I want to increment time in -rfc-3339 format and have used following command :
`date --date="(date --rfc-3339=seconds) + 5 minutes"`
but after increment current time by 5 minutes, it shows the incremented system time in standard format :
Thu Feb 2 20:06:30 IST 2017.
I want the desired output in --rfc-3339 format i.e
2017-02-02 20:06:30+05:30.
Thanks in advance.
This is pretty simple. The --rfc-3339 flag tells the date command how to format its output, but you're putting it inside the specifier for the input date — what date to format.
In your command
date --date="(date --rfc-3339=seconds) + 5 minutes"
the --date="stuff" part tells the command which date you want to show (as opposed to the default of "right now"). It looks like you might be doing some sort of math with "take the current date and add five minutes", but actually the part in () is not valid input and is ignored. You could say date --date="(colorless green ideas sleep furiously) + 5 minutes", and you'll get the same results — or just date --date="+5 minutes".
But of course, that's in the standard output format. To get it in RFC 3339 format, simply add that flag, outside of the input date string:
date --date="+5 minutes" --rfc-3339=seconds
and there you go.
Related
I'm doing some data transfers and using the 'time' command to measure how long it takes for the transfer to complete. The command is time bbcp source destination Unfortunately it doesn't show the start and end date, only the number of seconds the command took to run. Is there any way to do this without complicated scripts?
Please don't kill me for such a simple answer, but if I want to do something like this, I use this command for knowing the current time (accuracy in milliseconds date +"%T.%3N"), as follows:
Prompt> date +"%T.%3N"
Prompt> Launch_Whatever
Prompt> date +"%T.%3N"
This shows me two dates, which I can subtract from each other in order to know the duration. If your Launch_Whatever generates a lot of output, you might always capture the timestamps in a temporary file:
Prompt> date +"%T.%3N" >/tmp/datetime.txt
Prompt> Launch_Whatever
Prompt> date +"%T.%3N" >>/tmp/datetime.txt
The first one creates the datetime.txt file, the second one adds the next entry.
Simplest would be to call date, before and after, like:
date; bbcp source destination; date
with output like:
Tue Jun 21 14:48:32 CEST 2022
... the tool output ...
Tue Jun 21 14:48:34 CEST 2022
if you want a date format that is parsable you can do
date +%s; bbcp source destination; date +%s
with output like (the numbers are seconds since epoch):
1655815745
... the tool output ...
1655815747
Long Date
Fri Aug 07, 2020 05:12 UTC
Day | Date | Time |
Friday | 07:10:2020 | 05:12 |
to get the date from that string use:
=--MID(A1,5,LEN(A1)-8)
Put that in three cells. Then format the cells:
day cell: dddd
Date Cell: dd\:mm\:yyyy
Time Cell: HH:MM
You could use the following to put the text string into just a date time format that excel understands:
=DATEVALUE(RIGHT(LEFT(A1,LEN(A1)-4),LEN(A1)-8))
The above will not be affected by number of digits for year, day or hour; only that the bit it is using has 4 chars at the front (Fri and space) and 4 at the end (space and UTC).
Since you have two digits for both day and hour, this could be simplified with using MID, like:
=DATEVALUE(MID(A1,5,LEN(A1)-8))
To extract a date from a text string representing a date, you can use the DATE function e.g. =DATE([year],[month],[day]).
To get those parts from a text string you can use =MID([input text string], [start position], [number of characters]). So you could do MID for each component (year, month etc) needed, then put the results through the DATE function.
Date formats in Excel are distinguished between what is stored and what is displayed.
What is stored depends on your excel settings.
What is displayed depends on your excel settings as well as your "regional and language options" or equivalent settings in your computer's operating system.
To change what is displayed, set a custom format (ctrl-1, first tab, last list item) to something like Ddd Mmm dd, yyyy hh:mm.
Otherwise regional and language options, and program settings, can override how the date is shown (which can be important if you are sharing workbooks, especially with unknown future users).
Is the input UTC or do you need to convert?
