any one knows where I could find (or I could make) a function that calculate moon phase ? (AS3) - actionscript-3

I'm looking for a small fucntion in ActionScript 3 that could calculate Moon Phases.
I've tried to search on google. The only result is a website that gives this code but I think it's a wrong code.
//return frame number for moon phase display
function getMoonPhase(yr:Number, m:Number, d:Number):int{
//based on http://www.voidware.com/moon_phase.htm
//calculates the moon phase (frames 1-30 )
if (m < 3) { yr -= 1; m += 12; } //
m += 1;
var c:Number = 365.25*yr;
var e:Number = 30.6*m; //jd is total days elapsed
//divide by the moon cycle (29.53 days)
var jd:Number = (c+e+d-694039.09)/29.53; //subtract integer to leave fractional part
jd = jd - int(jd); //range fraction from 0-30 and round by adding 0.5
var frame:int = Math.round(jd*30 + 0.5);
return frame;
}
//test: september 23, 2002, not a full moon? //
Weirdly, sometimes the code is extremely accurate, but sometimes it's wrong..
Example : On the 16 september 2016 it's a full moon.
But if I enter this date in the code the result is "15" (16 is full moon)....
Any idea why ? or another way to calculate moon phases ?
Thx

In that algorithm, c is int, not Number, therefore an implicit round-down occurs here. Same with e. This can total to a one day error in calculation which in turn produces wrong phase.
Edit: Also since you request a 30-frame selection, but have a precision of 29, the calculation can skip a frame. Consider using 15 or 16 frames instead.

function julday(year, month, day) {
if (year < 0) { year ++; }
var jy = parseInt(year);
var jm = parseInt(month) +1;
if (month <= 2) {jy--; jm += 12; }
var jul = Math.floor(365.25 *jy) + Math.floor(30.6001 * jm) + parseInt(day) + 1720995;
if (day+31*(month+12*year) >= (15+31*(10+12*1582))) {
var ja = Math.floor(0.01 * jy);
jul = jul + 2 - ja + Math.floor(0.25 * ja);
}
return jul;
}
function moonphase(year,month,day) {
var n = Math.floor(12.37 * (year -1900 + ((1.0 * month - 0.5)/12.0)));
var RAD = 3.14159265/180.0;
var t = n / 1236.85;
var t2 = t * t;
var aas = 359.2242 + 29.105356 * n;
var am = 306.0253 + 385.816918 * n + 0.010730 * t2;
var xtra = 0.75933 + 1.53058868 * n + ((1.178e-4) - (1.55e-7) * t) * t2;
xtra += (0.1734 - 3.93e-4 * t) * Math.sin(RAD * aas) - 0.4068 * Math.sin(RAD * am);
var i = (xtra > 0.0 ? Math.floor(xtra) : Math.ceil(xtra - 1.0));
var j1 = julday(year,month,day);
var jd = (2415020 + 28 * n) + i;
return (j1-jd + 30)%30;
}
not that i fully understand it, but this should work fine. It's a (straight) port of some Basic code by Roger W. Sinnot from Sky & Telescope magazine, March 1985 ... talk about old school.
if you try with 16 sept 2016 it will give 15. same as 21 sept 2002, which was the actual day of the fullmoon

Related

Implementing Laguerre's root finding method

I'm trying to implement robust / stable Laguerre's method. My code works for most of polynomials but for few it "fails". Respectively I don't know how to properly handle "corner" cases.
Following code tries to find single root (F64 - 64bit float, C64 - 64bit complex):
private static C64 GetSingle( C64 guess, int degree, C64 [] coeffs )
{
var i = 0;
var x = guess;
var n = (F64) degree;
while( true )
{
++Iters;
var v0 = PolyUtils.Evaluate( x, coeffs, degree );
if( v0.Abs() <= ACCURACY )
break;
var v1 = PolyUtils.EvaluateDeriv1( x, coeffs, degree );
var v2 = PolyUtils.EvaluateDeriv2( x, coeffs, degree );
var g = v1 / v0;
var gg = g * g;
var h = gg - ( v2 / v0 );
var f = C64.Sqrt(( n - 1.0 ) * ( n * h - gg ));
var d0 = g - f;
var d1 = g + f;
var dx = d0.Abs() >= d1.Abs() ? ( n / d0 ) : ( n / d1 );
x -= dx;
// if( dx.Abs() <= ACCURACY )
// break;
if( ++i == ITERS_PER_ROOT ) // even after trying all guesses we didn't converted to the root (within given accuracy)
break;
if(( i & ( ITERS_PER_GUESS - 1 )) == 0 ) // didn't converge yet --> restart with different guess
{
x = GUESSES[ i / ITERS_PER_GUESS ];
}
}
return x;
}
At the end if it didn't found root it tries different guess, first quess (if not specified) is always 'zero'.
