfectly OK for elements written directly on the page. But it fails on elements that are only rendered via an @await Component.InvokeAsync() call, even though the actual jQuery code is still within $(document).ready() and $(window).load().

Why is this happening? Could it be because the jQuery code is executed before @await Component.InvokeAsync()? JIP | Talk 23:10, 12 August 2019 (UTC)[reply]

One thought is that it's returning the allocated length rather than the actual length. I have no idea why, however. SinisterLefty (talk) 02:49, 13 August 2019 (UTC)[reply]

It's hard to diagnose this without seeing it in action: can you make a small example that shows the behaviour and put it on CodePen or JSFiddle? AndrewWTaylor (talk) 06:26, 13 August 2019 (UTC)[reply]

I found out what was causing this. The elements that jQuery was failing at were inside a <div> that had a CSS class causing an animation lasting 0.15 seconds. That was apparently enough time for $(document).ready() and $(window).load() to fire before the elements had been fully rendered, causing jQuery to fail to find their exact widths. When I took the CSS class away the code started working. JIP | Talk 14:57, 13 August 2019 (UTC)[reply]

Glad you found it. SinisterLefty (talk) 15:23, 13 August 2019 (UTC)[reply]

Resolved

SinisterLefty (talk) 15:23, 13 August 2019 (UTC)[reply]

August 13

Unable to open a website

My colleague along with me run the Sherlock Holmes of India Website. I am unable to access it as it repeatedly says "www.sherlockholmessociety.in’s server IP address could not be found". I am not technically proficient. Can esteemed members suggest what should be done to get the site up and running?Sumalsn (talk) 07:52, 13 August 2019 (UTC)[reply]

The domain name www.sherlockholmessociety.in is registered at www.godaddy.com with Domain ID: D7108653-IN. You need the service of an ISP (Internet service provider - see article) to host your website. Free ISPs exist but cannot be relied on. DroneB (talk) 09:58, 13 August 2019 (UTC)[reply]

Thanks for your assistance. I am deeply obliged. Can you please tell me how I can recover that website now? Sumalsn (talk) 07:39, 14 August 2019 (UTC)[reply]

It appears that you do not understand the difference between a website and a domain name. The domain name "sherlockholmessociety.in" is a convenience service. It maps the words to a computer name, which is a number. Right now, as I type this, sherlockholessociety.in is being managed by INRegistry, but the service has lapsed. It is a yearly service to map the domain name to the computer's number. If you don't pay, the service is canceled and anyone in the world can take over use of the domain name. But, none of that has anything to do with the website. The website is a collection of files sitting on a server somewhere. The problem is that we have no means of identifying what the computer's number was. So, we can't go directly to the computer. At some point, someone created the website. That person SHOULD have kept a backup copy of the website. But, in the real world, nobody makes backups. So, you are left with two needs: First, register the domain name again and make sure you keep paying for it. Then, recreate the website. If you get the original person to remake it, tell him or her to make a backup. 135.84.167.41 (talk) 12:03, 14 August 2019 (UTC)[reply]

According to the last surviving copy of the website in archive dot com, the site looks like it's hosted by Blogger.

If you still have access to the Blogger account and you still own the domain, it may be a simple matter of making sure the domain is configured properly.

here is one of many guides on pointing your domain at a Blogger blog.

Hope this helps. ApLundell (talk) 17:20, 14 August 2019 (UTC)[reply]

August 14

Safe?

Is this website safe to use? I'm trying to convert a pdf file on my computer to html so I can put it on a webpage. --Puzzledvegetable^{Is it teatime already?} 15:18, 14 August 2019 (UTC)[reply]

Do you actually need to convert? I've just dropped a PDF in my DOCROOT and it was immediately readable. Web server: Apache/2.4.6, browser: Firefox Quantum 60.8.0esr. Regards, Martin of Sheffield (talk) 17:10, 14 August 2019 (UTC)[reply]

There are lots of website safety checkers (just Google website safe checker). Norton Safe Web found no issues with the site you mention.--Shantavira|^{feed me} 07:46, 15 August 2019 (UTC)[reply]

Upload pdf

I have a pdf stored as a file on my computer, and I would like to display it on a github page. How do I do this? Also note, I'm not that tech savvy so try to dumb it down for me. Thanks. --Puzzledvegetable^{Is it teatime already?} 21:44, 14 August 2019 (UTC)[reply]

If not very large, one option would be to do a screen grab (using the Print Screen button on keyboard, possibly with SHIFT, CTRL, or ALT), paste it (Edit + Paste ?) and trim the image with something like MS Paint, Save As a JPG image file, GIF, etc. (but not as a bitmap), and display that using HTML (presumably GitHub can help you upload an image and display it on the page). Exactly how large of an image you can grab depends on the screen resolution you can display. This could also be done if there are convenient page-sized chunks to break it into. Disadvantages include the difficulty in making modifications (you would need to modify the PDF and repeat the process), larger disk space used than text, and inability to make it sharper by zooming in beyond 100%. One possible advantage is that it makes it harder for others to copy and then modify the data.

