can plz someone help me to correct my Funktion
count number of cells with Date "Today" and Time "between 06:00 and 14:30"
=countifs('G:G">=Today() 06:00:00", 'G:G"<=Today() 14:30:00")
try:
=COUNTA(IFNA(FILTER('JIRA_Input SL'!G:G;
'JIRA_Input SL'!G:G>=(TODAY()+"06:00:00");
'JIRA_Input SL'!G:G<=(TODAY()+"14:30:00"))))
Related
Trying different things and combos, but can't get it right. Times and dates always get me.
I'm simply trying to compare now() and a date/time and show "in 2 days 6 hrs 30 mins" or "3 days 1 hr 20 mins ago" and I'm guessing someone has done this already?
So, let's say the date in column A is 2023-01-24 10:00 and now it is 2023-01-25 11:05, then Col B should say "1 day 1 hr 5 mins ago"
I've tried duration(), date/time formatting the cell, days(..), but I can't find something that works reliably.
Another thing you could try:
=INT(NOW()-A1)&" d "&TEXT(NOW()-A1,"h \hr m \min a\go")
=NOW()-A1 will actually calculate the difference between Now and that time, BUT consider that the time is updated only when you make a change, it's not a countdown second by second. You can add an updater per minute but not more than that. Go to File - Settings and set On change and every minute:
If you need a specific format like that you mentioned you should then do calculations to add them. If you only need days, hours, minutes and seconds, you can do the next checkings and concatenations:
=LAMBDA(dif,
IF(INT(dif),INT(dif)&" days ","")
&IF(HOUR(dif),HOUR(dif)&" hours ")
&IF(MINUTE(dif),MINUTE(dif)&" minutes ")
&IF(SECOND(dif),SECOND(dif)&" seconds")&" ago")
(Now()-A1)
please try:
=TRIM(LAMBDA(y,IF(NOW()-A1<1,"0 days",IF(y=1,y&" "&"day",y&" "&"days")))(DATEDIF(A1,NOW(),"D"))&" "&
LAMBDA(y,IF(y=1,y&" "&"hr",y&" "&"hrs"))(HOUR(NOW()-A1))&" "&
LAMBDA(y,IF(y=1,y&" "&"min",y&" "&"mins"))(MINUTE(NOW()-A1))&" ago")
try:
=INDEX(IF(ISDATE_STRICT(A2:A), TRIM(FLATTEN(QUERY(TRANSPOSE(
IFERROR(LAMBDA(a, LAMBDA(x, IF(x=0,,IF(x>1, x&a&"s", x&a)))
(DATEDIF(A2:A, NOW(), {"Y", "YM", "MD"})))({" year", " month", " day"}))),,9^9))), ))
I use Jiffy for calculate the difference in between 2 date in month.
But I don't the good result.
For end = 2/4/2023 and start = 1/4/2022 I have 12 months.
For end = 1/4/2023 and start = 1/4/2022 I have 11 months (error: expected 12).
Thanks,
num month = Jiffy(end).diff(Jiffy(start), Units.MONTH);
Right now it is checking 12 at midnight to 12 at midnight of the end date which is 11 months 30 days. Now if the end date is one second greater than the correct date it should work. So a hackey solution is to add one day duration to the end date and check
end.add(duration(day:1))
The value received from the Date Picker is in the format "02-06-2020" (formatted with mask = "DD-MM-YYYY" in q-date).
I need to convert it to "2020-06-02" format to send it to the server.
I tried the following, but I get undefined.
Example:
let myDate = "02-06-2020";
console.log(date.formatDate(myDate, "YYYY-MM-DD")) //undefined
I would appreciate suggestions for finding a solution.
Thanks.
You can use moment js.
moment(moment('13-01-2020', 'DD-MM-YYYY')).format('YYYY-MM-DD');
or
date.split("-").reverse().join("-");
I need to show last Sunday's date in a cell for a weekly report that I'm creating on google sheets. I've been googling to find a solution and the closest I found is this:
=TODAY()+(7-WEEKDAY(TODAY(),3))
But this gives next Monday's date. Any idea how to modify this to show last Sunday's date? Alternately, do you have another way to solve this problem?