For example for 'x^4 + x^3 + x + 1' it founds 1st root '-1'.
Deflates (divides) original poly by 'x + 1' so 2nd root search continues with polynomial 'x^3 + 1'.
Again it starts with 'zero' as initial guess... but now both 1st and 2nd derivates are 'zero' which leads to 'zero' in 'd0' and 'd1'... ending by division-by-zero (and NaNs in root).
Another such example is 'x^5 - 1' - while searching for 1st root we again ends with zero derivates.
Can someone tell me how to handle such situations?
Should I just try another guess if derivates are 'zero'? I saw many implementation on net but none
had such conditions so I don't know if I'm missing something.
Thank you

How to convert a simple timer into hour minute and second format in angular 7 [duplicate]

How can I convert seconds to an HH-MM-SS string using JavaScript?
You can manage to do this without any external JavaScript library with the help of JavaScript Date method like following:
const date = new Date(null);
date.setSeconds(SECONDS); // specify value for SECONDS here
const result = date.toISOString().slice(11, 19);
Or, as per #Frank's comment; a one liner:
new Date(SECONDS * 1000).toISOString().slice(11, 19);
Updated (2020):
Please use #Frank's one line solution:
new Date(SECONDS * 1000).toISOString().substring(11, 16)
If SECONDS<3600 and if you want to show only MM:SS then use below code:
new Date(SECONDS * 1000).toISOString().substring(14, 19)
It is by far the best solution.
Old answer:
Use the Moment.js library.
I don't think any built-in feature of the standard Date object will do this for you in a way that's more convenient than just doing the math yourself.
hours = Math.floor(totalSeconds / 3600);
totalSeconds %= 3600;
minutes = Math.floor(totalSeconds / 60);
seconds = totalSeconds % 60;
Example:
let totalSeconds = 28565;
let hours = Math.floor(totalSeconds / 3600);
totalSeconds %= 3600;
let minutes = Math.floor(totalSeconds / 60);
let seconds = totalSeconds % 60;
console.log("hours: " + hours);
console.log("minutes: " + minutes);
console.log("seconds: " + seconds);
// If you want strings with leading zeroes:
minutes = String(minutes).padStart(2, "0");
hours = String(hours).padStart(2, "0");
seconds = String(seconds).padStart(2, "0");
console.log(hours + ":" + minutes + ":" + seconds);
I know this is kinda old, but...
ES2015:
var toHHMMSS = (secs) => {
var sec_num = parseInt(secs, 10)
var hours = Math.floor(sec_num / 3600)
var minutes = Math.floor(sec_num / 60) % 60
var seconds = sec_num % 60
return [hours,minutes,seconds]
.map(v => v < 10 ? "0" + v : v)
.filter((v,i) => v !== "00" || i > 0)
.join(":")
}
It will output:
toHHMMSS(129600) // 36:00:00
toHHMMSS(13545) // 03:45:45
toHHMMSS(180) // 03:00
toHHMMSS(18) // 00:18
As Cleiton pointed out in his answer, moment.js can be used for this:
moment().startOf('day')
.seconds(15457)
.format('H:mm:ss');
Here's a simple function for converting times that might help
function formatSeconds(seconds) {
var date = new Date(1970,0,1);
date.setSeconds(seconds);
return date.toTimeString().replace(/.*(\d{2}:\d{2}:\d{2}).*/, "$1");
}
This does the trick:
function secondstotime(secs)
{
var t = new Date(1970,0,1);
t.setSeconds(secs);
var s = t.toTimeString().substr(0,8);
if(secs > 86399)
s = Math.floor((t - Date.parse("1/1/70")) / 3600000) + s.substr(2);
return s;
}
(Sourced from here)
var timeInSec = "661"; //even it can be string
String.prototype.toHHMMSS = function () {
/* extend the String by using prototypical inheritance */
var seconds = parseInt(this, 10); // don't forget the second param
var hours = Math.floor(seconds / 3600);
var minutes = Math.floor((seconds - (hours * 3600)) / 60);
seconds = seconds - (hours * 3600) - (minutes * 60);
if (hours < 10) {hours = "0"+hours;}
if (minutes < 10) {minutes = "0"+minutes;}
if (seconds < 10) {seconds = "0"+seconds;}
var time = hours+':'+minutes+':'+seconds;
return time;
}
alert("5678".toHHMMSS()); // "01:34:38"
console.log(timeInSec.toHHMMSS()); //"00:11:01"
we can make this function lot shorter and crisp but that decreases the readability, so we will write it as simple as possible and as stable as possible.