Note that PDFs include both text and images, so the data you want may be in either form initially. If it is stored as text internally, it may also be possible to extract that text, while if it is already an image, then converting it to text would require OCR.

It would help if you could describe the contents you want to grab from the PDF. Try zooming way in to tell if it is actual text, or an image of text. (If it gets sharper with every further zoom, then it is actual text.) Also, if you can highlight the text a character at a time with mouse, then it is actual text and you may be able to just cut and paste it directly to any text editor, but likely losing any formatting in the process. SinisterLefty (talk) 00:22, 15 August 2019 (UTC)[reply]

August 16

Comparing computers

I was planning to improve (rewrite) Shakuntala Devi and landed upon this news-piece which documents her battle with an UNIVAC-1108. The latter took more than a minute to calculate the 23rd root of a random 201 digit number, thus losing to Devi, who took ~50 seconds.

My question is that, if an Apple II were used (which launched the same year), what would have been the time taken by it? Any approximation? ∯WBG^converse 13:45, 16 August 2019 (UTC)[reply]

P.S:- The model number has been also reported to be UNIVAC 1110 in some media. At any case, how did these compare to other popular models of those days? Any model, which might have beaten Devi comfortably? ∯WBG^converse 13:50, 16 August 2019 (UTC)[reply]

One issue in answering this question is that standard computers can't process 201-digit numbers natively, so they need to use bignum software to do this sort of calculation. It's possible that different versions of the software exist, or that they may be configured to store different numbers of digits, which would affect the time. (The number of digits matters because "the 23rd root of a random 201-digit number" would be irrational, so it must be computed to a certain number of decimal places.) --76.69.116.4 (talk) 14:57, 16 August 2019 (UTC)[reply]

Thanks; that's a point! I am assuming 9/10 decimal places, FWIW. ∯WBG^converse 16:02, 16 August 2019 (UTC)[reply]

The answer listed in the news report is 546372891, so perhaps that's the accuracy to which it was provided. By the way, the more important issue than the computational power of the machine used is the matter of the algorithm. I guess, for example, that Shakuntala's mental calculation was based upon

{\sqrt[{23}]{N}}=\exp({\frac {1}{23}}\log N)

(she would surely have the logarithmic tables memorized, at least to some precision, and division by 23 would be trivial for her), which does not necessitate the person/computer to store or work with the actual 201 digits to get the answer to the ~10 significant digits; double precision arithmetic, which UNIVACs could do since the 1960s, would suffice. Not sure if the computer used such approximations, and what numerical algorithms and databases it had access to. Of course, knowing what methods and approximations to use is part of the human advantage in such competitions. Abecedare (talk) 16:59, 16 August 2019 (UTC)[reply]

If the answer was 546372891, then it wasn't a random 201-digit number. There are tricks that lightning calculators can use for some such roots (mainly odd-numbered roots of integers), which wouldn't be programmed into the computer. And if they had entered the number into the computers double-precision floating-point, it should have been able to calculate it in a tiny fraction of a second. Bubba73 ^{You talkin' to me?} 20:55, 16 August 2019 (UTC)[reply]

And in fact, reading the badly scanned article, it doesn't say it was a random number. Obviously the number was 916,748,676,920,039,158,098,660,927,585,380,162,483,106,680,144,308,622,407,126,516,427,934,657,040,867,096,593,279,205,767,480,806,790,022,783,016,354,924,852,380,335,745,316,935,111,903,596,577,547,340,075,681,688,305,620,821,016,129,132,845,564,805,780,158,806,771, which is a 23rd power, and we can forget about irrationals and decimal places. --76.69.116.4 (talk) 21:06, 16 August 2019 (UTC)[reply]

And that 201-digit number is written on the blackboard in the photo in the article. They probably used multiple-precision integers on the computer. Bubba73 ^{You talkin' to me?} 21:09, 16 August 2019 (UTC)[reply]

And if the number was not an odd power of an integer, the problem is a lot harder, i.e. an even power or not the power of an integer. Bubba73 ^{You talkin' to me?} 22:45, 16 August 2019 (UTC)[reply]

Here and here is some discussion. Bubba73 ^{You talkin' to me?} 22:51, 16 August 2019 (UTC)[reply]

