How to get number of week for specific date in cakephp 3?
$this->mydate->i18nFormat('yyyy-MM-dd') will display year-month-date.
But what about string format for week number?
Thanks.
Have a closer look at the docs for Cake\I18n\Time::i18nFormat(), it tells you what you can pass as the format, and where to find a list of the accepted formatting patterns:
[...] You can either pass IntlDateFormatter constants as the first
argument of this function, or pass a full ICU date formatting string
as specified in the following resource:
https://unicode-org.github.io/icu/userguide/format_parse/datetime/#datetime-format-syntax.
Cookbook > Date & Time > Formatting > Cake\I18n\Time::i18nFormat()
So long story short, the pattern letter for the week of the year is w, or ww if you want a leading zero, and for the week of the month it's W or WW.
Related
In Google Sheets, I need to create a date in the following format:
2016-06-15T12:00-0800/2016-10-16T12:00-0800
(Yesterday's date / today's date + 120 days)
Using =NOW()-1, I get yesterday's date.
Using Format - Date - More - Year(1930)-Month(05)-Day(01)T:Hour(01):00-0800, I get the proper format for the 1st part of the date range (2016-06-15T12:00-0800).
Repeating the same process with =NOW()+120.
Got the 2nd part of the date range (2016-10-16T12:00-0800).
PROBLEM: Trying to =CONCATENATE(A2,"/",B2), results in this:
42901.6965777315/43022.6965777315
...and no matter what I do - change the format, try to use =CONCATENATE(=TEXT(A2),"/",=TEXT(B2)), or other tricks I know, I either get a blank cell, an error message, or an even worse mess.
All I want is to combine 2 date cells into 1, with a slash in between. How can this be accomplished?
Try join instead of concatenate:
=join("/",A1,B1)
Maybe:
=text(now()-1,"yyyy-mm-ddThh:mm")&"-0800/"&text(now()+120,"yyyy-mm-ddThh:mm")&"-0800"
TEXT with only a date as argument returns the serial number corresponding to that date,
First you should confer the result of you formulas that involves NOW to a formatted text by using TEXT with the second argument. Then you could concatenate the result of that.
The above be donde on a single formula but you maybe should start by doing each step on separate cells in order to make it easy to check the result of each part.
Can anyone tell me why if I type in MATLAB
datestr('17-03-2016','dd-mmmm-yyyy')
I get
06-September-0022
From the datestr docs
DateString = datestr(___,formatOut) specifies the format of the output text using formatOut. You can use formatOut with any of the input arguments in the above syntaxes.
So in your example the 'dd-mmmm-yyyy' is specifying the output format, not the input format.
Also
DateString = datestr(DateStringIn) converts DateStringIn to text in the format, day-month-year hour:minute:second. All dates and times represented in DateStringIn must have the same format.
where
'dd-mm-yyyy' is not in the list of allowed DateStringIn formats AND the documentation explicitly recommends using datenum to ensure correct behaviour. (Note: I underlined the wrong must in the sentence, it's the second must I wanted to emphasise)
So Sandar_Usama's answer of
datestr(datenum('17-03-2016','dd-mmmm-yyyy'))
is the officially correct method straight out of the docs.
Bottom line, always read the documentation.
Use this instead: datestr(datenum('17-03-2016','dd-mmmm-yyyy'))
To address the last unanswered point in this question, why does datenum behave like this?
>> datestr(datenum('17-03-2016'))
ans =
06-Sep-0022
Without explicitly telling datestr and datenum how it should treat the input, it will try to match against the expected formats. Since none of the documented formats match (see #dan's answer), it fails.
Although what it does next is undocumented, at least up to whatever version of Matlab we are running, it falls into a "last resource" attempt to give you a date number.
Matlab will try to parse different month names from your input, remove non-numeric characters, and then timedate elements from the string. In your case, they are 17, 03, and 2016. The first is expected to be either month or year. Since there's no 17th month, it is treated as year. Then 03 is the month, and 2016 is the day.
Now, March 2016th, 17 is not a valid date, but Matlab will give it a slack and read as 1985 days past March 31st, 17. And that gives us September 6th, 22.
Because Matlab's timestamp is a floating number for the number of days since its epoch, you can trigger that answer, using valid dates, like so:
>> datestr(datenum('0017-03-31') + 1985)
ans =
06-Sep-0022
As an absolute beginner to SAS I quickly ran into problems with date formatting.
I have a dataset containing transaction with three types of dates: BUSDATE, SPOTDATE, MATURITY. Each transaction is represented on two lines, and I want BUSDATE and SPOTDATE from line 1 but MATURITY from line 2.
In the original set, the dates are in YYYY-MM-DD format.
DATA masterdata;
SET sourcedata(rename(BUSDATE=BUSDATE2 SPOTDATE=SPOTDATE2 MATURITY=MATURITY2));
BUSDATE=BUSDATE2;
SPOTDATE=SPOTDATE2;
IF TRANS_TYPE='Swap' THEN;
MATURITY=SPOTDATE;
RUN;
Problem is, this returns something like 17169 (which I guess is the number of days from a certain date).
