Hello all my question is that i have two dates Like $date1 = 2021-05-04 15:52:32 and $date2 = 2021-05-05 15:52:32 how to calculate difference in milliseconds between this dates in php please help.
Related
I am trying to convert EPOC time to date time and need to extract the time only from that
I am doing below
$min = $Time_Start | measure -Minimum
$max = $Time_End | measure -Maximum
[datetime]$oUNIXDatemin=(Get-Date 01.01.1970)+([System.TimeSpan]::fromseconds($min.Minimum))
$oUNIXDatemin_1 = $oUNIXDatemin.ToString("HH:mm:ss")
[datetime]$oUNIXDatemax=(Get-Date 01.01.1970)+([System.TimeSpan]::fromseconds($max.Maximum))
$oUNIXDatemax_1 = $oUNIXDatemax.ToString("HH:mm:ss")
Problem is while converting I am getting $oUNIXDatemin_1 and $oUNIXDatemax_1 value like
$oUNIXDatemin_1
12 October 2021 07:46:46
$oUNIXDatemax_1
12 October 2021 21:16:04
My EPOC values are
$min.Minimum
1634024806
$max.Maximum
1634073364
Please let me know what is wrong here. Need to find the difference in HH:mm:ss format.
In PowerShell, you'd usually use a format string. Subtracting two PowerShell datetimes returns a value of type Timespan, which is well-behaved over a span of more than 24 hours.
([datetime]"12 October 2021 21:16:04" - [datetime]"12 October 2021 07:46:46") -f "HH:mm:ss"
13:29:18
Be careful here. Both intervals (durations) and time (of day) have the same format, but different meanings. For example, it makes sense to multiply the interval "01:00:00" (1 hour) by 3 to get three hours; it doesn't make sense to multiply the time "01:00:00" (1 o'clock AM) by 3.
I'm sure the overall calculation can be simplified, but it's too early for me.
can anyone explain how can I get difference between two dates in calendar days, not in whole 24-hour periods. There is a good solution here: Getting the difference between two NSDates in (months/days/hours/minutes/seconds) -- but it doesn't work for me as, for example, it gives the difference between 23:00 today and 1:30 tomorrow as 0 days despite of calendar dates differ already by 1.
Use the normal way to calculate the difference in days with one change - convert both of your dates to midnight.
let d1 = ... // your first date
let d2 = ... // your second date
let cal = Calendar.current
let days = cal.dateComponents([.day], from: cal.startOfDay(for: d1), to: cal.startOfDay(for: d2)).day!
This will give an answer of 1 for "yesterday at 23:00" and "today at 1:30", for example.
I have 2 times stored in character arrays in MATLAB.
a = '11:00 PM'
b = '07:30 AM'
I want to find the difference in hours between the 2 times, which should be 8.5 hours in this example. Is there any short method to do that? I can datenum both numbers, subtract them, datevec the difference, extract the hours and minutes from the vector, and convert them into hours, but this takes a lot of lines. Is there a more efficient way of doing this or is there an existing function?
You can do this by converting each string using datetime, taking the difference, then converting the result with hours:
numHours = hours(diff(datetime({a; b}, 'InputFormat', 'hh:mm a')));
numHours = numHours + 24.*(numHours < 0)
numHours =
8.5000
The second line accounts for the condition in your example, where the second time has to occur on the next day for the time difference to be positive, so 24 hours are added to the (negative) difference.
add a date to the time
like
a = '1/1/2000 11:00 PM'
b = '1/1/2000 07:30 AM'
the convert the string to datetime
x=str2num(strrep(a,':',''))
y=str2num(strrep(b,':',''))
then fine the difference between 2 dates
e = etime(x,y)
this will give you number of seconds between both times
I am looking for helping doing time conversions from UTC time to string using MATLAB.
I am trying to extract time from a data file collected at the end of October 2010.
The data file says it is reporting in UTC time and the field is an integer string value in milliseconds that is around 3.02e11. I would like to convert this to a string but am have some trouble.
I figured out that the units are most definitely in milliseconds so I convert this to fractions of days to be compatible with datenum format.
If the data was collected at the end of October (say, October 31, 2010) then I can guess what kind of number I might get. I thought that January 1, 2001 would be a good epoch and calculated what sort of number (in days) I might get:
suspectedDate = datenum('October 31, 2010')
suspectedEpoch = datenum('January 1, 2001')
suspectedTimeInDays = suspectedDate - suspectedEpoch
Which comes out as 3590.
However, my actual time, in days, comes out with the following code
actualTime = 3.02e11
actualTimeInDays = 3.02e11/1000/24/3600
as 3495.4.
