I am using bootstrap datetimepicker. I want the datetimepicker format as short month name eg. Jan. My code is below and it shows the full month name now as January. How to make it Jan?
$('.datepickersren').datetimepicker({
format: 'MMMM',
viewMode: "months",
})
Please note that the format:'M' is not working in datetimepicker. It will work on the datepicker only. If we give format:'M' in datetimepicker, the month format is month number eg. 5, 6, 7 etc. Not the month names like Jan, Feb, etc.
$('.datepickersren').datetimepicker({
format: 'MMM',
viewMode: "months",
})
The above code is working for me and the format: 'MMM' displays the three letter month name as Jan, Feb etc. Thank you all for the response.
A single M is "Month name short"
$('.datepickersren').datetimepicker({
format: 'dd-M-yy',
viewMode: "months",
})
Related
I have i have 2 input fields that have text and the other dropdown (select option) in html. The user types the text and select the month from the dropdown (select option) then its submitted to Database. The dropdown have four months March, June, September and December. I want to create a query from database that selects between months like:
SELECT * FROM `returndocs` WHERE Year='2020' and Month BETWEEN 'March' AND 'June'
My database is:
I want to select all the data in the database posted starting from March to June. How do i do that?
in your drop down put the date instead of month name as value
like
03 => March
06 => June
then make it 2020-03-01 and 2020-06-01
then run your query with
SELECT * FROMreturndocsWHERE Year='2020' and Month BETWEEN '2011-03-01' AND '2011-06-01'
Since you defined month in text type you will need to type all between month manually :
SELECT *
FROM `returndocs`
WHERE Year = 2020 AND
Month IN ('March', 'April', 'May','June');
I'm trying to create three columns based on date in seconds format.
My user.updated_at = 1521533490
I would like to get year, month and day separately and put these formatted values to columns for example:
year -> 2018, month -> 11, day -> 23
Does someone know how can I do that in pgSQL?
I would like to get year, month and day separately and put these formated values to columns
Don't do that.
Use a single column of type date or timestamp, depending on your application. Not every combination of your three columns will be a valid date. But every value in a single column of type date will be a valid date.
If you need the parts of a date separately, use PostgreSQL's date/time functions.
Try this approche to get differents arguments, then you can do whatever you want:
SELECT to_timestamp(1521533490); //2018-03-20T08:11:30.000Z
SELECT to_char(to_timestamp(1521533490), 'HH'); // 08 Hour
SELECT to_char(to_timestamp(1521533490), 'MI'); // 11 Minutes
SELECT to_char(to_timestamp(1521533490), 'SS'); // 30 Seconds
SELECT to_char(to_timestamp(1521533490), 'DD'); // 20 Day
SELECT to_char(to_timestamp(1521533490), 'Mon'); // MAR Month
SELECT to_char(to_timestamp(1521533490), 'YYYY'); // 2018 Year
Use the EXTRACT function.
SELECT to_timestamp(updated_at) "Date",
EXTRACT(YEAR FROM (to_timestamp(updated_at))) "Year",
EXTRACT(MONTH FROM (to_timestamp(updated_at))) "Month",
EXTRACT(DAY FROM (to_timestamp(updated_at))) "Day"
FROM users
Output
Date Year Month Day
2018-03-20T08:11:30Z 2018 3 20
SQL Fiddle: http://sqlfiddle.com/#!15/afe0e/15/0
More information on the EXTRACT function.
I need to print the last sunday of the year from which adding 7 to it will give me all the sundays of next year.
I have the code to print all sundays for a particular year if i have a start date but i need the user to put the year so that last sunday of the previous year will be generated and 7 will be added to get first sunday of that year and so on till it reaches last sunday of next year
For example input year is 2017 it will check the last sunday of 2016 and add 7 to it to get first sunday of 2017 which is 1-1-2017 and it will go on printing all sundays till it reaches 31st december 2017
The function next_day() takes two arguments: a date and the name of a day of the week. It returns the closest "next" day (following the date argument) that matches the given day of the week. So the result is between one and seven days forward. (If you want 'Tuesday' and the input date is a Tuesday, the function returns the date seven days later.)
If you want the last Sunday of a year, it will be between Dec. 25 and Dec. 31. So if you call the next_day() function with the arguments Dec. 24 (!!) and 'Sunday' you'll get what you want.
The result will have the same time-of-day as the date argument, so if you give a date without a time-of-day, so will be the output (which is probably what you want). So:
select next_day(date '2016-12-24', 'Sunday') from dual;
NEXT_DAY(D
----------
2016-12-25
Added: If you take an input from your user, as a bind variable, you can do something like this:
select next_day(to_date(:input_year - 1 || '-12-24', 'yyyy-mm-dd'), 'Sunday') from dual;
If you provide 2017 as input (whatever mechanism your interface has for bind variables), the output will be 2016-12-25 (in DATE data type, so don't ask "in what format" - dates don't have a format!)
