Calculate date from numeric value - language-agnostic

The number 71867806 represents the present day, with the smallest unit of days.
Sorry guy's, caching owned me, it's actually milliseconds!
How can I
calculate the currente date from it?
(or) convert it into an Unix timestamp?
Solution shouldn't use language depending features.
Thanks!

This depends on:
What unit this number represents (days, seconds, milliseconds, ticks?)
When the starting date was
In general I would discourage you from trying to reinvent the wheel here, since you will have to handle every single exception in regards to dates yourself.

If it's truly an integer number of days, and the number you've given is for today (April 21, 2010, for me as I'm reading this), then the "zero day" (the epoch) was obviously enough 71867806 days ago. I can't quite imagine why somebody would pick that though -- it works out to roughly 196,763 years ago (~194,753 BC, if you prefer). That seems like a strange enough time to pick that I'm going to guess that there's more to this than what you've told us (perhaps more than you know about).
It seems to me the first thing to do is verify that the number does increase by one every 24 hours. If at all possible keep track of the exact time when it does increment.

First, you have only one point, and that's not quite enough. Get the number for "tomorrow" and see if that's 71867806+1. If it is, then you can safely bet that +1 means +1 day. If it's something like tomorrow-today = 24, then odds are +1 means +1 hour, and the logic to display days only shows you the "day" part. If it's something else check to see if it's near (24*60, which would be minutes), (24*60*60, which would be seconds), or (24*60*60*1000, which would be milliseconds).
Once you have an idea of what kind of units you are using, you can estimate how many years ago the "start" date of 0 was. See if that aligns with any of the common calendar systems located at http://en.wikipedia.org/wiki/List_of_calendars. Odds are that the calendar you are using isn't a truly new creation, but a reimplementation of an existing calendar. If it seems very far back, it might be an Julian Date, which has day 0 equivalent to BCE 4713 January 01 12:00:00.0 UT Monday. Julian Dates and Modified Julian dates are often used in astronomy calculations.
The next major goal is to find Jan 1, 1970 00:00:00. If you can find the number that represents that date, then you simply subtract it from this foreign calendar system and convert the remainder from the discovered units to milliseconds. That will give you UNIX time which you can then use with the standard UNIX utilities to convert to a time in any time zone you like.
In the end, you might not be able to be 100% certain that your conversion is exactly the same as the hand implemented system, but if you can test your assumptions about the calendar by plugging in numbers and seeing if they display as you predicted. Use this technique to create a battery of tests which will help you determine how this system handles leap years, etc. Remember, it might not handle them at all!

What time is: 71,867,806 miliseconds from midnight?
There are:
- 86,400,000 ms/day
- 3,600,000 ms/hour
- 60,000 ms/minute
- 1,000 ms/second
Remove and tally these units until you have the time, as follows:
How many days? None because 71,867,806 is less than 86,400,000
How many hours? Maximum times 3,600,000 can be removed is 19 times
71,867,806 - (3,600,000 * 19) = 3,467,806 ms left.
How many minutes? Maximum times 60,000 can be removed is 57 times.
3,467,806 - (60,000 * 57) = 47,806 ms left
How many seconds? Maximum times 1,000 can be removed is 47 times.
47,806 - (1,000 * 47) = 806
So the time is: 19:57:47.806

It is indeed a fairly long time ago if the smallest number is in days. However, assuming you're sure about it I could suggest the following shell command which would be obviously not valid for dates before 1st Jan. 1970:
date -d "#$(echo '(71867806-71853086)*3600*24'|bc)" +%D
or without bc:
date -d "#$(((71867806 - 71853086) * 3600 * 24))" +%D

Sorry again for the messy question, i got the solution now. In js it looks like that:
var dayZero = new Date(new Date().getTime() - 71867806 * 1000);

Related

converting strange timestamp from json to normal data time

I'm getting some very odd looking timestamps from netflix's video API, they look like this ;
527637852762
I assume its a timestamp as in the json it looks like this,
"time": 527780548207
How can I convert this ? it should equate to around some day in September 2017.
So far I've tired dividing by 100, 1000 and no luck.
Thanks
Seconds since 1.1.1970?
It is a common unix time stamp
Referring to the Netflix API documentation they say:
For times since the epoch both seconds and milliseconds are supported because both are in common use and it helps to avoid confusion when copy and pasting from another source.
Milliseconds Timestamp to Date
I figured it out, its a 2001 timestamp. Seconds since 2001. So take that figure / 1000 and that converts correctly :)
epochconverter.com/coredata

Why is MySQL's maximum time limit 838:59:59?