This is troubling as the difference is only 94.6 -- not a full year. This would mean either the documentation for the file is wrong or the epoch is close to April 1-5, 2001:
calculatedEpoch = suspectedDate - actualTimeInDays
calculatedEpochStr = datestr(calculatedEpoch)
Alternately, if the epoch is January 1, 2001 then the actual date in the file is from the end of July.
ifEpochIsJanuaryDate = suspectedEpoch + actualTimeInDays
ifEpochIsJanuaryDateStr = datestr(ifEpochIsJanuaryDate)
Is this a known UTC format and can anyone give suggestions on how to get an October date from 3.02e11 magnitude number?
Unix time today is about 13e11, and is measured in ms since 1970.
If your time is ~3e11, then it's probably since year 2000.
>> time_unix = 1339116554872; % example time
>> time_reference = datenum('1970', 'yyyy');
>> time_matlab = time_reference + time_unix / 8.64e7;
>> time_matlab_string = datestr(time_matlab, 'yyyymmdd HH:MM:SS.FFF')
time_matlab_string =
20120608 00:49:14.872
Notes:
1) change 1970 into 2000 if your time is since 2000;
2) See the definition of matlab's time.
3) 8.64e7 is number of milliseconds in a day.
4) Matlab does not apply any time-zone shifts, so the result is the same UTC time.
5) Example for backward transformation:
>> matlab_time = now;
>> unix_time = round(8.64e7 * (matlab_time - datenum('1970', 'yyyy')))
unix_time =
1339118367664
You can't just make up your own epoch. Also datenum returns things in days. So the closeness you got with doing your math was just a coincidence.
Turns out that
>> datenum('Jan-1-0000')
ans =
1
and
>> datenum('Jan-1-0001')
ans =
367
So Matlab should be returning things in days since Jan. 1, 0000. (Not a typo)
However, I'd look carefully at this 3.02e11 number and find out exactly what it means. I'm pretty sure it's not standard Unix UTC, which should be seconds since January 1, 1970. It's way too big. It's close to GMT: Mon, 1 Jan 11540 08:53:20 UTC.
I have a dataset for which I have extracted the date at which an event occurred. The date is in the format of MMDDYY although MatLab does not show leading zeros so often it's MDDYY.
Is there a method to find the mean or median (I could use either) date? median works fine when there is an odd number of days but for even numbers I believe it is averaging the two middle ones which doesn't produce sensible values. I've been trying to convert the dates to a MatLab format with regexp and put it back together but I haven't gotten it to work. Thanks
dates=[32381 41081 40581 32381 32981 41081 40981 40581];
You can use datenum to convert dates to a serial date number (1 at 01/01/0000, 2 at 02/01/0000, 367 at 01/01/0001, etc.):
strDate='27112011';
numDate = datenum(strDate,'ddmmyyyy')
Any arithmetic operation can then be performed on these date numbers, like taking a mean or median:
mean(numDates)
median(numDates)
The only problem here, is that you don't have your dates in a string type, but as numbers. Luckily datenum also accepts numeric input, but you'll have to give the day, month and year separated in a vector:
numDate = datenum([year month day])
or as rows in a matrix if you have multiple timestamps.
So for your specified example data:
dates=[32381 41081 40581 32381 32981 41081 40981 40581];
years = mod(dates,100);
dates = (dates-years)./100;
days = mod(dates,100);
months = (dates-days)./100;
years = years + 1900; % set the years to the 20th century
numDates = datenum([years(:) months(:) days(:)]);
fprintf('The mean date is %s\n', datestr(mean(numDates)));
fprintf('The median date is %s\n', datestr(median(numDates)));
In this example I converted the resulting mean and median back to a readable date format using datestr, which takes the serial date number as input.
Try this:
dates=[32381 41081 40581 32381 32981 41081 40981 40581];
d=zeros(1,length(dates));
for i=1:length(dates)
d(i)=datenum(num2str(dates(i)),'ddmmyy');
end
m=mean(d);
m_str=datestr(m,'dd.mm.yy')
I hope this info to be useful, regards
Store the dates as YYMMDD, rather than as MMDDYY. This has the useful side effect that the numeric order of the dates is also the chronological order.
Here is the pseudo-code for a function that you could write.
foreach date:
year = date % 100
date = (date - year) / 100
day = date % 100
date = (date - day) / 100
month = date
newdate = year * 100 * 100 + month * 100 + day
end for
Once you have the dates in YYMMDD format, then find the median (numerically), and this is also the median chronologically.
You see above how to present dates as numbers.
I will add no your issue of finding median of the list. The default matlab median function will average the two middle values when there are an even number of values.
But you can do it yourself! Try this:
dates; % is your array of dates in numeric form
sdates = sort(dates);
mediandate = sdates(round((length(sdates)+1)/2));