or you can check this working here:
Try this:
function toTimeString(seconds) {
return (new Date(seconds * 1000)).toUTCString().match(/(\d\d:\d\d:\d\d)/)[0];
}
I think the most general (and cryptic) solution could be this
function hms(seconds) {
return [3600, 60]
.reduceRight(
(pipeline, breakpoint) => remainder =>
[Math.floor(remainder / breakpoint)].concat(pipeline(remainder % breakpoint)),
r => [r]
)(seconds)
.map(amount => amount.toString().padStart(2, '0'))
.join('-');
}
Or to copy & paste the shortest version
function hms(seconds) {
return [3600, 60]
.reduceRight(
(p, b) => r => [Math.floor(r / b)].concat(p(r % b)),
r => [r]
)(seconds)
.map(a => a.toString().padStart(2, '0'))
.join('-');
}
Some example outputs:
> hms(0)
< "00-00-00"
> hms(5)
< "00-00-05"
> hms(60)
< "00-01-00"
> hms(3785)
< "01-03-05"
> hms(37850)
< "10-30-50"
> hms(378500)
< "105-08-20"
How it works
Algorithm
To get hours you divide total seconds by 3600 and floor it.
To get minutes you divide remainder by 60 and floor it.
To get seconds you just use the remainder.
It would also be nice to keep individual amounts in an array for easier formatting.
For example given the input of 3785s the output should be [1, 3, 5], that is 1 hour, 3 minutes and 5 seconds.
Creating pipeline
Naming the 3600 and 60 constants "breakpoints" you can write this algorithm into function as this
function divideAndAppend(remainder, breakpoint, callback) {
return [Math.floor(remainder / breakpoint)].concat(callback(remainder % breakpoint));
}
It returns an array where first item is the amount for given breakpoint and the rest of the array is given by the callback.
Reusing the divideAndAppend in callback function will give you a pipeline of composed divideAndAppend functions. Each one of these
computes amount per given breakpoint and append it to the array making your desired output.
Then you also need the "final" callback that ends this pipeline. In another words you used all breakpoints and now you have only the remainder.
Since you have already the answer at 3) you should use some sort of identity function, in this case remainder => [remainder].
You can now write the pipeline like this
let pipeline = r3 => divideAndAppend(
r3,
3600,
r2 => divideAndAppend(
r2,
60,
r1 => [r1]));
> pipeline(3785)
< [1, 3, 5]
Cool right?
Generalizing using for-loop
Now you can generalize with a variable amount of breakpoints and create a for-loop that will compose individial divideAndAppend functions into
the pipeline.
You start with the identity function r1 => [r1], then use the 60 breakpoint and finally use the 3600 breakpoint.
let breakpoints = [60, 3600];
let pipeline = r => [r];
for (const b of breakpoints) {
const previousPipeline = pipeline;
pipeline = r => divideAndAppend(r, b, previousPipeline);
}
> pipeline(3785)
< [1, 3, 5]
Using Array.prototype.reduce()
Now you can rewrite this for-loop into reducer for shorter and more functional code. In other words rewrite function composition into the reducer.
let pipeline = [60, 3600].reduce(
(ppln, b) => r => divideAndAppend(r, b, ppln),
r => [r]
);
> pipeline(3785)
< [1, 3, 5]
The accumulator ppln is the pipeline and you are composing it using the previous version of it. The initial pipeline is r => [r].
You can now inline the function divideAndAppend and use Array.prototype.reduceRight which is the same as [].reverse().reduce(...) to make the breakpoints
definitions more natural.
let pipeline = [3600, 60]
.reduceRight(
(ppln, b) => r => [Math.floor(r / b)].concat(ppln(r % b)),
r => [r]
);
Which is the final form. Then you just appy mapping to string with padded 0's on left and join the strings with : separator;
More generalizations
Wrapping the reducer into function
function decompose(total, breakpoints) {
return breakpoints.reduceRight(
(p, b) => r => [Math.floor(r / b)].concat(p(r % b)),
r => [r]
)(total);
}
> decompose(3785, [3600, 60])
< [1, 3, 5]
you now have very general algorithm you can work with. For example:
Convert easily (the weird) us length standards
Given the standards
Unit
Divisions
1 foot
12 inches
1 yard
3 feet
1 mile
1760 yards
> decompose(123_456, [1760 * 3 * 12, 3 * 12, 12])
< [1, 1669, 1, 0]
123456 in = 1 mi, 1669 yd, 1 feet and 0 in
Or you can somewhat convert to decimal or binary representations
> decompose(123_456, [100_000, 10_000, 1000, 100, 10])
< [1, 2, 3, 4, 5, 6]
> decompose(127, [128, 64, 32, 16, 8, 4, 2])
< [0, 1, 1, 1, 1, 1, 1, 1]
Works also with floating point breakpoints
Since Javascript supports mod operator with floating point numbers, you can also do
> decompose(26.5, [20, 2.5])
< [1, 2, 1.5]
The edge case of no breakpoints is also naturally covered
> decompose(123, [])
< [123]
Here is an extension to Number class. toHHMMSS() converts seconds to an hh:mm:ss string.
