I have a command where I can enter a specific date for start date, please see below a part of the command I am using.
and when I am calling it in the test I need to enter a dynamic date like today <=(+30 days)
cy.create123((new Date().getDate() - 1), '2023-08-07')
ofc it did not work, but I have no idea how can I do it. How I can setup to cy.command to get always today-1 as startDate!
My issue is to make the dynamic enter date to work on Cypress.Commands()
TLDR
Install dayjs and use
const startDate = dayjs().add(-1, 'day').format('YYYY-MM-DD')
cy.create123(startDate, '2023-08-07')
The Custom Command and the cy.request() inside it are expecting the date as a string type.
Your calculated dynamic date (new Date().getDate() - 1) is giving you a number type.
But .toISOString() only works on Date types, not number types.
So after doing math on the Date(), you get a number which must be converted into a Date and then into a string.
const today = new Date()
const yesterday = new Date(today.setDate(today.getDate() -1))
const startDate = yesterday.toISOString()
But even that's not the end of the problems, because the timezone might give you invalid dates.
I recommend using dayjs as shown above.
You can do something like this. Instead of subtracting 1, I am decreasing the day with 24 hours.
cy.create123(new Date(Date.now() - (3600 * 1000 * 24)).getUTCDate(), '2023-08-07')
Considering today is 29 Aug, this will give the output as 28.
To get the date in the format yyyy-mm-dd use:
new Date(Date.now() - ( 3600 * 1000 * 24)).toISOString().slice(0, 10)
I have to variables: year and month, both of type integer.
E.g:
year = 2016
month = 1
I want to, in my select statement, return a timestamp given those two variables.
I've had a look at the documentation, specifically to_timestamp and to_date, but all the examples I've come across show a string being converted into a timestamp.
I do not really want to convert my year and month into a string, such as:
to_timestamp(to_char(int,year) + ' ' + to_char(int, month),YYYY/MM/DD HH24:MI:SS)
So, how can I (if it is possible) convert my year and month into a timestamp?
If you are using Postgres 9.4 or later, you could try using make_timestamp():
make_timestamp(2016, 1, 1, 0, 0, 0.0)
This would create a timestamp for January 1, 2016 at midnight. We need to specify values for the other components, even if they end up not being relevant to your query/calculation (e.g. you only need the date).
I try to understand why
print(pd.Timestamp("2015-01-01") - pd.DateOffset(day=1))
does not result in
pd.Timestamp("2014-12-31")
I am using Pandas 0.18. I run within the CET timezone.
You can check pandas.tseries.offsets.DateOffset:
*kwds
Temporal parameter that add to or replace the offset value.
Parameters that add to the offset (like Timedelta):
years
months
weeks
days
hours
minutes
seconds
microseconds
nanoseconds
Parameters that replace the offset value:
year
month
day
weekday
hour
minute
second
microsecond
nanosecond
print(pd.Timestamp("2015-01-01") - pd.DateOffset(days=1))
2014-12-31 00:00:00
Another solution:
print(pd.Timestamp("2015-01-01") - pd.offsets.Day(1))
2014-12-31 00:00:00
Also it is possible to subtract Timedelta:
print(pd.Timestamp("2015-01-01") - pd.Timedelta(1, unit='d'))
pd.DateOffset(day=1) works (ie no error is raised) because "day" is a valid parameter, as is "days".
Look at the below one: "day" resets the actual day, "days" adds to the original day.
pd.Timestamp("2019-12-25") + pd.DateOffset(day=1)
Timestamp('2019-12-01 00:00:00')
pd.Timestamp("2019-12-25") + pd.DateOffset(days=1)
Timestamp('2019-12-26 00:00:00')
Day(d) and DateOffset(days=d) do not behave exactly the same when used on timestamps with timezone information (at least on pandas 0.18.0). It looks like DateOffset add 1 day while keeping the hour information while Day adds just 24 hours of elapsed time.