I have to get/create date (date of first week) from the user input of day name
Examples-
if input saturday then date should be 7 (for current month 1st saturday)
if input sunday then date 1 (current month 1st sunday)
I am having to use lot of logic to get the date but couldn't get exact output
any suggestions on how to come up with the SQL query for such a function ?
If you are using SQL*plus in Oracle then the code will be as :
select next_day(sysdate,'&d')-7 from dual;
If any update required please do inform.
okay i have a link hereby where you can understand this and try to do
Your answer!!!
Try this.
SELECT DATEPART(dw,DATEADD(m, DATEDIFF(m, 0, GETDATE()), 0))
input: if getdate() returns date in jan 2017
output: 1
input:if getdate() returns date in jan 2017
output: 4
I have an issue with my vbnet extented datetime picker.
When the element pass to new year (2016), the week number displayed on the left is wrong.
I have a "datetimepicker" which is not the default component, it was downloaded here :
http://www.codeproject.com/Articles/17063/ExtendedDateTimePicker-control-with-week-numbers
I don't understand why the calendar pass from 53 to 2 and not 53 to 1.
Maybe one of you has the same error.
Thanks for your time.
I don't understand why the calendar pass from 53 to 2 and not 53 to 1
It is pretty much working as expected. The way it is counting weeks, those first 3 days of 2016 count as the first week of 2016.
Note that the control doesnt do anything calendar or display related. It is simply changing the display style of the calendar window provided by Windows. The code seen on the CP page is all there is and mainly it just sets a style flag to tell Windows to add the week numbers:
style = style | MCS_WEEKNUMBERS;
The MSDN entry for it indicates:
Week 1 is defined as the first week that contains at least four days.
Since Jan 1-3 is not 4 days, it would seem that there is either an error, a different calendar being used or MSDN is out of date.
From comments:
From what i understood, what's wrong is "date format". Maybe it's not a 8601
No, it is more than that: ISO8601 is a different calendar which neither Windows nor NET implements. Wikipedia notes:
The first week of a year is the week that contains the first Thursday of the year (and, hence, always contains 4 January). ISO week year numbering therefore slightly deviates from the Gregorian for some days close to 1 January.
This is what you see in the calendar drop down.
Alternative
But the ISO8601 Week Of Year is easy to calculate:
Start with the code for GetISOWeekOfYear() from my answer to a very similar question. You can use that to display the ISO8601 week of year for the selected date in a label or something next to the DTP.
Print the first and last week numbers for 2011 To 2021:
Dim cal As Calendar = CultureInfo.CurrentCulture.DateTimeFormat.Calendar
For n As Int32 = 2011 To 2017 '2021
dt = New DateTime(n, 12, 21)
Console.WriteLine(" ***** {0} *****", n)
For j = 0 To 3
Dim NetWk = cal.GetWeekOfYear(dt, CalendarWeekRule.FirstDay, firstD)
Console.WriteLine("Invariant Date: {0} ISO #:{1:00} NET #:{2:00}",
dt.ToString("MM/dd/yyyy"), GetISOWeekOfYear(dt), NetWk)
dt = dt.AddDays(7)
Next
Next
The result for 2015/2016 portion:
***** 2015 *****
Invariant Date: 12/21/2015 ISO #:52 NET #:52
Invariant Date: 12/28/2015 ISO #:53 NET #:53
Invariant Date: 01/04/2016 ISO #:01 NET #:02
Invariant Date: 01/11/2016 ISO #:02 NET #:03
***** 2016 *****
Invariant Date: 12/21/2016 ISO #:51 NET #:52
Invariant Date: 12/28/2016 ISO #:52 NET #:53
Invariant Date: 01/04/2017 ISO #:01 NET #:01
Invariant Date: 01/11/2017 ISO #:02 NET #:02
Unless you are willing to write your own control from scratch or license one which can be set to a different calendar (and has a definition for ISO8601), that may be the best you can do.
The Bottomline: The Week number is not wrong. It using a different calendar than you expect/want.
References:
Get or convert NET GetWeekOfYear() to ISO week
MSDN: Month Calendar Control Styles
DateTimePicker in Reference Source
The control is working fine.
When the year changes over the final few days of the year are in week 53 - but it's not a full week. Similarly the first few days of the year are in week 1, but the control takes the system's "first day of the week" setting to determine when week 2 begins - so it is possible for the first week of the year to have any where from 1 to 7 days in it.
This means that the image you've shown is showing Week 53 because you're in December and Week 2 because the 2nd week of January does start on the 4th.
If you navigate to January it would display week 1 for the row starting on December 28.
The bottom-line is that the first week in January only has 3 days in it.
This is just the normal and correct behaviour of this control.