The formula you're looking for would be:
=DATE(YY, MM, DD) - (WEEKDAY(DATE(YY, MM, DD)) - 1) - 7
// OR
=TODAY() - (WEEKDAY(TODAY()) - 1) - 7
Depending on what you take to be "last Sunday," you can simplify/factor this:
If you take "last Sunday" to mean, "the Sunday which has most recently happened:"
=DATE(A2,B2,C2) - WEEKDAY(DATE(A2,B2,C2)) + 1
If you take "last Sunday" to mean, "the Sunday which took place during the previous week:"
=DATE(A4,B4,C4) - WEEKDAY(DATE(A4,B4,C4)) - 6
Working through it:
=TODAY() - (WEEKDAY(TODAY()) - 1) - 7
TODAY()
// get today's date, ie. 22/11/2019
WEEKDAY(TODAY())
// get the current day of the week, ie. 6 (friday)
-
// take the first date and subtract that to rewind through the week,
// ie. 16/11/2019 (last saturday, since saturday is 0)
- 1
// rewind back one less than the entire week
// ie. 17/11/2019 (this sunday)
- 7
// rewind back to the sunday before this
// sunday, ie. 10/11/2019
Hopefully this explanation explains what the numbers at the end are doing. + 1 takes you from the Saturday of last week to the Sunday of the current week (what I would call "this Sunday"), and - 6 takes you back through last week to what I would call "last Sunday."
See here:
try:
=ARRAYFORMULA(TO_DATE(FILTER(ROW(INDIRECT(VALUE(TODAY()-6)&":"&VALUE(TODAY()))),
TEXT(ROW(INDIRECT(VALUE(TODAY()-6)&":"&VALUE(TODAY()))), "ddd")="Sun")))
if today is Sunday and you want still the last Sunday then:
=ARRAYFORMULA(IF(TEXT(TODAY(), "ddd")="Sun", TODAY()-6,
TO_DATE(FILTER(ROW(INDIRECT(VALUE(TODAY()-6)&":"&VALUE(TODAY()))),
TEXT(ROW(INDIRECT(VALUE(TODAY()-6)&":"&VALUE(TODAY()))), "ddd")="Sun"))))
Using monday as 1 index, this will return the previous sunday or today if it is sunday
=if((7-WEEKDAY(today(),2))>0,today()-(7-(7-WEEKDAY(today(),2))),today()+(7-WEEKDAY(today(),2)))
One can select for other days of the week by changing the number "7" directly before "-WEEKDAY(today(),2)" in the three places that pattern exists.
We can use fbusdate to get the first business day of a month:
Date = fbusdate(Year, Month);
However, how do we get the first business day of a week?
As an example, during the week that I'm posting this, Monday 09/07/2017 was a holiday in the US:
isbusday(736942) % = 0
How do I determine that the first business day for this week would be the next day 736943?
I'm not aware of a builtin function that returns the first working day of a week, but you can obtain it by requesting the next working day after Sunday:
busdate(736941); % 736941 = Sunday 09/03/2017
Your desired fbusdateweek function can be done in one line using just the function weekday to get the first Sunday of the week then busdate to get the next business day after that:
dn = 736942; % Date number for any day in a week
Date = busdate(dn-weekday(dn)+1);
Note: busdate uses the function holidays by default to get all holidays and special nontrading days for the New York Stock Exchange. If necessary, you can define an alternate set of holidays for busdate to use as follows:
holidayArray = ...; % Some set of date numbers, vectors, or datetimes
Date = busdate(dn-weekday(dn)+1, 1, holidayArray);
This way you can define a set of localized holidays.
Solved it. Here is a function that is based on the answer of #m7913d:
function Busday = fbusdateweek(date)
% Return the first business day after Sunday
% 'date' is a datenum input
dperiod = date-6:date;
sundays = weekday(dperiod)==1;
sunday = find(sundays==1,1,'first');
datesunday = dperiod(sunday);
% -->
Busday = busdate(datesunday);
end