Number.prototype.toHHMMSS = function() {
var hours = Math.floor(this / 3600) < 10 ? ("00" + Math.floor(this / 3600)).slice(-2) : Math.floor(this / 3600);
var minutes = ("00" + Math.floor((this % 3600) / 60)).slice(-2);
var seconds = ("00" + (this % 3600) % 60).slice(-2);
return hours + ":" + minutes + ":" + seconds;
}
// Usage: [number variable].toHHMMSS();
// Here is a simple test
var totalseconds = 1234;
document.getElementById("timespan").innerHTML = totalseconds.toHHMMSS();
// HTML of the test
<div id="timespan"></div>
Easy to follow version for noobies:
var totalNumberOfSeconds = YOURNUMBEROFSECONDS;
var hours = parseInt( totalNumberOfSeconds / 3600 );
var minutes = parseInt( (totalNumberOfSeconds - (hours * 3600)) / 60 );
var seconds = Math.floor((totalNumberOfSeconds - ((hours * 3600) + (minutes * 60))));
var result = (hours < 10 ? "0" + hours : hours) + ":" + (minutes < 10 ? "0" + minutes : minutes) + ":" + (seconds < 10 ? "0" + seconds : seconds);
console.log(result);
This function should do it :
var convertTime = function (input, separator) {
var pad = function(input) {return input < 10 ? "0" + input : input;};
return [
pad(Math.floor(input / 3600)),
pad(Math.floor(input % 3600 / 60)),
pad(Math.floor(input % 60)),
].join(typeof separator !== 'undefined' ? separator : ':' );
}
Without passing a separator, it uses : as the (default) separator :
time = convertTime(13551.9941351); // --> OUTPUT = 03:45:51
If you want to use - as a separator, just pass it as the second parameter:
time = convertTime(1126.5135155, '-'); // --> OUTPUT = 00-18-46
See also this Fiddle.
Chiming in on this old thread -- the OP stated HH:MM:SS, and many of the solutions work, until you realize you need more than 24 hours listed. And maybe you don't want more than a single line of code. Here you go:
d=(s)=>{f=Math.floor;g=(n)=>('00'+n).slice(-2);return f(s/3600)+':'+g(f(s/60)%60)+':'+g(s%60)}
It returns H+:MM:SS. To use it, simply use:
d(91260); // returns "25:21:00"
d(960); // returns "0:16:00"
...I tried to get it to use the least amount of code possible, for a nice one-liner approach.
For the special case of HH:MM:SS.MS (eq: "00:04:33.637") as used by FFMPEG to specify milliseconds.
[-][HH:]MM:SS[.m...]
HH expresses the number of hours, MM the number of minutes for a
maximum of 2 digits, and SS the number of seconds for a maximum of 2
digits. The m at the end expresses decimal value for SS.
/* HH:MM:SS.MS to (FLOAT)seconds ---------------*/
function timerToSec(timer){
let vtimer = timer.split(":")
let vhours = +vtimer[0]
let vminutes = +vtimer[1]
let vseconds = parseFloat(vtimer[2])
return vhours * 3600 + vminutes * 60 + vseconds
}
/* Seconds to (STRING)HH:MM:SS.MS --------------*/
function secToTimer(sec){
let o = new Date(0)
let p = new Date(sec*1000)
return new Date(p.getTime()-o.getTime())
.toISOString()
.split("T")[1]
.split("Z")[0]
}
/* Example: 7hours, 4 minutes, 33 seconds and 637 milliseconds */
const t = "07:04:33.637"
console.log(
t + " => " +
timerToSec(t) +
"s"
)
/* Test: 25473 seconds and 637 milliseconds */
const s = 25473.637 // "25473.637"
console.log(
s + "s => " +
secToTimer(s)
)
Example usage, a milliseconds transport timer:
/* Seconds to (STRING)HH:MM:SS.MS --------------*/
function secToTimer(sec){
let o = new Date(0)
let p = new Date(sec*1000)
return new Date(p.getTime()-o.getTime())
.toISOString()
.split("T")[1]
.split("Z")[0]
}
let job, origin = new Date().getTime()
const timer = () => {
job = requestAnimationFrame(timer)
OUT.textContent = secToTimer((new Date().getTime() - origin) / 1000)
}
requestAnimationFrame(timer)
span {font-size:4rem}
<span id="OUT"></span>
<br>
<button onclick="origin = new Date().getTime()">RESET</button>
<button onclick="requestAnimationFrame(timer)">RESTART</button>
<button onclick="cancelAnimationFrame(job)">STOP</button>
Example usage, binded to a media element
/* Seconds to (STRING)HH:MM:SS.MS --------------*/
function secToTimer(sec){
let o = new Date(0)
let p = new Date(sec*1000)
return new Date(p.getTime()-o.getTime())
.toISOString()
.split("T")[1]
.split("Z")[0]
}
VIDEO.addEventListener("timeupdate", function(e){
