Iam reading out a Json file with a date in it.
jsonValue->GetObject()->GetNamedObject("board")->GetNamedNumber("date")
That date is saved in Unix code format:
"date":1347973494
But I need to get it in a normal format like "19.09.2012".
I cant find the right function to solve that problem.
I already tried the DateTimeFormatter class but I think that was not the correct way to make this.
So anyone knows how to change the DateTime from Unix timestamp to a normal format like "19.09.2012"?
A Unix timestamp is seconds since 1970, so add the seconds to 1970-01-01.
int unixTimestamp = 1347973494;
System::DateTime timestamp = System::DateTime(1970, 1, 1).AddSeconds(unixTimestamp);
Then format the DateTime into whatever string format you like, or use it as a DateTime.
System::String^ formatted = timestamp.ToString("dd.MM.yyyy")
I solved the problem with the Calendar class which you can find here
int unixTimestamp = (int)jsonValue->GetObject()->GetNamedNumber("date");
Windows::Globalization::Calendar^ cal = ref new Windows::Globalization::Calendar();
cal->Year = 1970;
cal->Month = 1;
cal->Day = 1;
cal->Minute = 0;
cal->Hour = 0;
cal->Second = 0;
cal->AddSeconds(unixTimestamp);
mainDate -> Text = cal->DayAsString() + ". " + cal->MonthAsString() + " " + cal->YearAsString();
Related
I am attempting to convert the following 2HR time in String to a DateTime, so that I manage it appropriately.
"1800"
This question came close, but no one answered how to convert the String into a valid DateTime.
How to convert time stamp string from 24 hr format to 12 hr format in dart?
Attempt 1:
DateTime.parse("1800") -
Invalid Date Format
Attempt 2:
DateTime.ParseExact("1800") -
This doesn't seem to exist, although it shows up on various
Still no luck and need a second pair of eyes to point out the obvious to me.
The time by itself is not a datetime so you could do something like:
DateTime myTime(DateTime baseDate, String hhmm) {
assert(hhmm.length == 4, 'invalid time');
final _hours = int.parse(hhmm.substring(0, 2));
final _mins = int.parse(hhmm.substring(2, 2));
return DateTime(baseDate.year, baseDate.month, baseDate.day, _hours, _mins);
}
DateTime.parse expects to parse dates with times, not just times. (You can't even create a DateTime object without a date!)
Some people generate a dummy date string, but in your case you could trivially parse it with int.parse and then apply appropriate division and remainder operations:
var rawTime = int.parse('1800');
var hour = rawTime ~/ 100;
var minute = rawTime % 100;
Also see How do I convert a date/time string to a DateTime object in Dart? for more general DateTime parsing.
Im trying to convert the first two columns of a cell into a Matlab time. First column {1,1} is the date in YYYY-MM-DD format and the second is the time in HH:MM format.
Any ideas where I'm going wrong? My code:
file = 'D:\Beach Erosion and Recovery\Bournemouth\Bournemouth Tidal
Data\tidal_data_jtide.txt'
fileID = fopen(file);
LT_celldata = textscan(fileID,'%D%D%D%D%d%[^\n\r]','delimiter',',');
formattime = 'yyyy-mm-dd HH:MM'
date = LT_celldata{1,1};
time = LT_celldata{1,2};
date_time = datenum('date','time'); code
Screenshot below is LT_celldata{1,1} :
You can combine variables date and time with the following code:
date = datetime(LT_celldata{1,1},'InputFormat','yyyy-MM-dd');
time = datetime(LT_celldata{1,2},'InputFormat','HH:mm:ss','Format','HH:mm:ss');
myDatetime = datetime(date + timeofday(time),'Format','yyyy-MM-dd HH:mm:ss');
The code uses timeofday function to combine date and time information from the two different variables. You may find more information and examples at this documentation page.
I'm having trouble finding a good way of formatting a UTC-time stamp with this format: yyyyMMdd-HH:mm:ss.<three additional digits>
I wasn't able to find any character that represents milliseconds/hundredths, I'm not even sure this is possible, to parse that format that is.
Ideally I'd like to use the parseToStringDate that's part of the Date library.
My plan b is to convert yyyyMMdd-HH:mm:ss to milliseconds and then add the three last digits to that number.
Use yyyyMMdd-HH:mm:ss.SSS
This will get you milliseconds as well.
Test Code:
def now = new Date()
println now.format("yyyyMMdd-HH:mm:ss.SSS", TimeZone.getTimeZone('UTC'))
I would convert it like that:
def now = new Date()
println now.format("YYYYMMdd-HH:mm:ss")
You can try this:
TimeZone.getTimeZone('UTC')
Date date = new Date()
String newdate = date.format("YYYY-MM-dd HH:mm:ss.Ms")
log.info newdate
I am using ColdFusion 9.0.1 and some database that I cannot change.
I am accessing a database that stores a date as an eight digit numeric with zero decimal places like this:
YYYYMMDD
I need to be able to read the date, add and subtract days from a date, and create new dates. I am looking for a ColdFusion solution to efficiently (not much code) to convert the date to our standard format, which is
MM/DD/YYYY
And then convert it back into the database's format for saving.
I need to code this in such a way that non-ColdFusion programmers can easily read this and use it, copy and modify it for other functions (such as adding a day to a date). So, I am not looking for the most least amount of code, but efficient and readable code.