I've run into the limit myself, but despite lots of chatter online, I've never seen an explanation for why the upper and lower limit for the TIME data type is what it is. The official reference at http://dev.mysql.com/doc/refman/5.7/en/time.html says
TIME values may range from '-838:59:59' to '838:59:59'. The hours part may be so large because the TIME type can be used not only to represent a time of day (which must be less than 24 hours), but also elapsed time or a time interval between two events (which may be much greater than 24 hours, or even negative).
But I'm wondering not why the hours part is allowed to be "so large", but why it's cut off where it is. There doesn't seem to be any significance to that many hours in regards to days, or if I try to imagine possible cutoffs for how many seconds could be stored as an integer. So why the range?
The TIME values were always stored on 3 bytes in MySQL. But the format changed on version 5.6.4. I suspect this was not the first time when it changed. But the other change, if there was one, happened long time ago and there is no public evidence of it. The MySQL source code history on GitHub starts with version 5.5 (the oldest commit is from May 2008) but the change I am looking for happened somewhere around 2001-2002 (MySQL 4 was launched in 2003)
The current format, as described in the documentation, uses 6 bits for seconds (possible values: 0 to 63), 6 bits for minutes, 10 bits for hours (possible values: 0 to 1023), 1 bit for sign (add the negative values of the already mentioned intervals) and 1 bit is unused and labelled "reserved for future extensions".
It is optimized for working with time components (hours, minutes, seconds) and doesn't waste much space. Using this format it's possible to store values between -1023:59:59 and +1023:59:59. However MySQL limits the number of hours to 838, probably for backward compatibility with applications that were written a while ago, when I think this was the limit.
Until version 5.6.4, the TIME values were also stored on 3 bytes and the components were packed as days * 24 * 3600 + hours * 3600 + minutes * 60 + seconds. This format was optimized for working with timestamps (because it was, in fact, a timestamp). Using this format it would be possible to store values in the range of about -2330 to +2330 hours. While having this big range of values available, MySQL was still limiting the values to -838 to +838 hours.
There was bug #11655 on MySQL 4. It was possible to return TIME values outside the -838..+838 range using nested SELECT statements. It was not a feature but a bug and it was fixed.
The only reason to limit the values to this range and to actively change any piece of code that produces TIME values outside it was backward compatibility.
I suspect MySQL 3 used a different format that, due to the way the data was packed, limited the valid values to the range -838..+838 hours.
By looking into the current MySQL's source code I found this interesting formula:
#define TIME_MAX_VALUE (TIME_MAX_HOUR*10000 + TIME_MAX_MINUTE*100 + TIME_MAX_SECOND)
Let's ignore for the moment the MAX part of the names used above and let's remember only that TIME_MAX_MINUTE and TIME_MAX_SECOND are numbers between 00 and 59. The formula just concatenates the hours, minutes and seconds in a single integer number. For example, the value 170:29:45 becomes 1702945.
This formula raises the following question: given that the TIME values are stored on 3 bytes with sign, what is the maximum positive value that can be represented this way?
The value we are looking for is 0x7FFFFF that in decimal notation is 8388607. Since the last four digits (8607) should be read as minutes (86) and seconds (07) and their maximum valid values is 59, the greatest value that can be stored on 3 bytes with sign using the formula above is 8385959. Which, as TIME is +838:59:59. Ta-da!
Guess what? The fragment of C code listed above was extracted from this:
/* Limits for the TIME data type */
#define TIME_MAX_HOUR 838
#define TIME_MAX_MINUTE 59
#define TIME_MAX_SECOND 59
#define TIME_MAX_VALUE (TIME_MAX_HOUR*10000 + TIME_MAX_MINUTE*100 + TIME_MAX_SECOND)
I am sure this is how MySQL 3 used to keep the TIME values internally. This format imposed the limitation of the range, and the backward compatibility requirement on the subsequent versions propagated the limitation to our days.
DATETIME is stored based on a base of 10, see Date and Time Data Type Representation:
DATETIME: Eight bytes: A four-byte integer for date packed as YYYY×10000 + MM×100 + DD and a four-byte integer for time packed as HH×10000 + MM×100 + SS
For convinience and some other reasons, the (old) time format was encoded in the same way, using 3 bytes:
Hours * 10000 + Minutes * 100 + Seconds
This means:
3 bytes = 2^24 = 16.777.216
with sign: 2^23 = 8.388.608
Using the encoding, this represents the magical 838 hours. And max. 8608 seconds for the minutes and seconds (without overflow), which results in the largest valid time 838:59:59. One nice thing about this is that the integer representation of that time, 8385959, is easily readable to a human. But this encoding of course leaves gaps, invalid (unused) integer values (like 8309999).
As of MySQL 5.6.4, time format changed its encoding to
1 bit sign (1= non-negative, 0= negative)
1 bit unused (reserved for future extensions)
10 bits hour (0-838)
6 bits minute (0-59)
6 bits second (0-59)
---------------------
24 bits = 3 bytes
Even though it could now store more hours, for compatibility it still just allows 838 hours.
Obviously, it's hard to answer these types of questions without getting direct feedback from the designers of the database.
But there is some documentation regarding how the different data types are stored internally, and, to an extent, it can help us understand this a little bit.
For, instance, regarding the TIME data type, notice how it's stored internally according to the documentation:
TIME encoding for nonfractional part:
1 bit sign (1= non-negative, 0= negative)
1 bit unused (reserved for future extensions)
10 bits hour (0-838)
6 bits minute (0-59)
6 bits second (0-59)
---------------------
24 bits = 3 bytes
So, as you can see, the goal is to fit the information within 3 bytes. And, of those 3 bytes, 10 bits are reserved for the hours, which pretty much determines the overall range.
That said, 10 bits does allow values up to 1023, so I guess, technically, without any changes to the storage size, the range could have been -1023:59:59 to 1023:59:59. Why they didn't do that and they chose 838 as the cutoff, I have no idea.