>>> # 30/10/2016 02:00+02:00 is the hour before the DST change
>>> print(pd.Timestamp("2016-10-30 02:00+02:00", tz="Europe/Brussels") + pd.offsets.Day(1))
2016-10-31 01:00:00+01:00
>>> print(pd.Timestamp("2016-10-30 02:00+02:00", tz="Europe/Brussels") + pd.DateOffset(days=1))
2016-10-31 02:00:00+01:00
I'm looking to write a pretty straightforward aggregation query in MongoDB, but struggling with a certain part.
What I would like to do is pull the sum of all records within the last 7 days grouped by day. It's easy enough to define the date 7 days ago as UTC, but I'd like to do it programmatically so I don't need to work out the UTC date everytime. For instance instead of 1341964800 i'd like to specify something like date() - 7 days.
Here's the current aggregation function I have which works:
db.visits_calc.group(
{ key:{date:true},
cond:{date:{$gt:1341964800}},
reduce:function(obj,prev) {prev.csum += obj.total_imp},
initial:{csum:0}
});
Thanks in advance!
You can perform arithmetic on the milliseconds timestamp returned by Date.now() to find the appropriate timestamp for 7 days ago. You need to subtract the number of milliseconds in 7 days (1000ms/s, 60 s/min, 60min/hr, 24 hrs/dy, 7dys/wk).
var weekAgo = Date.now() - (1000*60*60*24*7);
db.visits_calc.group(
{ key:{date:true},
cond:{date:{$gt:weekAgo}},
reduce:function(obj,prev) {prev.csum += obj.total_imp},
initial:{csum:0}
});
I am writing code to convert from a Gregorian date to a JDE (J.D.Edwards) Julian date.
Note: a JDE Julian date is different from the normal usage of the term Julian date.
As far as I can work out from Googling, the definition of a JDE Julian date is:
1000*(year-1900) + dayofyear
where year is the 4-digit year (e.g. 2009), and dayofyear is 1 for 1st January, and counts up all year to either 365 or 366 for 31st December (depending whether this is a leap year).
My question is this: are years before 1900 supported? If so, does the above formula still hold, or should it be this:
1000*(year-1900) - dayofyear
(note minus instead of plus.)
or something else?
Does anyone have a link to the official documentation for this date format?
The JDE Julian date consists of CYYDDD which is Century, Year, Day of year.
Century is zero for 20th e.g. 19XX and one for 21st e.g. 20XX.
The year is two digits.
So 101001 is 1 January 2001
As you can see this will not support dates before 1900.
See this Oracle page for a simple and official explanation: About the Julian Date Format
The "JDE Julian Date Converter" does return a negative value for:
1809/07/23 : -90635
As opposed to the classical Julian Date:
The Julian date for CE 1809 July 23 00:00:00.0 UT is
JD 2381986.50000
Here is a example of JD EDWARDS (AS/400 software) Julian Date, but that is not an "official" documentation and it does not seems to support dates before 1900...
Note: this "ACC: How to Convert Julian Days to Dates in Access and Back" does not support date before 1900 either... as it speaks about an "informal" Julian day, commonly used by government agencies and contractors.
The informal Julian day format used in this article is the ordinal day of a year (for example, Julian day 032 represents February 1st, or the 32nd day of the year).
Variations on informal Julian day formats include using a preceding two-digit year (for example 96032 for 2/1/96) and separating the year with a dash (for example 96-032).
Another, less popular, Julian day format uses a one digit year (for example 6-032). These additional formats do not uniquely identify the century or decade. You should carefully consider the consequences when using these formats; for example, the Julian day 00061 can be interpreted as 3/1/2000 or 3/2/1900.