OUT.textContent = secToTimer(e.target.currentTime)
}, false)
span {font-size:4rem}
<span id="OUT"></span><br>
<video id="VIDEO" width="400" controls autoplay>
<source src="https://www.w3schools.com/html/mov_bbb.mp4" type="video/mp4">
</video>
Outside the question, those functions written in php:
<?php
/* HH:MM:SS to (FLOAT)seconds ------------------*/
function timerToSec($timer){
$vtimer = explode(":",$timer);
$vhours = (int)$vtimer[0];
$vminutes = (int)$vtimer[1];
$vseconds = (float)$vtimer[2];
return $vhours * 3600 + $vminutes * 60 + $vseconds;
}
/* Seconds to (STRING)HH:MM:SS -----------------*/
function secToTimer($sec){
return explode(" ", date("H:i:s", $sec))[0];
}
After looking at all the answers and not being happy with most of them, this is what I came up with. I know I am very late to the conversation, but here it is anyway.
function secsToTime(secs){
var time = new Date();
// create Date object and set to today's date and time
time.setHours(parseInt(secs/3600) % 24);
time.setMinutes(parseInt(secs/60) % 60);
time.setSeconds(parseInt(secs%60));
time = time.toTimeString().split(" ")[0];
// time.toString() = "HH:mm:ss GMT-0800 (PST)"
// time.toString().split(" ") = ["HH:mm:ss", "GMT-0800", "(PST)"]
// time.toTimeString().split(" ")[0]; = "HH:mm:ss"
return time;
}
I create a new Date object, change the time to my parameters, convert the Date Object to a time string, and removed the additional stuff by splitting the string and returning only the part that need.
I thought I would share this approach, since it removes the need for regex, logic and math acrobatics to get the results in "HH:mm:ss" format, and instead it relies on built in methods.
You may want to take a look at the documentation here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Date
below is the given code which will convert seconds into hh-mm-ss format:
var measuredTime = new Date(null);
measuredTime.setSeconds(4995); // specify value of SECONDS
var MHSTime = measuredTime.toISOString().substr(11, 8);
Get alternative method from Convert seconds to HH-MM-SS format in JavaScript
var time1 = date1.getTime();
var time2 = date2.getTime();
var totalMilisec = time2 - time1;
alert(DateFormat('hh:mm:ss',new Date(totalMilisec)))
/* ----------------------------------------------------------
* Field | Full Form | Short Form
* -------------|--------------------|-----------------------
* Year | yyyy (4 digits) | yy (2 digits)
* Month | MMM (abbr.) | MM (2 digits)
| NNN (name) |
* Day of Month | dd (2 digits) |
* Day of Week | EE (name) | E (abbr)
* Hour (1-12) | hh (2 digits) |
* Minute | mm (2 digits) |
* Second | ss (2 digits) |
* ----------------------------------------------------------
*/
function DateFormat(formatString,date){
if (typeof date=='undefined'){
var DateToFormat=new Date();
}
else{
var DateToFormat=date;
}
var DAY = DateToFormat.getDate();
var DAYidx = DateToFormat.getDay();
var MONTH = DateToFormat.getMonth()+1;
var MONTHidx = DateToFormat.getMonth();
var YEAR = DateToFormat.getYear();
var FULL_YEAR = DateToFormat.getFullYear();
var HOUR = DateToFormat.getHours();
var MINUTES = DateToFormat.getMinutes();
var SECONDS = DateToFormat.getSeconds();
var arrMonths = new Array("January","February","March","April","May","June","July","August","September","October","November","December");
var arrDay=new Array('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday');
var strMONTH;
var strDAY;
var strHOUR;
var strMINUTES;
var strSECONDS;
var Separator;
if(parseInt(MONTH)< 10 && MONTH.toString().length < 2)
strMONTH = "0" + MONTH;
else
strMONTH=MONTH;
if(parseInt(DAY)< 10 && DAY.toString().length < 2)
strDAY = "0" + DAY;
else
strDAY=DAY;
if(parseInt(HOUR)< 10 && HOUR.toString().length < 2)
strHOUR = "0" + HOUR;
else
strHOUR=HOUR;
if(parseInt(MINUTES)< 10 && MINUTES.toString().length < 2)
strMINUTES = "0" + MINUTES;
else
strMINUTES=MINUTES;
if(parseInt(SECONDS)< 10 && SECONDS.toString().length < 2)
strSECONDS = "0" + SECONDS;
else
strSECONDS=SECONDS;
switch (formatString){
case "hh:mm:ss":
return strHOUR + ':' + strMINUTES + ':' + strSECONDS;
break;
//More cases to meet your requirements.