How can I make it output in YYYY-MM-DD format - or is this approach wrong; should I first convert the date variables to some SAS date format?
if you have valid SAS dates, just add a FORMAT statement to your DATA STEP.
Format busdate spotdate maturity yymmdd10. ;
SAS dates are numeric variables. They represent the number of days since 1/1/1960. You use a FORMAT to display dates.
Adding to CarolinaJay's answer, you normally want to keep them as numeric format, since you can do math (like "# of days since date X") with them. However, if for some reason you need a character variable, you can do this:
date_As_char=put(datevar,YYMMDD10.);
Incidentally, YYMMDD10 will actually give you YYYY-MM-DD, as you asked for; if you want a different separator, see http://support.sas.com/documentation/cdl/en/lrdict/64316/HTML/default/viewer.htm#a000589916.htm (YYMMDDxw. format) - if you put a letter after the last D, for certain letters, you get a different separator. Like, YYMMDDn10. gives you no separator, or YYMMDDs10. gives you slashes. YYMMDDd10. gives you dashes, just like omitting the letter would. This concept also applies to MMDDYY formats, and I think a few others.
Using Crystal Reports 2008, I need to extract a date from a text field. This date is usually in the format dd/mm/yy, but could also be entered as d/m/yy, dd/m/yyyy, etc.
This date could appear anywhere within the string.
At the moment I am relying on the fact that the date is placed at the end of the string, without a following fullstop, and using LEFT/RIGHT to extract each date part. These parts are then passed to another formula to create a full date:
Dim AllocationDate() as Date
If Not(IsNull({Command.Notes})) then
Formula = DateValue ((ToNumber ({#Year})), (ToNumber ({#Month})), (ToNumber ({#Day})))
However, if anyone uses a variation of format, adds a fullstop or more notes after the date the whole report keels over.
Is there any way I could extract this date by looking for a pattern? I'm guessing I could the use TRIM to get around the inconsistencies in format.
tyvmia
You may want to consider using a regular expression.
Crystal Reports doesn't have native support for regular expressions, so you'll need to add a UFL: crystal reports : is there a way to regex in crystal reports?
You should be able to adapt the pattern in this question for your needs: Javascript date regex DD/MM/YYYY
Finally, you can test the pattern on your text using regexpal.com.
** edit **
Create a SQL-expression field (Oracle 10 syntax) to extract date string and convert it to a date field:
// {%Allocation Date}
(
-- match date-like string, then convert to date type; is no dates are found, NULL is returned
TO_DATE( REGEXP_SUBSTR( TABLE.FIELD, '\d{1,2}\/\d{1,2}\/\d{2,4}',1 ,1), 'dd/mm/yyyy')
)
While you could
And, as a last resort, you can try converting the multi-line string into a long string by replacing the special characters that represent CR, LF, etc. Replacing them with spaces or another innocuous character, and then treat the resultant string as if it were just a regular string (with the date in the middle).
You would still have to make some assumptions to make this possible: ONE date per string, all special characters are known (or you have to test for all possible special characters), the date has SOME conformity to the format, etc.
I have a very strange problem, Zend_Date is converting my timestamp to a year earlier.
In my action:
// Timestamp
$intTime = 1293922800;
// Zend_Date object
$objZendDate = new Zend_Date($intTime);
// Get date
echo date('Y-m-d',$intTime).'<br>';
echo $objZendDate->get('YYYY-MM-dd');
This outputs:
2011-01-02
2010-01-02
Can anyone tell me what i'm doing wrong?
From the ZF issue tracker it seems this is a known issue:
Recently a lot of ZF users are filing a bug that Zend_Date returns the wrong year, 2009 instead of 2008. This is however expected behaviour, and NOT A BUG!
From the FAQ:
When using own formats in your code you could come to a situation where you get for example 29.12.2009, but you expected to get 29.12.2008.
There is one year difference: 2009 instead of 2008. You should use the lower cased year constant. See this example:
$date->toString('dd.MM.yyyy');
instead of
$date->toString('dd.MM.YYYY');
From the manual
Note that the default ISO format differs from PHP's format which can be irritating if you have not used in previous. Especially the format specifiers for Year and Minute are often not used in the intended way.
For year there are two specifiers available which are often mistaken. The Y specifier for the ISO year and the y specifier for the real year. The difference is small but significant. Y calculates the ISO year, which is often used for calendar formats. See for example the 31. December 2007. The real year is 2007, but it is the first day of the first week in the week 1 of the year 2008. So, if you are using 'dd.MM.yyyy' you will get '31.December.2007' but if you use 'dd.MM.YYYY' you will get '31.December.2008'. As you see this is no bug but a expected behaviour depending on the used specifiers.
For minute the difference is not so big. ISO uses the specifier m for the minute, unlike PHP which uses i. So if you are getting no minute in your format check if you have used the right specifier.
To add to zwip's answer, what happens behind the scenes is that your date format YYYY-MM-dd is actually translated into o\-m\-d, which is then passed to PHP's date() function internally with the timestamp you provided.
Like mentioned in the other answer, and in the documentation for the o format on the date format page, the calculation of the year based on the ISO week can sometimes result in the year being one different to the value that you expect.