Update: Sorry, JDE is probably something else. But for reference:
The JDE I know is different. From page 59 in the book
"Astronomical algorithms" (Jean Meeus, ISBN 0-943396-35-2):
"If the JD corresponds to an instant
measured in the scale of Dynamical
Time (or Ephemeris Time), the
expression Julian Ephemeris Day
(JDE) is generally used. (Not JED as
it is sometimes written. The 'E' is a
sort of index appended to 'JD')"
JD and JDE (for the same point in time) are close in value
as the difference UT and ET is on the order of minutes. E.g. ET-UT was 56.86 seconds in 1990 and -2.72 seconds in 1900.
There is also MJD (Modified Julian Day):
MJD = JD - 2400000.5
Zero point for MJD is 1858-11-17, 0h UT.
Note that JD as Julian date is a misnomer. It is
Julian day. The JD has nothing to do with the Julian
calendar. (This is in disagreement with the Wikipedia article, this
is from the author of the book mentioned above, Jean Meeus - a Belgian astronomer specializing in celestial mechanics.)
Maybe off from the question, you can convert in Excel using the following formula:
Convert Julian to Date in Excel
In Cell A2 place a Julian date, like 102324
in Cell B2 place this formula: (copy it in)
=DATE(YEAR("01/01/"&TEXT(1900+INT(A2/1000),0)),MONTH("01/01/"&TEXT(1900+INT(A2/1000),0)),DAY("01/01/"&TEXT(1900+INT(A2/1000),0)))+MOD(A2,1000)-1
The date 11/20/02 date will appear in cell B2
Convert Date to Julian in Excel
In Cell C2 copy this formula:
=(YEAR(B2)-2000+100)*1000+B2-DATE(YEAR(B2),"01","01")+1
This will convert B2 back to 102324
Save the below source code in a source member called JDEDATES. Use the runsqlstm on the first line to create the functions. You can then do things like
select jde2date(A1UPMJ), f.* from f00095 f
and see a real date.
Source:
--RUNSQLSTM SRCFILE(qtxtsrc) SRCMBR(JDEDATES) COMMIT(*NONE) NAMING(*SQL)
-- jde 2 date
create function QGPL/jde2date ( d decimal(7,0))
returns date
language sql
deterministic
contains sql
SET OPTION DATFMT=*ISO
BEGIN
if d=0 then return null;
else
return date(digits(decimal(d+1900000,7,0)));
end if;
end; -- date 2 jde
create function QGPL/date2jde ( d date)
returns decimal(7,0)
language sql
deterministic
contains sql
SET OPTION DATFMT=*ISO
BEGIN
if d is null then return 0;
else
return (YEAR(D)-1900)*1000+DAYOFYEAR(D);
end if;
end ;
Several years late to the party, but for other folks like me that find yourselves working with legacy systems like this, I hope some of my java snippets can help. I'm leveraging the fact that you can convert this CYYDDD format into yyyyDDD format and parse based on that.
import java.util.Date;
import java.util.GregorianCalendar;
import java.util.Calendar;
import java.util.SimpleDateFormat;
String jdeJulianDate = "099365"; //Testing with December 31, 1999
// Compile what the year number is
int centIndex = Integer.parseInt(jdeJulianDate.substring(0,1));
int yearIndex = Integer.parseInt(jdeJulianDate.substring(1,3));
int yearNumber = 1900 + (100 * centIndex) + yearIndex;
// Put the year number together with date ordinal to get yyyyDDD format
String fullDate = String.valueOf(yearNumber) + jdeJulianDate.substring(3,6);
// Date parsing, so need to wrap in try/catch block
try {
Date dt = new SimpleDateFormat("yyyyDDD").parse(fullDate);
// Validate it parses to a date in the same year...
Calendar cal = new GregorianCalendar();
cal.setTime(dt);
if (cal.get(Calendar.YEAR) != yearNumber) {
// Cases happen where things like 121366 (should be invalid) get parsed, yielding 2022-01-01.
// Throw exception or what-not here.