}
}
I just wanted to give a little explanation to the nice answer above:
var totalSec = new Date().getTime() / 1000;
var hours = parseInt( totalSec / 3600 ) % 24;
var minutes = parseInt( totalSec / 60 ) % 60;
var seconds = totalSec % 60;
var result = (hours < 10 ? "0" + hours : hours) + "-" + (minutes < 10 ? "0" + minutes : minutes) + "-" + (seconds < 10 ? "0" + seconds : seconds);
On the second line, since there are 3600 seconds in 1 hour, we divide the total number of seconds by 3600 to get the total number of hours. We use parseInt to strip off any decimal. If totalSec was 12600 (3 and half hours), then parseInt( totalSec / 3600 ) would return 3, since we will have 3 full hours. Why do we need the % 24 in this case? If we exceed 24 hours, let's say we have 25 hours (90000 seconds), then the modulo here will take us back to 1 again, rather than returning 25. It is confining the result within a 24 hour limit, since there are 24 hours in one day.
When you see something like this:
25 % 24
Think of it like this:
25 mod 24 or what is the remainder when we divide 25 by 24
None of the answers here satisfies my requirements as I want to be able to handle
Large numbers of seconds (days), and
Negative numbers
Although those are not required by the OP, it's good practice to cover edge cases, especially when it takes little effort.
It's pretty obvious is that the OP means a NUMBER of seconds when he says seconds. Why would peg your function on String?
function secondsToTimeSpan(seconds) {
const value = Math.abs(seconds);
const days = Math.floor(value / 1440);
const hours = Math.floor((value - (days * 1440)) / 3600);
const min = Math.floor((value - (days * 1440) - (hours * 3600)) / 60);
const sec = value - (days * 1440) - (hours * 3600) - (min * 60);
return `${seconds < 0 ? '-':''}${days > 0 ? days + '.':''}${hours < 10 ? '0' + hours:hours}:${min < 10 ? '0' + min:min}:${sec < 10 ? '0' + sec:sec}`
}
secondsToTimeSpan(0); // => 00:00:00
secondsToTimeSpan(1); // => 00:00:01
secondsToTimeSpan(1440); // => 1.00:00:00
secondsToTimeSpan(-1440); // => -1.00:00:00
secondsToTimeSpan(-1); // => -00:00:01
Simple function to convert seconds into in hh:mm:ss format :
function getHHMMSSFromSeconds(totalSeconds) {
if (!totalSeconds) {
return '00:00:00';
}
const hours = Math.floor(totalSeconds / 3600);
const minutes = Math.floor(totalSeconds % 3600 / 60);
const seconds = totalSeconds % 60;
const hhmmss = padTo2(hours) + ':' + padTo2(minutes) + ':' + padTo2(seconds);
return hhmmss;
}
// function to convert single digit to double digit
function padTo2(value) {
if (!value) {
return '00';
}
return value < 10 ? String(value).padStart(2, '0') : value;
}
Here is a function to convert seconds to hh-mm-ss format based on powtac's answer here
jsfiddle
/**
* Convert seconds to hh-mm-ss format.