Great finds Bubba73! Abecedare (talk) 02:02, 18 August 2019 (UTC)[reply]

And this talks about easy ways of taking the 13th root of 100-digit numbers. No doubt that the 23rd root of a 201-digit number is something along these lines. Bubba73 ^{You talkin' to me?} 03:19, 18 August 2019 (UTC)[reply]

The 13th root of a 201 digit number to 10-ish decimal places should take almost no time on an Apple II or even a pocket calculator. You'd use logarithms, not bignums. E.g. if the number is 3.4567e200, its common logarithm is 200+log10(3.4567) = 200.53866168963842, and 1/13th of that is 15.426050899202956, so the answer is 1e15 * 10^{0.426050899202956} = 2.6671712379959103e15. At least 10 places of that should be accurate using Univac 1108 double precision or some 64-bit software floating point (pre-IEEE) on the Apple II. Computation time on the Apple might have been in double (or maybe triple) digit milliseconds vs much faster on the 1108 and sub-microsecond on today's PC's. Applesoft BASIC extended precision floating point apparently had a 1-byte exponent and 31-bit mantissa, so that would be about 9 decimal digits of precision, which might satisfy you. Otherwise I'm sure higher precision software existed though a quick web search doesn't find an easy link. The Univac had 36 bit words, so 72 bit double precision, for 20+ decimal places of precision. It looks from this that VisiCalc, the Apple II killer app spreadsheet, did decimal arithmetic to around 10 or 11 digits (guessing that the US govt budget was around 1e11 dollars in that era). Anyway math of that precision was no big deal even back then. Here is a 13-digit desk calculator from the 1960s. 173.228.123.207 (talk) 04:34, 18 August 2019 (UTC)[reply]

I had an Apple II, and I don't remember for sure, but I doubt that it had 64-bit floating point. But Apple Pascal had 36-digit integers. You should be able to take the most-significant digits of the 201-digit number and take the 23rd root and account for the location of the decimal point (like they did with fixed-point back in the old days), and get the answer. My guess is under 1 second on the Apple II. Bubba73 ^{You talkin' to me?} 05:34, 18 August 2019 (UTC)[reply]

Of course, a story about a computer crushing a mental calculator isn't going to make the news. Bubba73 ^{You talkin' to me?} 19:56, 18 August 2019 (UTC)[reply]

Entering the first few digits into my calculator in scientific notation, taking the log, dividing by 23, and then take the exponential immediately gives 546372891.025, which rounded off to an integer gives the correct answer. Bubba73 ^{You talkin' to me?} 20:01, 18 August 2019 (UTC)[reply]

Google Fuchsia embedded device requirements.

What are the minimum and maximum requirements for smartphones (such as RAM etc..)?Ajax-x86 (talk) 15:12, 16 August 2019 (UTC)[reply]

According to our article, Google Fuchsia is "...designed to scale to any application from embedded RTOS (Real-Time Operating Systems) to mobile and desktop devices of all kinds". That implies that it would run with very minimal RAM, but with a reduced feature set. How low, and at what reduction of features, I do not know. SinisterLefty (talk) 18:01, 16 August 2019 (UTC)[reply]

Making columns responsive

There could be two-four columns of my articles, depending on the screen size. I used media query for my article class of title. However, nothing seems to happen when the site is displayed on different resolutions.

I tried adding @media ONLY screen instead of @media screen but this does not work neither.

@media screen and (min-width: 576px) {

 .skill {
   float: left;
   width: 50%;
 }

} @media screen and (min-width: 1200px) {

 .skill {
   width: 25%;
 }

}

What should I change in my code in order to get two or four columns of articles on 576px and 1200px screens respectively? Vs⁶ ₅⁰⁷ 20:39, 16 August 2019 (UTC)[reply]

Try using the CSS3 columns property:

@media screen and (min-width: 576px) {
.skill {
	columns: 2;
 }

@media screen and (min-width: 1200px) {
.skill {
	columns: 4;
}

--Canley (talk) 12:49, 17 August 2019 (UTC)[reply]

August 17

SQL update

This is from a practical problem but part of the motivation is trying to get better at SQL in general, which should frame any perceived XY problem. I.e. there are only a few hundred rows involved so I could easily write a script that does it row by row, but I'm trying to figure out if it's possible to write a sane SQL update that does the following.