}
}
catch (Exception e) {
// Date parsing error handling here
}
A sample of VBA code to convert back and forth between JDE Julian Date and Gregorian:
Public Const Epoch = 1900
Public Const JDateMultiplier = 1000
Public Const FirstJan = "01/01/"
Public Function Julian2Date(ByVal vDate As Long) As Date
Dim Year As Long
Dim Days As Long
Dim SeedDate As Date
' Day Number
Days = vDate - (Int(vDate / JDateMultiplier) * JDateMultiplier) - 1
' Calendar Year
Year = ((vDate - Days) / JDateMultiplier) + Epoch
' First Day of Calendar Year
SeedDate = CDate(FirstJan + CStr(Year))
' Add Number of Days to First Day in Calendar Year
Julian2Date = DateAdd("d", Days, SeedDate)
End Function
Public Function Date2Julian(ByVal vDate As Date) As Long
Dim JYear As String
Dim BeginDate As Date
Dim JDays As Long
' Calendar Year
JYear = Format(Year(vDate), "0000")
' First Day of Calendar Year
BeginDate = CDate(FirstJan + JYear)
' Day Number
JDays = DateDiff("d", BeginDate, vDate) + 1
' Add Number of Days to Year Number
Date2Julian = ((CLng(JYear) - Epoch) * JDateMultiplier) + JDays
End Function
I have tried to make it as clear and simple as possible, and to this end I have intentionally left out any error trapping. However, you should be able to add the code to a VBA module and call them directly from your own code.
I also include some useful snippets of T-SQL:
Todays Date as JDE Julian Date:
(datepart(yy,getdate())-1900) * 1000 + datepart(dy, getdate())
Convert JDE Julian Date to Gregorian (DD/MM/YYYY), replace XXXXXX with the column name containing the JDE Julian Date:
convert (varchar, dateadd (day,convert (int, right(XXXXXX,3)) - 1, convert (datetime, ('1/1/' + convert ( varchar, (cast(left(right(XXXXXX+1000000,6),3) as varchar) + 1900))))),103)
If you require a different Gregorian format, replace the 103 value (right at the end) with the applicable value found here: https://msdn.microsoft.com/en-us/library/ms187928.aspx
I have an easy way for C using time now and epoch 1970, 01, 01 midnight if anybody is interested.
But this is for Julian Day Numbers which is not the same as JDE but they are similar in respect to using math to compute days and I'm sure this idea could be adapted for JDE. Sometimes people just confuse the two like I do. Sorry. But still this is an example of using a time reference which should always be done and since most computers use this it would be just as easy for us not to get too bogged down in dates and just use days before or after this epoch.
Since JDE is now owned by Oracle, they also now support Julian_Day. see:
https://docs.oracle.com/javase/8/docs/api/java/time/temporal/JulianFields.html
#include <stdio.h>
#include <time.h>
#define EPOCH (double) 2440587.5 /* Julian Day number for Jan. 01, 1970 midnight */
int main ()
{
double days = time(0)/86400.0;
printf ("%f days since January 1, 1970\n", days);
printf ("%f\n", days + EPOCH);
return 0;
}
Wow, there's a lot of complicated code in some of these answers just to convert to and from JDE julian dates. There are simple ways in Excel and VBA to get there.
FROM JULIAN
Excel (assuming julian date is in A1):
=DATE(1900+LEFT(A1,LEN(A1)-3),1,RIGHT(A1,3))
VBA (from julian date, j, stored as String):
d = DateSerial(1900 + Left$(j, Len(j) - 3), 1, Right$(j, 3))
VBA (from julian date, j, stored as Long):
d = DateSerial(1900 + Left$(j, Len(CStr(j)) - 3), 1, Right$(j, 3))
TO JULIAN
Excel (assuming date is in A1):
=(YEAR(A1)-1900)*1000+A1-DATE(YEAR(A1),1,0)
VBA (to a Long, j):
j = (Year(d) - 1900) * 1000 + DatePart("y", d)