* #param {number} totalSeconds - the total seconds to convert to hh- mm-ss
**/
var SecondsTohhmmss = function(totalSeconds) {
var hours = Math.floor(totalSeconds / 3600);
var minutes = Math.floor((totalSeconds - (hours * 3600)) / 60);
var seconds = totalSeconds - (hours * 3600) - (minutes * 60);
// round seconds
seconds = Math.round(seconds * 100) / 100
var result = (hours < 10 ? "0" + hours : hours);
result += "-" + (minutes < 10 ? "0" + minutes : minutes);
result += "-" + (seconds < 10 ? "0" + seconds : seconds);
return result;
}
Example use
var seconds = SecondsTohhmmss(70);
console.log(seconds);
// logs 00-01-10
There are lots of options of solve this problem, and obvious there are good option suggested about, But I wants to add one more optimized code here
function formatSeconds(sec) {
return [(sec / 3600), ((sec % 3600) / 60), ((sec % 3600) % 60)]
.map(v => v < 10 ? "0" + parseInt(v) : parseInt(v))
.filter((i, j) => i !== "00" || j > 0)
.join(":");
}
if you don't wants formatted zero with less then 10 number, you can use
function formatSeconds(sec) {
return parseInt(sec / 3600) + ':' + parseInt((sec % 3600) / 60) + ':' + parseInt((sec % 3600) % 60);
}
Sample Code http://fiddly.org/1c476/1
In one line, using T.J. Crowder's solution :
secToHHMMSS = seconds => `${Math.floor(seconds / 3600)}:${Math.floor((seconds % 3600) / 60)}:${Math.floor((seconds % 3600) % 60)}`
In one line, another solution that also count days :
secToDHHMMSS = seconds => `${parseInt(seconds / 86400)}d ${new Date(seconds * 1000).toISOString().substr(11, 8)}`
Source : https://gist.github.com/martinbean/2bf88c446be8048814cf02b2641ba276
var sec_to_hms = function(sec){
var min, hours;
sec = sec - (min = Math.floor(sec/60))*60;
min = min - (hours = Math.floor(min/60))*60;
return (hours?hours+':':'') + ((min+'').padStart(2, '0')) + ':'+ ((sec+'').padStart(2, '0'));
}
alert(sec_to_hms(2442542));
Have you tried adding seconds to a Date object?
Date.prototype.addSeconds = function(seconds) {
this.setSeconds(this.getSeconds() + seconds);
};
var dt = new Date();
dt.addSeconds(1234);
A sample:
https://jsfiddle.net/j5g2p0dc/5/
Updated:
Sample link was missing so I created a new one.
You can also use below code:
int ss = nDur%60;
nDur = nDur/60;
int mm = nDur%60;
int hh = nDur/60;
For anyone using AngularJS, a simple solution is to filter the value with the date API, which converts milliseconds to a string based on the requested format. Example:
<div>Offer ends in {{ timeRemaining | date: 'HH:mm:ss' }}</div>
Note that this expects milliseconds, so you may want to multiply timeRemaining by 1000 if you are converting from seconds (as the original question was formulated).
I ran into the case some have mentioned where the number of seconds is more than a day. Here's an adapted version of #Harish Anchu's top-rated answer that accounts for longer periods of time:
function secondsToTime(seconds) {
const arr = new Date(seconds * 1000).toISOString().substr(11, 8).split(':');
const days = Math.floor(seconds / 86400);
arr[0] = parseInt(arr[0], 10) + days * 24;
return arr.join(':');
}
Example:
secondsToTime(101596) // outputs '28:13:16' as opposed to '04:13:16'
String.prototype.toHHMMSS = function () {
var sec_num = parseInt(this, 10); // don't forget the second param
var hours = Math.floor(sec_num / 3600);
var minutes = Math.floor((sec_num - (hours * 3600)) / 60);
var seconds = sec_num - (hours * 3600) - (minutes * 60);
if (hours < 10) {hours = "0"+hours;}
if (minutes < 10) {minutes = "0"+minutes;}
if (seconds < 10) {seconds = "0"+seconds;}
return hours+':'+minutes+':'+seconds;
}
Usage Example
alert("186".toHHMMSS());

How can I make Math.random call a different value everytime it is called in AS3?

Let's say I had something like what's below in a function, the first time it gets called, I have a random value, but every other time it's called, I get the same number as it spit out the first time. What can I do to ensure it is randomized each time it is called?
if (Cosmo.hitTestObject(Asteroid5))
{
Asteroid5.y = (Math.random() * 20 - 5);
Asteroid5.x = (Math.random() * 20 - 15);
Asteroid5.x = Asteroid5.x + (Math.random() * 20 - 15);
Asteroid5.y = Asteroid5.y + (Math.random() * 20 - 5);
}
[Edited, based on comments below]
The AS3 reference page for Math.random() says:
Returns a pseudo-random number n, where 0 <= n < 1. The number
returned is calculated in an undisclosed manner, and is
"pseudo-random" because the calculation inevitably contains some
element of non-randomness.
If you want to use a seed-based pseudo-random number generator, this should help you:
http://www.kirupa.com/forum/showthread.php?365564-AS3-Seeded-Pseudo-Random-Number-Generator
in your 1st 2 statements, it looks like you're using Asteroid5.y & Asteroid5.x as variables. so one idea would be to use variables of the correct type (i'm assuming it's Number):
var rndX:Number = (Math.random() * 20 - 15);
var rndY:Number = (Math.random() * 20 - 5);
var rndX2:Number = rndX + (Math.random() * 20 - 15);
var rndY2:Number = rndY + (Math.random() * 20 - 5);
Asteroid5.x = rndX2;
Asteroid5.y = rndY2; or
var rndX:Number;
var rndY:Number;
for ( var i:uint=0; i<2; i++ )
{
rndX += (Math.random() * 20 - 15);
rndY += (Math.random() * 20 - 5);
}
Asteroid5.x = rndX;
Asteroid5.y = rndY;
now you can step through your code in debug mode and see exactly what going on as the values of the variables change. also maybe search your code for 'Asteroid5.x =' & 'Asteroid5.y =' & 'Asteroid5.y=' & 'Asteroid5.x='

as3 getting the "hours" in milliseconds in mp3

In As3 the code below gets the minutes and the seconds:
var minutes:uint = Math.floor(PrayPrayer.position / 1000 / 60);
var seconds:uint = Math.floor(PrayPrayer.position / 1000) % 60;
But what if your listening to an audio talk that goes over the hour mark?