Can you suggest anything that would make this code block more flexible, readable, or more efficient (less code)?
<cfscript>
// FORMAT DB DATE FOR BROWSER
DateFromDB = "20111116";
DatedToBrowser = createBrowserDate(DateFromDB);
writeOutput(DatedToBrowser);
function createBrowserDate(ThisDate) {
ThisYear = left(ThisDate, 4);
ThisMonth = mid(ThisDate, 4, 2);
ThisDay = right(ThisDate, 2);
NewDate = createDate(ThisYear, ThisMonth, ThisDay);
NewDate = dateFormat(NewDate, "MM/DD/YYYY");
return NewDate;
}
// FORMAT BROWSER DATE FOR DB
DateFromBrowser = "11/16/2011";
DateToDB = createDBDate(DateFromBrowser);
writeDump(DateToDB);
function createDBDate(ThisDate) {
ThisYear = year(ThisDate);
ThisMonth = month(ThisDate);
ThisDay = day(ThisDate);
NewDate = "#ThisYear##ThisMonth##ThisDay#";
return NewDate;
}
</cfscript>
First find who ever did the database and kick them in the nads...
Personally I'd Convert with sql so my code only dealt with date objects.
Select Convert(DateTime, Convert(VarChar(8),DateTimeInventedByIdjitColumn))
From SomeTable
As stated by our peers, store dates as dates.
'08/06/2011' could be 8th of june of the 6th of August depending on locale.
20111643 is a valid integer..
Not using a proper date type is just a massive collection of features and bugs that at best are waiting to happen.
You can actually rewrite each function into 1 line of code.
function createBrowserDate(ThisDate) {
return mid(ThisDate,4,2) & "/" & right(ThisDate,2) & "/" & left(ThisDate,4);
}
and
function createDBDate(ThisDate) {
return dateFormat( ThisDate, "YYYYMMDD" );
}
Don't keep dates as strings - keep dates as dates and format them when you need to.
If you can't correct the database to use actual date columns (which you should if you can), then you can use these two functions to convert to/from YYYYMMDD and a date object:
function parseYMD( YYYYMMDD )
{
if ( ! refind('^\d{8}$' , Arguments.YYYYMMDD ) )
throw "Invalid Format. Expected YYYYMMDD";
return parseDateTime
( Arguments.YYYYMMDD.replaceAll('(?<=^\d{4})|(?=\d{2}$)','-') );
}
function formatYMD( DateObj )
{
return DateFormat( DateObj , 'yyyymmdd' );
}
By using date objects it means that any level of developer can work with them, without needing to care about formatting, via built-in functions like DateAdd, DateCompare, and so on.
I'm not a regular expression fan since it's not that readable to me.
Since you're using CF9, I'd typed the argument and specify the returntype of the functions to be even more readable for the next person picking up your code.
First, right after I read the date from DB, I'd parse it to a Date object using parseDBDate()
Date function parseDBDate(required String dbDate)
{
var yyyy = left(dbDate, 4);
var mm = mid(dbDate, 4, 2);
var dd = right(dbDate, 2);
return createDate(yyyy , mm, dd);
}
Once you have the date object, you can use all those built-in Date functoin like DateAdd() or DateDiff().
Call browserDateFormat() right before you need to display it.
String function browserDateFormat(required Date date)
{
return dateFormat(date, "MM/DD/YYYY");
}
Call dBDateFormat() inside <cfqueryparam value=""> when it's time to persist to DB
String function dBDateFormat(required Date date)
{
return dateFormat(date, "YYYYMMDD");
}
One liner :)
myDateString = "20110203";
myCfDate = createObject("java","java.text.SimpleDateFormat").init("yyyyMMdd").parse(myDateString,createObject("java","java.text.ParsePosition").init(0*0));
If you want to parse different patterns, change "yyyyMMdd" to any other supported pattern.
http://download.oracle.com/javase/1.5.0/docs/api/java/text/SimpleDateFormat.html
The ParsePosition is used to say where to start parsing the string.
0*0 is shorthand for JavaCast("int",0) - in the Adobe cf engine, 0 is a string, until you apply math to it, then it becomes a Double, which the ParsePosition constructor supports. Technically, it constructs with an int, but cf is smart enough to downgrade a Double to an int.
http://download.oracle.com/javase/1.5.0/docs/api/java/text/ParsePosition.html
I have the following code returning the time in a 24 hour format like "13:10:23"
DateTime.DateFormat = "h:mm a"
now = DateTime.Now
t = DateTime.Time(now)
lblCurrentTime.Text = t
But I'd like to have the format in hours:minutes:seconds:milliseconds
Unfortunately, the link to the supported formats from this page is broken..
http://www.basic4ppc.com/android/wiki/index.php/DateTime
Could anyone please suggest how to use DateTime.Dateformat to return a time value to millisecond resolution?
Thank you, but I'm afraid it's still returning time in the format: "23:11:09"
Perhaps my variables are incorrect?
Sub Time_tick
Dim now As Long
Dim t As String
DateTime.DateFormat = "HH:mm:ss:SS"
now = DateTime.Now
t = DateTime.Time(now)
lblCurrentTime.Text = t
End Sub
Here is the correct link.
Try: "HH:mm:ss:SS"
Note that you should use DateTime.TimeFormat and not DateFormat.
Can you try the following ?
DateTime.DateFormat = "HH:mm:ss:SS"