I have a collection, that I would like to achieve group by the year and month.
2014
jan
2nd
5th
...
feb
11th
12th
...
mar
...
2013
jan
feb
mar
...
2012
jan
feb
mar
...
this is my current collection but only by year. I couldn't get what I really wanted with this.
$collection =$model->orderBy('date', 'desc')->get()->groupBy(function($item) {
return $item->date->format('Y');
});
Do a Group By on DB::raw('MONTH(your date variable)')
For example, try this:
$model->orderBy('date')->groupBy(DB::raw('MONTH(date)'))->selectRaw('DATE_FORMAT(date, "%M %Y") as Month)->get();
Related
I'm writing a validation and I have an hash with this structure
elements.map{ |e| [e.id,e.coverable.published_at] }.to_h
=> {305=>Fri, 17 Apr 2020 15:23:00 CEST +02:00,
306=>Fri, 17 Apr 2020 13:00:00 CEST +02:00,
307=>Fri, 17 Apr 2020 09:20:00 CEST +02:00,
308=>Fri, 17 Apr 2020 12:59:00 CEST +02:00,
309=>Fri, 17 Apr 2020 11:39:00 CEST +02:00}
I have a reference date...
published_at
=> Mon, 04 May 2020 23:51:00 CEST +02:00
I have to check if any of the element has a published_at datetime value bigger than my published_at.
Is there a short way to do that?
Try something like this
elements.any? { |e| e.coverable.published_at > your_published_at }
In case you need the element which passes the condition use find
element = elements.find { |e| e.coverable.published_at > your_published_at }
# if element is not nil such element is present
I want to display the short name for each month, 12 months back from the previous month, but with the below I get an error on mon1 and mon2 and I guess since that is last year?
mon1=MonthName(Month(Now())-11,1)
mon2=MonthName(Month(Now())-10,1)
mon3=MonthName(Month(Now())-9,1)
mon4=MonthName(Month(Now())-8,1)
mon5=MonthName(Month(Now())-7,1)
mon6=MonthName(Month(Now())-6,1)
mon7=MonthName(Month(Now())-5,1)
mon8=MonthName(Month(Now())-4,1)
mon9=MonthName(Month(Now())-3,1)
mon10=MonthName(Month(Now())-2,1)
mon11=MonthName(Month(Now())-1,1)
mon12=MonthName(Month(Now()),1)
So how can I display now,dec,jan,feb,mar,apr,may,jun,jul,aug,sep,oct
Thanks.
The problem here is the Month() function returns an integer between 1 and 12 to represent each month. Instead you want to minus the number of months from the Now() value before wrapping it with Month().
Below is an example that does this using a For loop and a single dimension Array.
Dim dt: dt = Now()
Dim i, mon(12)
Const numOfMonths = 12
For i = 1 To numOfMonths
mon(i) = MonthName(Month(DateAdd("m", i - numOfMonths, dt)), True)
Next
Call Response.Write(Join(mon, vbCrLf))
Output:
Sep
Aug
Jul
Jun
May
Apr
Mar
Feb
Jan
Dec
Nov
Oct
my data looks like
JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC
22.60 24.60 30.60 34.60 36.20 35.70 32.10 30.20 31.40 31.60 28.00 24.80
25.40 27.60 32.40 34.60 36.50 38.10 31.70 31.40 30.30 30.20 27.00 23.90
and there are like hundreds of rows! I want to find a maximum value in each row and write it in different column next to data along with month
so my out put will be
36.20 MAY
38.10 JUN
.
.
I want to use maxloc function, but i have no idea how to use it!
Try
index = maxloc(myTable(3,:))
print *, myTable((/1,3/), index)
It should select the highest value from the third row and display the first and third value at this index.
I would like my watcher to run from Monday to Friday only. So I'm trying to use this schedule:
"trigger": {
"schedule" : { "cron" : "0 0 0/4 * * MON-FRI" }
},
"input": {
...
However, I'm getting
Error
Watcher: [parse_exception] could not parse [cron] schedule
when I'm trying to save the watcher. Removing MON-FRI does helps but I need it.
This expression works:
0 0 0/4 ? * MON-FRI
But I'm not sure I understand why ? is required for either the day_of_week or day_of_month
Thank you!
I believe this is what you are looking for:
"0 0 0/4 ? * MON-FRI"
You can use croneval to check your cron expressions 1:
$ /usr/share/elasticsearch/bin/x-pack/croneval "0 0 0/4 ? * MON-FRI"
Valid!
Now is [Mon, 20 Aug 2018 13:32:26]
Here are the next 10 times this cron expression will trigger:
1. Mon, 20 Aug 2018 09:00:00
2. Mon, 20 Aug 2018 13:00:00
3. Mon, 20 Aug 2018 17:00:00
4. Mon, 20 Aug 2018 21:00:00
5. Tue, 21 Aug 2018 01:00:00
6. Tue, 21 Aug 2018 05:00:00
7. Tue, 21 Aug 2018 09:00:00
8. Tue, 21 Aug 2018 13:00:00
9. Tue, 21 Aug 2018 17:00:00
10. Tue, 21 Aug 2018 21:00:00
For the first expression you'll get following java exception:
java.lang.IllegalArgumentException: support for specifying both a day-of-week AND a day-of-month parameter is not implemented.
You can also use Crontab guru to get human readable descriptions like:
At every minute past every 4th hour from 0 through 23 on every day-of-week from Monday through Friday.
The question mark means 'No Specific value'. From the documentation on Quartz's website:
? (“no specific value”) - useful when you need to specify something in one of the two fields in which the character is allowed, but not the other. For example, if I want my trigger to fire on a particular day of the month (say, the 10th), but don’t care what day of the week that happens to be, I would put “10” in the day-of-month field, and “?” in the day-of-week field. See the examples below for clarification.
http://www.quartz-scheduler.org/documentation/quartz-2.x/tutorials/crontrigger.html
I suppose since you want your schedule to run every 4 hours, mon-fri, the actual day of the month is irrelevant, so the ? specifies that. * on teh other hand would be 'all values' which would not make sense since you are specifying only mon-fri for day of the week.
Hope that helps!
I create a hash with months as keys and timelaps as values
biens_delai[bien_date.mon] = b.delai
I get this result without month parsing
{Wed, 18 Jan 2017=>3.0, Sat, 25 Feb 2017=>2.0, Fri, 17 Mar 2017=>3.0, Sat, 25 Mar 2017=>5.0, Tue, 18 Apr 2017=>2.0, Thu, 29 Jun 2017=>2.0}
In March i have 2 values but when i parse by month i get the most high value and i want a addition of 2 values for March not the most high
{1=>3.0, 2=>2.0, 3=>5.0, 4=>2.0, 6=>2.0}
That's not the high value which you are getting, the values are getting overwritten, try the following
biens_delai[bien_date.mon] = biens_delai[bien_date.mon].to_f + b.delai