If you need to convert, do you have a fixed offset in hours? If so, it is a simple formula. For example: =A1+3/24 for UTC-3 hours.
To change what is stored, first understand whether the number stored represents the number of days since 1 Jan 1901, or 1 Jan 1904, or some other convention (such as 1 Jan 1970).
Then decide if you want to store it as a date or a text string.
Then decide if you want to store it as 3 copies of the same value (each displaying a different aspect: day, date, time) or if you just want each value to be its own part not the whole date "hidden" and the display set.
To store it as a text string, use =TEXT(A1,"Dddd"), =TEXT(A1,"MM:DD:YYYY") and =TEXT(A1,"HH:MM").
Watch out for 24 or 12 hour time: the difference is whether "AM/PM" is appended. Your input is likely 24h time but check another example from the dataset to be sure.
Trying to convert 1504865618099.00 Unix time into a readable date time.
I tried this:
=(UNIX + ("1/1/1970"-"1/1/1900"+1)*86400) / 86400
But it's not working.
To convert the epoch(Unix-Time) to regular time like for the below timestamp
Ex: 1517577336206
First convert the value with the following function like below
=LEFT(A1,10) & "." & RIGHT(A1,3)
The output will be like below
Ex: 1517577336.206
Now Add the formula like below
=(((B1/60)/60)/24)+DATE(1970,1,1)
Now format the cell like below or required format(Custom format)
m/d/yyyy h:mm:ss.000
Now example time comes like
2/2/2018 13:15:36.206
The three zeros are for milliseconds
=A1/(24*60*60) + DATE(1970;1;1) should work with seconds.
=(A1/86400/1000)+25569 if your time is in milliseconds, so dividing by 1000 gives use the correct date
Don't forget to set the type to Date on your output cell. I tried it with this date: 1504865618099 which is equal to 8-09-17 10:13.
TLDR
=(A1/86400)+25569
...and the format of the cell should be date.
If it doesn't work for you
If you get a number you forgot to format the output cell as a date.
If you get ##### you probably don't have a real Unix time. Check your
timestamps in https://www.epochconverter.com/. Try to divide your input by 10, 100, 1000 or 10000**
You work with timestamps outside Excel's (very extended) limits.
You didn't replace A1 with the cell containing the timestamp ;-p
Explanation
Unix system represent a point in time as a number. Specifically the number of seconds* since a zero-time called the Unix epoch which is 1/1/1970 00:00 UTC/GMT. This number of seconds is called "Unix timestamp" or "Unix time" or "POSIX time" or just "timestamp" and sometimes (confusingly) "Unix epoch".
In the case of Excel they chose a different zero-time and step (because who wouldn't like variety in technical details?). So Excel counts days since 24 hours before 1/1/1900 UTC/GMT. So 25569 corresponds to 1/1/1970 00:00 UTC/GMT and 25570 to 2/1/1970 00:00.
Now if you also note that we have 86400 seconds per day (24 hours x60 minutes x60 seconds) and you will understand what this formula does: A1/86400 converts seconds to days and +25569 adjusts for the offset between what is zero-time for Unix and what is zero-time for Excel.
By the way DATE(1970,1,1) will helpfully return 25569 for you in case you forget all this so a more "self-documenting" way to write our formula is:
=A1/(24*60*60) + DATE(1970,1,1)
P.S.: All these were already present in other answers and comments just not laid out as I like them and I don't feel it's OK to edit the hell out of another answer.
*: that's almost correct because you should not count leap seconds
**: E.g. in the case of this question the number was milliseconds since the the Unix epoch.
If you have ########, it can help you:
=((A1/1000+1*3600)/86400+25569)
+1*3600 is GTM+1
in case the above does not work for you. for me this did not for some reasons;
the UNIX numbers i am working on are from the Mozilla place.sqlite dates.
to make it work : i splitted the UNIX cells into two cells : one of the first 10 numbers (the date) and the other 4 numbers left (the seconds i believe)
Then i used this formula, =(A1/86400)+25569 where A1 contains the cell with the first 10 number; and it worked
You are seeing the date as ######## most likely because by definition the EPOCH times is in seconds - https://en.wikipedia.org/wiki/Unix_time. This means the number should be 10 characters long. Your number has 13 characters (see 1504865618099) and it is most likely in milliseconds (MS). In order to fix the formula just divide the number by 1000. Just keep in mind this way you'll loose the MS precision, but in most cases this is OK. So the final formula should be:
=A1/(86400 * 1000) + DATE(1970,1,1)
Just point and shoot.
