I am trying to understand the websocketTraffic data exported from my Chrome dev tools. An example looks like this:
{
'type': 'receive',
'time': 1640291138.212745,
'opcode': 1,
'data': '<r xmlns=\'urn:xmpp:sm:3\'/>',
}
I see a "time" field but I actually cant find anything about what it means except this from the spec (http://www.softwareishard.com/blog/har-12-spec/):
time [number] - Total elapsed time of the request in milliseconds. This is the sum of all timings available in the timings object (i.e. not including -1 values) .
Is this really milliseconds, down to the millionth of a millisecond? I am trying to see how much time has elapsed between two WS events, so any insight would be very helpful. Thanks
Disclaimer:
This answer is not backed by official docs. However, I studied this problem for quite some time now, and my solution seems to make sense.
Answer:
Move the dot 3 places to the right, (i.e 1640291138.212745 -> 1640291138212.745) and you will get the actual time. Try to run this
new Date(1640291138212.745).toISOString()
and see if it fits your startedDateTime in the parent WebSocket entry in your har.
Probably Chrome saves the "time" field as seconds since epoch, instead of milliseconds since epoch. So "moving the dot 3 places to the right" actually means to multiply by a 1000 and that means converting to milliseconds.
I've got a small fun puzzle for you.
It has me 100% stumped! After 2-hours of trial-and-error + Google searches, I'm asking for help :)
Timestamp:1393878630947498
What is the excel formula to convert that to a Date/Time?
I know for a FACT the actual Date/Time is: May 9, 2017, 10:21 PM
Many thanks!
Take the big integer and multiply it by:
0.000000000030752269457537
then format the product as date/time
NOTE:
This simple relationship is only valid if the two time/date forms have the same starting point.
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).
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);
I'm currently storing various metadata about videos and one of those bits of data is the length of a video.
So if a video is 10 minutes 35 seconds long, it's saved as "10:35" in the database.
But what I'd like to do is retrieve a listing of videos by length (longest first, shortest last).
The problem I'm having is that if a video is "2:56", it's coming up as longest because the number 2 is more than the number 1 in.
So, how can I order data based on that length field so that "10:35" is recognized as being longer than "2:56" (as per my example)?
SELECT * FROM table ORDER BY str_to_date(meta_time,'%l:%i')
You can find the specific formatters on the MySQL Website.
For example:
%k -> Hour (0..23)
%l -> Hour (1..12)
The easiest choice is to store a integer (seconds) or a float (minutes) instead of a string. So 10:35 would be 635 in seconds or 10.583 in minutes. You can sort by these numerically very easily. And you can output them in the format you'd like with some simple math and string functions.
Some options:
Save it as an integer representing the total number of seconds. "10:35" => 635
Save it as a timestamp object with no date component. "10:35" => MAKETIME(0, 10, 34)
Save it with leading decimals or spaces. "2:25" => " 2:25"
My preference would be for the first option.
You could try to see if
ORDER BY TIME_TO_SEC(timefield)
would parse it correctly, however it is not an optimal approach to store time as strings in the database, and I suggest that you store them as TIME if you are able to. Then you can use standard formatting functions to present them as you like.
I had the same problem - storing videos length in database.
I solved it by using TIME mysql type - it solves all ordering and converting issues.