I have a table OLD with columns NAME and ADDRESS, and some of the people in it have moved, so I have another table NEW which also has NAME and ADDRESS (giving the new addresses of the people who have moved). Most of the people in OLD have not moved, so they don't have entries in NEW. I want to update OLD to replace the addresses of people who have moved with their new addresses. This is wrong SQL but is conceptually something like what I'm looking for:

UPDATE old SET address = new.address FROM old JOIN new ON old.name = new.name;

Any suggestions? Thanks. 173.228.123.207 (talk) 01:42, 17 August 2019 (UTC)[reply]

That's pretty much right, but you don't need to put the FROM clause in because you are calling the old table in the UPDATE statement, and you also need to disambiguate the address field, like this:

UPDATE old JOIN new ON old.name = new.name SET old.address = new.address;

--Canley (talk) 02:23, 17 August 2019 (UTC)[reply]

Wow, thanks! Sqlite3 complains of a syntax error near "JOIN" when I try that, but I'll keep loking into that approach (didn't know it was possible to update anything but a named table). Meanwhile I've messed up the table and have to rebuild it from the data sources, which should be easy except the scripts that build it have gotten disorganized so I have to straighten those out first. 173.228.123.207 (talk) 02:54, 17 August 2019 (UTC)[reply]

Oh OK, I didn't know you were using SQLite, that doesn't use JOINs the same way, so I think you would have to use a nested query, like this:

UPDATE old SET address = (SELECT address FROM new WHERE name = old.name) WHERE EXISTS (SELECT address FROM new WHERE name = old.name);

--Canley (talk) 04:15, 17 August 2019 (UTC)[reply]

Thanks! That worked, though I'll have to study it to figure out what it did. It hadn't occurred to me that you could use a select on the rhs of an = like that. It makes sense though. There must be a good place to read up on this stuff: it seems like every sql db has its own sql dialect and it's not really clear how anything works. I'll keep at it though. 173.228.123.207 (talk) 06:33, 17 August 2019 (UTC)[reply]

The term to read up on is "subquery". I find a lot of SQL material online to be very difficult to read, but this is a decent overview. This is an overview based on SQLite. Matt Deres (talk) 12:29, 17 August 2019 (UTC)[reply]

Getting the exact URL of a PDF file I want to archive on megalodon.jp

Hi, guys! At https://www.webcitation.org/6XfufE6Jb?url=http://ci.nii.ac.jp/els/110000469807.pdf?id%3DART0000851255%26type%3Dpdf%26lang%3Den%26host%3Dcinii%26order_no%3D%26ppv_type%3D0%26lang_sw%3D%26no%3D1428634639%26cp%3D there is a PDF file that I want to make a second archive of, on http://megalodon.jp

However the archive seems to only show a blank page https://megalodon.jp/2019-0818-0337-28/https://www.webcitation.org:443/6XfufE6Jb ... I'd like a find a way to archive the actual PDF. However I need the actual URL of the PDF itself, and:

When I try to view only the PDF frame the resulting URL is "https://www.webcitation.org/mainframe.php" which obviously won't result in the actual PDF if I archive it.
And when I try "View Page Info" in Firefox, the PDF does not show up in the file list (otherwise it would as "embed").

How do I get the exact URL of the embedded PDF file? Thanks, WhisperToMe (talk) 18:41, 17 August 2019 (UTC)[reply]

It looks messy, i.e. the server is sending the pdf to mainframe.php based on a cookie set in the outer page or something like that. Simplest might be to set up your browser preferences for application/pdf to save the pdf to disk instead of previewing it. Then re-upload it to wherever you want your 2nd archive to be. The file is about 1.4MB so the download/upload is not too bad. 67.164.113.165 (talk) 08:21, 18 August 2019 (UTC)[reply]

August 18

Wichita linemen

Wikimedia's ISP 2620:0:862:ED1A::1 geolocates to 37° 45′ 3.60″ N latitude, 97° 49′ 19.20″ W longitude, as does that of astronomer Simon Cassidy (209.179.253.xxx). Both are in San Francisco but the geolocation is Wichita, Kansas. Is this one of the "default" locations discussed here recently where Google doesn't have enough information to geolocate accurately? This issue was discussed previously at Wikipedia:Reference desk/Archives/Computing/2011 December 1#Geolocating an IP. 92.31.141.118 (talk) 14:33, 18 August 2019 (UTC)[reply]

August 19

connecting to a private server to play Painkiller multiplayer (PS3)

The instruction on the internet says:

This means you can still connect to them via console and play. Just open your console and write /connect (ip address)

This information is for me useless. If I google for it how to open the console on the ps3, I have something like a "terminal Command" in my mind, but I dont find there any console in the multiplayer menu and the search result are only showing me how to open the console itself with a screwdriver. What a waste of time.
Can anybody please explain to me how to find / open this console bar / menu to connect to a private server? --46.167.62.33 (talk) 05:14, 19 August 2019 (UTC)[reply]