Replace the C2 with your cell no. No need to format your Excel cell.
Also, you can use this unixtimestamp website to verify your data.
International format (ISO 8601):
=TEXT(C2/(1000*60*60*24)+25569,"YYYY-MM-DD HH:MM:SS")
2022-10-20 00:04:22
2022-10-20 00:05:20
2022-10-20 00:14:58
US format:
=TEXT(C2/(1000*60*60*24)+25569,"MM/DD/YYYY HH:MM:SS")
10/20/2022 00:04:22
10/20/2022 00:05:20
10/20/2022 00:14:58
Europe format:
=TEXT(C2/(1000*60*60*24)+25569,"DD.MM.YYYY HH:MM:SS")
20.10.2022 00:04:22
20.10.2022 00:05:20
20.10.2022 00:14:58
If you only need the date, remove the 'HH:MM:SS'.
=TEXT(C2/(1000*60*60*24)+25569,"YYYY-MM-DD")
This question already has answers here:
Convert decimal day to HH:MM
(3 answers)
Closed 3 years ago.
I have a dataset that encodes a date-time into two separate variables. Normally, I'd just paste them together inside of an as.POSIXct and carry on. However, the date is provided as a string, and the time of day as a fraction of 24 hours - e.g., 12pm is 0.5, 9:30am is 0.1458333, etc.
It doesn't seem all that tricky to convert the fractional days into clock hours, but I'd prefer to use a pre-existing function if possible. Does something like that exist in base R? A package?
If it's any use, this is an Excel (xlsx) time field imported into R through RODBC.
EDIT
Oddly enough, upon revisiting this problem, the times are now read in as POSIXct. Not sure what to make of that.
The R News 4/1 Help Desk article has a section on reading Excel dates in R.
POSIXct values are simply the number of seconds since midnight GMT 1970-01-01. (So you need to pay attention to your offset from UTC.) You can use the date part and add the number of days times 24*3600 (as.Date(dtval) to your time value * 24*3600. Gabor pointed to the article in R News (which he wrote, thank you, Gabor.)
You didn't give an example of the string. If you are getting your date as a string, then as.Date(strDate) will convert a variable "strDate" to Date class when it is in either "YYYY-MM-DD" or "YYYY/MM/DD" format. Otherwise the formatting codes are on the ?strptime page.
Once you have a POSIXct-classed variable you can just add the number of seconds. This example add 30 minutes to midnight today Feb 1, 2011 (in my time zone which is UTC-5):
> as.POSIXct(as.Date("2011-02-01")) +30*60
[1] "2011-01-31 19:30:00 EST"
And this is your time value added to midnight my time:
> as.POSIXct(as.Date("2011-02-01 00:00", tzone="UTC"))+3600*5 + 3600*24*timeval
[1] "2011-02-01 03:29:59 EST"
I have a time field of this format HH:MM:SS in many excel cells (when i import in as a csv).
i would like to Attached the date field from another worksheet in and when I tried to concatenate =U10&" "&V10, let's said
U10 is 22-Dec
V10 is 18:14:01
will become
40169 0.759733796296296
so how can i convert a number like 0.759733796296296 back into 18:14:01 ?
Is your date field recognized as date?
I tried the following (actually adding date and time, not concatenating) and it worked:
22.12.2010 | 18:14:01
=A1+B1 => 22.12.2010 18:14:01
If you want the result as text only (not date), then you can format your time-field as Text (not Time), and concatenating should work the way you describe it.
Try changing the format of the cell back to DATE/TIME.
Note that 18:14:01 (18 hours, 14 minutes and 1 second) is about 75.97% of a 24 hour day.