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Extracting only date portion from LastLogonDate
(3 answers)
Closed 3 years ago.
I'm running a powershell script and exporting the output for a rapport.
Tried different format possibilities to change dd-MM-YYYY HH-MM-SS to just showing dd-MM-YYYY but can't figure it out
#{Name="Date"; Expression={[DateTime]::FromFileTime($_.LastLogonTimestamp)}}
I'm getting 30-10-2017 10:07:10 and only want to get the output 30-10-2017
Just convert them to the required format using the below statement.
#{Name="Date"; Expression={[DateTime]::FromFileTime($_.LastLogonTimestamp).ToString('dd-MM-yyyy')}}
Check this documentation for more reference!
https://learn.microsoft.com/en-us/dotnet/standard/base-types/custom-date-and-time-format-strings
https://learn.microsoft.com/en-us/dotnet/standard/base-types/standard-date-and-time-format-strings
If you already have the value as a DateTime, e.g.
$date = [DateTime]::FromFileTime($_.LastLogonTimestamp)
You can use one of the ToString overloads provided by the object type, e.g.
$formattedDate = $date.ToString("dd-MM-yyyy")
You can get further information about formatting options on this link.
Related
I've looked for help on the internet for the following, but I could not find a satisfying answer: for an assignment, I need to plot the time series of a certain variable (the term spread in percentages), with years on the x-axis.
However, we use daily data. Does anybody know a convenient way in which this can be done? The 'date' variable that I've got is formulated in the following way: 20111017 represents the 17th of October 2011.
I tried to extract the first 4 numbers of the variable 'date', by using the substr(date, 1, 4) command, but the message 'type mismatch' popped up. Also, I'm not quite sure if it gives the right information if I only use the years to plot daily data (over the years). It now gives the following graph, which doesn't look that nice.
Answering the question in your title.
The date() function expects a string. If your variable with value 20111017 is in a numeric format you can convert it like this: tostring datenum , gen(datestr).
Then when using the date() function you must provide a mask that tells Stata what format the date string is in. Below is a reproducible example you can run to see how this works.
* Example generated by -dataex-. For more info, type help dataex
clear
input float datenum
20111016
end
* Convert numberic varaible to string
tostring datenum , gen(datestr)
* Convert string to date
gen date = date(datestr, "YMD")
* Display date as date
format date %td
If this does not help you, try to provide a reproducible example.
This adds some details to the helpful answer by #TheIceBear.
As he indicates, one way to get a Stata daily date from your run-together date variable is convert it to a string first. But tostring is just one way to do that and not essential. (I have nothing against tostring, as its original author, but it is better suited to other tasks.)
Here I use daily() not date(): the results are identical, but it's a good idea to use daily(): date() is all too often misunderstood as a generic date function, whereas all it does is produce daily dates (or missings).
To get a numeric year variable, just divide by 10000 and round down. You could convert to a string, extract the first 4 characters, and then convert to numeric, but that's more operations.
clear
set obs 1
gen long date = 20111017
format date %8.0f
gen ddate = daily(strofreal(date, "%8.0f"), "YMD")
format %td ddate
gen year = floor(date/10000)
list
+-----------------------------+
| date ddate year |
|-----------------------------|
1. | 20111017 17oct2011 2011 |
+-----------------------------+
* date is in %td format
gen date1 = real(string(mofd(daily(date, "DMY")), "%tmCYN"))
* type mismatch error
tostring date, gen(dt)
gen date1 = real(string(mofd(daily(dt, "DMY")), "%tmCYN"))
* the code runs but generates no results
tostring date, gen(dt)
gen date2=date(dt, "YMD")
* the code runs but generates no results
If a date variable has a display format %td it must be numeric and stored as some kind of integer. The display format is, and is only, an instruction to Stata on how to display such integers. Confusions about conversion often seem to hinge on a misunderstanding about what format means, as format is an overloaded word in computing, referring variously to file format (as in graphics file format, .png or jpg or whatever); data layout (as in wide or long layout, structure or format); variable or storage type; and (here) display format. There could well be yet other meanings.
A date displayed as 30jan2015 is stored as an integer, namely
. display mdy(1, 30, 2015)
20118
and a glance at help data types shows that your variable date could be stored as an int, float, long or double. All would work, although int is least demanding of memory. You would need (e.g.) to run describe date to find out which type is being used in your case, but nothing to come in this answer depends on knowing that type. Note that finding out what Stata is doing and thinking can be illuminated by running display with simple, single examples.
Your question is ambiguous.
Want to change display format? If you wish merely to see your dates in a display format exemplified by 20150130 then consulting help datetime display formats shows that the display format is as tested here with display, which can be abbreviated all the way down to di
. di %tdCCYYNNDD 20118
20150130
so
format date %tdCCYYNNDD
is what you need. That instructs Stata to change the display format, but the numbers stored remain precisely as they were.
Want such dates as variables held as integers? If you want the dates to be held as integers like 20150130 then you could convert it to string using the display format above, and then to a real value. A minimal sandbox dataset shows this:
. clear
. set obs 1
Number of observations (_N) was 0, now 1.