JIRA REST API Sprint Issue route - timeSpent and timeSpentSeconds don't match

I'm currently using the JIRA REST API to retrieve the amount of time spent on a ticket in a project using the following route:
/rest/agile/1.0/board/{boardId}/sprint/{sprintId}/issue
Within one of the issues, I receive the following data relating to time:
"timetracking": {
"remainingEstimate": "9w 1d",
"timeSpent": "4d",
"remainingEstimateSeconds": 1324800,
"timeSpentSeconds": 115200
},
When converting the seconds to days, it results in 1.33333333 days, where as the timeSpent specifies 4 days.
Why does timeSpent and timeSpentSeconds not add up, and is there a way to resolve the issue to ensure that they do match?
I figured this out. As it turns out, it does not work on a 24 hour period, i.e. 1d does not equal 86400, but rather it works on hours spent on the project per day, in my case, 8h. Therefore the seconds is 4 * 8 which equals 1.33333... days.

Getting correct sampling frequency, how to?

I am trying to fully understand the sampling frequency concept. I got a useful answer from PaxRomana99 (I need help setting up the correct frequency vector) and it seems like the way to get the correct frequency vector is when you set up the time interval to be related to fs:
t = 0:1/fs:10; % 10 seconds of data sampled at fs
y = sin(2*pi*100.*t);
However, in my case, I have no control on how my time interval is set up, I am given 2 arrays, one is the time interval and one is the sinusoidal wave resulting from that time interval. So should I got through the time array and extract fs? which should be the spacing between the time steps? am I in the wrong path here?
Thanks for any hints and any help!
Yes, as you suggest, you get the sampling frequency from the time array. It's just 1/(time sample interval).

MySQL storing duration time - datatype?

I need to calculate the time a user spends on site. It is difference between logout time and login time to give me something like "Mr X spent 4 hours and 43 minutes online". So to store the4 hours and 43 minutes i declared it like this:
duration time NOT NULL
Is this valid or a better way to store this? I need to store in the DB because I have other calculations I need to use this for + other use cases.
Storing it as an integer number of seconds will be the best way to go.
The UPDATE will be clean and simple - i.e. duration = duration + $increment
As Tristram noted, there are limitations to using the TIME field - e.g. "TIME values may range from '-838:59:59' to '838:59:59'"
The days/hours/minutes/seconds display formatting won't be hardcoded.
The execution of your other calculations will almost surely be clearer when working with an integer "number of seconds" field.
I wouldn't use time as you would be limited to 24 hours. The easiest would just to store an integer in minutes (or seconds depending on the resolution you need).
Consider storing both values as a UNIX-epoch-delta.
I generally prefer to use a signed (64b) bigint (for secondly resolution), or a (signed) (64b) double (if fractional seconds are needed), or a signed (32b) int (if scaled down to minutely or hourly).
Make the unit explicit in the name of the column, for example with a suffix like "_epoch_minutely", for example: "started_epoch_minutely", "finished_epoch_minutely".