How I will change this date format to in a day or hours
"modifiedTime": "2021-01-30T12:15:14.896Z[GMT]",
"modifiedTime": "2021-01-29T13:58:37.328Z[Etc/UCT]",
DateTime.parse(aString) should handle a Z suffix for timezone or an explicit offset (-05:00). To parse other named timezones, you'll need the https://pub.dev/packages/timezone package, which unfortunately has to get updated at least a few times a year because politicians keep messing with things like DST.
Related
I have an todo application and some tasks have deadlines. If deadline is bigger or equal than current date, I want to display them. However, there is a local datetime problem i believe.
This is the current date using 'print(Date())':
2021-07-27 18:53:03 +0000
This is the beginning of that date using 'print(Calendar.current.startOfDay(for: Date()))':
2021-07-26 21:00:00 +0000
As you see, 27th day becomes 26th. I tried timezone operations also. It seems true but day is still starting at 21.00
When I convert them as string and print them, it is ok it show the expected results but without string type it becomes like above. Because of I want to get the bigger or equal days compared to current date , string solution is not working. It is the 27th of day but because of date object it is shown as 26th day and my app showing yesterdays tasks etc.
Can you provide a help for me?
Thanks
There's the subtract() method but the documentation says it's not aware of daylight savings which makes it pretty much useless in this case, or in any other case except where the programmer doesn't know how many milliseconds there are in 24 hours.
I'm thinking of two ways:
get the day of the month, and then subtract N from it and if it's less than 1 then subtract the month and the year if appropriate and set the day for the last day of whichever month it turns out to be
OR
subtract N days from the noon of the current day and then get start of the day for the resulting day
Is there some easier/better way to do this?
You should probably try to convert both DateTimes to UTC (standardize), then call difference(). That converts it to a nice, easy Duration, which you can convert as necessary to hours, days, months, or whatever else.
DateTime one = somedatetime.toUtc();
DateTime two = someotherdatetime.toUtc();
Duration diff = one.difference(two);
//Then just convert...
return diff.inDays;
I am using the below query with date filtering, but I am getting wrong result.
SELECT * FROM TRANSACTIONSHISTORY
WHERE DATE > "29-01-2015 12:00:00"
AND DATE < "30-01-2015 00:00:00" AND USERID=abc
I am getting result with date column with value of 29-Jan-2016 records, what am I missing here, can any one help me to get out of this value.
The date format in your SQL will not work because SQLite doesn't have a native datetime type, so it's generally stored either as a string, in YYYY-MM-DD HH:MM:SS.SSS format, or as an numeric value representing the number of seconds since 1970-01-01 00:00:00 UTC. See date and time types on SQLite.org. Note that if you're using the string representation that the sequence is year, month, day (which, when sorting/querying this string field, the this alphanumeric string will sort correctly by year first, then month, and then day, which is critical when doing queries like yours).
If you really stored dates in the database as a string in the DD-MM-YYYY HH:MM:SS format, you should consider changing the format in which you saved the values into one of the approved date formats. It will make the date interactions with the database much, much easier, allowing queries like the one you asked for (though, obviously, with DD-MM-YYYY replaced with YYYY-MM-DD format).
You have cast your string to Date
SELECT * FROM TRANSACTIONSHISTORY WHERE DATE between Datetime('29-01-2015 12:00:00') and Datetime('30-01-2015 00:00:00') AND USERID=abc
The first answer is exactly what you need. What you did in your code would be comparing strings using ASCII values.
I would recommend you to use the linux time stamps like: 1453818208, which is easier to save and compare. In addition, it can always be translated to human-readable dates like: 29-01-2015 12:00:00.
SELECT * FROM TRANSACTIONSHISTORY
WHERE DATE > "29-01-2015 12:00:00"
AND DATE < "30-01-2015 00:00:00" AND USERID=abc
I hope this helps you :)
Try this first try without Time,after that try date and time both , Hope i will work for you
SELECT TRANSACTIONSHISTORY
FROM SHIPMENT
WHERE DATE
BETWEEN '11-15-2010'
AND '30-01-2015'
// you can try this one also
SELECT * FROM TRANSACTIONSHISTORY WHERE DATE BETWEEN "2011-01-11" AND "2011-8-11"
I want to calculate time in Lua.
For example: I want to get the current date (incl. at least seconds) and then I want to add seconds to this date that I get a date which is in the future.
Try this:
now=os.time()
print(os.date("%c",now))
print(os.date("%c",now+10000))
I want to store date in my Postgres database.
The only problem is that this date can have optional day or even month.
Example:
User provides time period when he was employed - not necessary full date (day + month + year), but only start year and end year.
However there are users, who worked only from may to october in the same year so month have to be provided too.
How to handle this kind of optional date parts?
Use a proper date type anyway. Do not store text or multiple columns. That would be more expensive and less reliable.
Use the function to_date(), which is fit to deal with your requirements out of the box. For instance, if you call it with a pattern 'YYYYMMDD' and the actual string is missing characters for day, or month and day, it defaults to the first month / day of the year / month:
db=# SELECT to_date('2001', 'YYYYMMDD');
to_date
------------
2001-01-01
db=# SELECT to_date('200103', 'YYYYMMDD');
to_date
------------
2001-03-01
You could store a precision flag indicating year / month / day in addition if you need that.
While the accepted answer is a good one, there is another alternative.
ISO 8601
The ISO 8601 standard defines sensible formats for textual representations of various kinds of date-time values.
A year is represented in the obvious manner, a four-digit number: 2014
A year-month is represented with a required hyphen: 2014-01Note that in other ISO 8601 formats, the hyphen is optional. But not for year month, to avoid ambiguity.
A full date is similar: 2014-08-21 or without optional hyphens: 20140821. I recommend keeping the hyphens.
So you could store the values as text. The length of text would tell you whether it is year-only, year-month, or date.