. gen date = 20118
. gen wanted = real(strofreal(date, "%tdCCYYNNDD"))
. format wanted %8.0f
. l
+------------------+
| date wanted |
|------------------|
1. | 20118 20150130 |
+------------------+
A display format such as %8.0f is needed to see such values directly.
Another method is to generate a large integer directly. You need to be explicit about a suitable storage type and (as just mentioned) need to set an appropriate format, but it can be got to work:
. gen long also = 10000 * year(date) + 100 * month(date) + day(date)
. format also %8.0f
Want such dates as variables held as strings? This is the previous solution, but leave off the real(). The default display format will work fine.
. gen WANTED = strofreal(date, "%tdCCYYNNDD")
. l
+-----------------------------+
| date wanted WANTED |
|-----------------------------|
1. | 20118 20150130 20150130 |
+-----------------------------+
I have not used tostring here but as its original author I have no bias against it. The principles needed here are better illustrated using the underlying function strofreal(). The older name string() will still work.
Turning to your code,
tostring date, gen(dt)
will just put integers like 20118 in string form, so "20118", but there is no way that Stata can understand that alone to be a daily date. You could have run tostring with a format argument, which would have been equivalent to the code above. The advantage of tostring would only be if you had several such variables you wished to convert at once, as tostring would loop over such variables for you.
I can't follow why you thought that conversion to a monthly date or use of a monthly date display format was needed or helpful, as at best you'd lose the information on day of the month. Thus at best Stata can only map a monthly date back to the first day of that month, and at worst a monthly date (here 660) could not be understood as anything you want.
. di mofd(20118)
660
. di %td mofd(20118)
22oct1961
. di %td dofm(mofd(20118))
01jan2015
There is no shortcut to understanding how Stata thinks about dates that doesn't involve reading the needed parts of help datetime and help datetime display formats.
Yet more explanation and examples can be found at https://www.stata-journal.com/article.html?article=dm0067
I'm trying to get DAYID as string in format YYYYMMDD, however its not working correctly.
I have a timestamp field, I take first 10 characters and convert it into date (working correctly)
toDate(substring(EventTimestamp,1,10))
-- 2021-03-24
However when I try to convert to string using below expression, I;m getting wrong answer. I;m getting the Day as 83.
toString(toDate(substring(EventTimestamp,1,10)),'YYYYMMDD')
--20210383
Date format characters are case-sensitive as per this answer here, so use yyyyMMdd instead.
An example using formatDateTime:
#formatDateTime(utcnow(), 'yyyyMMdd')
I need to find the previous hour time with formatting with PS:
I need the below format:
"yyyy-MM-dd-HH:mm:sstt")
I can use the below code to get the date and time in this format:
(Get-Date -Format "yyyy-MM-dd-HH:mm:sstt")
2019-09-17-08:45:27AM
I need to get the previous hour time but in the above format
I know how to get the last hour time :
(Get-Date).AddHours(-1)
How can i get the previous hour time with a combination of the above
Format?
Using -f, the format operator, as shown in Ivan Mirchev's helpful answer is definitely an option, and -f is a great general-purpose formatting option to know about, for any data type.
However, in your particular case there is a simpler solution, because the .ToString() method of [datetime] instances directly accepts a format string:
(Get-Date).AddHours(-1).ToString('yyyy-MM-dd-HH:mm:sstt')
You may try using the format operator:
"{0:yyyy-MM-dd-HH:mm:sstt}" -f (get-date).AddHours(-1)
more details: https://learn.microsoft.com/en-us/powershell/module/microsoft.powershell.core/about/about_operators?view=powershell-6#format-operator--f
Hope it helps! :)
At the moment I am using xlsread to open a set of data that I have in excel with given timestamps. But when these values are placed in matlab it changes the formatting of the timestamp.
In excel it is:
dd/mm/yyyy HH:MM
but when it puts it into matlab it changes it to
mm/dd/yyyy HH:MM
which ruins my other code. I have tried using formatIn and specifying it, but then it returns an error if no value for midnight is given.
Any help would be appreciated.
You can use datenum and datestr to convert the format to what you want. In the following example I'm assuming your timestamps are contained in a cell array of strings, but it also works if it's a char matrix:
>> timestamps = {'08/25/2014 13:14'; '08/26/2014 14:15'} %// mm/dd/yyyy HH:MM
>> result = datestr(datenum(timestamps, 'mm/dd/yyyy HH:MM'), 'dd/mm/yyyy HH:MM')
result =
25/08/2014 13:14
26/08/2014 14:15
What Luis recommended should help you to get any format that you like. However there is something important to realize here:
Excel does not 'have' the date in your format. It has the date stored as a number like 123546.123 and presents it to you in a certain way.
If you want to get the date in exactly the way that excel presents it, the trick is to avoid importing the relevant column as a date, but just import it as text instead.
How to do this depends on your import method, but it should not be very hard.