What is the math needed to get the hours from an mp3 talk?
var hours:uint = Math.floor(PrayPrayer.position / 1000) % 60 & (((???????)));
this is my conversion method:
public static var MINUTE:Number = 60;
public static var HOUR:Number = 60 * MINUTE;
public static var DAY:Number = 24 * HOUR;
/**
* returns string created from seconds value in following format hours:minutes:seconds, i.e. 121 seconds will be displayed as 00:02:01
* #param seconds <i>Number</i>
* #return <i>String</i>
*/
public static function secondsToHMS(seconds:Number, doNotRound:Boolean = false):String
{
var _bNegative:Boolean = seconds < 0;
seconds = Math.abs(seconds);
var time:Number = (doNotRound) ? seconds:Math.round(seconds);
var ms:Number;
var msec:String;
if (doNotRound)
{
ms = seconds - (seconds | 0);
msec = prependZeros((ms * 1000) | 0, 3);
}
var sec:Number = (time | 0) % MINUTE;
var min:Number = Math.floor((time / MINUTE) % MINUTE);
var hrs:Number = Math.floor(time / HOUR);
//
return (_bNegative ? "-":"") +
((hrs > 9) ? "":"0") + hrs + ":" +
((min > 9) ? "":"0") + min + ":" +
((sec > 9) ? "":"0") + sec +
(doNotRound ? "." + msec:"");
}
prependZeros is another utility to add "0" in front of given string.
So PrayPrayer.position is in milliseconds. Your minutes line is dividing by 1000 to get seconds, then dividing by 60 to go from seconds to minutes. Your seconds line is looking at the remainder.
What you started in your hours line is using %, so will look at the remainder - you're using seconds there. % is the modulo operator. It gives you the remainder of integer division. So your line
var seconds:uint = Math.floor(PrayPrayer.position / 1000) % 60;
is finding the number of seconds (PrayPrayer.position / 1000), which could be something big like 2337, dividing by 60 and just keeping the remainder. 2337/60 = 38 remainder 57, so 2337%60 will be 57.
An easy way to find hours is to use the same trick with your minutes.
var minutes:uint = Math.floor(PrayPrayer.position / 1000 / 60);
var seconds:uint = Math.floor(PrayPrayer.position / 1000) % 60;
var hours:uint = Math.floor(minutes / 60);
minutes %= 60; // same as minutes = minutes % 60. Forces minutes to be between 0 and 59.

increase chances of even number

If I am getting a random number, how do I increase my chances of making that random number to be even. I am not looking to make it even every time. I am just looking to generate a random number say... %70 of the time or %90 of the time.
private function randNum (high, low) {
return Math.floor(Math.random() * (1+high-low)) + low;
}
Would I pass in a greater range of numbers to this function? Or would I have to scrap this function altogether?
Thank you
private function randNum (high : Number, low : Number) : int
{
var even : Boolean = Math.random() < 0.7; //set probability of even number here
var rand : int = Math.floor(Math.random() * (1+high-low)) + low;
if (even)
while (rand % 2 != 0)
rand = Math.floor(Math.random() * (1+high-low)) + low;
else
while (rand % 2 != 1)
rand = Math.floor(Math.random() * (1+high-low)) + low;
return rand;
}
Test:
var even : int = 0;
var odd : int = 0;
for (var i : int = 0; i < 100000; i++)
{
var a : int = randNum(1, 20);
if (a % 2 == 0)
even++;
else
odd++;
}
trace(even, odd);
Output:
[trace] 69869 30131
A little too late ;) but another one with no loop and using bit masking operation :
ret & -2 will make your number even, then depending on the result of (Math.random() >= evenProbability) you set the lower bit to be 1 to give an odd number
function randomRange(low:int, high:int, evenProbability:Number = 0.5):int{
var ret:int = int( Math.random() * ( 1 + high - low ) ) + low
return ( ret & -2 ) | int( Math.random() >= evenProbability )
}
Here a live test with wonderfl : http://wonderfl.net/c/9IHx