My project is supposed to fetch specific values from multiple hashes a put those values in a text file. Ideally what I need my code to do is to have every date for the employees be seven days apart, so the text file would look something like this:
"Rachel Thorndike
2017-10-09-T04:29:46-05:00
Stacie Smith
2017-10-16-T04:29:46-05:00"
What this is supposed to do is fetch employee's names and put the time of their "handoff" on the line under them. I looked online and found the DateTime that Ruby features but it looks like whatever I do isn't working. My code is this:
require 'date'
jsonUser["users"].each do |user|
somefile.puts user["user"]["summary"]
print 'Handoff Date + Time: '
parsed = DateTime.strptime(jsonUser["start"], '%d-%m-%Y %H:%M')
utc = parsed.next_day(7).strftime('%d-%m-%Y %H:%M')
puts utc
end
But terminal returns this code with an error 'strptime': invalid date (ArgumentError). Would anybody help me get this code to work the way I want to? Anything that points me to the right direction? With explanations, if it isn't too much.
Thank you so much!
Update
I was able to get the iso8601 to appear under their name. My new code is
require 'date'
jsonUser["users"].each do |user|
somefile.puts user["user"]["summary"]
print 'Handoff Date + Time: '
parsed = DateTime.iso8601(jsonUser["start"])
utc = parsed.next_day(7).iso8601
somefile.puts utc
end
BUT the .next_day method isn't increasing by 7 days that I want too. Thought? I have only got one value that is appearing and that is going under every line. Its the value of jsonUser["start"] + 7 days...so `2017-
10-16T04:29:46-05:00`
This is what parse.next_day(7) gives me.
"Sr Chid
Handoff Date + Time: 2017-10-16T04:29:46-05:00
Ash A
Handoff Date + Time: 2017-10-16T04:29:46-05:00
Ven D
Handoff Date + Time: 2017-10-16T04:29:46-05:00
Abhi S
Handoff Date + Time: 2017-10-16T04:29:46-05:00"
The value of jsonUser["start"] is 2017-10-09T04:29:46-05:00 so the good thing is that it did increase by 7 but it only did it once.
Update for Amadan
require 'date'
date = DateTime.iso8601(jsonUser["start"])
jsonUser["users"].each do |user|
if user["user"]["self"] == nil
nil
else
somefile.puts user["user"]["summary"].gsub(/\w+/, &:capitalize).gsub(/[.]/, ' ')
somefile.print 'Handoff Date + Time: '
date = date.next_day(7)
somefile.puts date.iso8061
end
end
I have been trying for a while for saving time in my database in timeformat like 10:00:00.
What I want is simply pick up time from timepicker and save data(time) in timeformat.
Here is what I have done:
I have used timepicker as it gets data in format 10:52 AM.
I have used accessor methods to save time as follows:
public function setRTimeAttribute($value)
{
$this->attributes['r_time'] = date('h:i:A', strtotime($value));
}
My post controller in case needed as follows:
public function postSessionReservation(Request $request,$profile_slug)
{
$restaurant = Restaurant::where('slug_profile', $profile_slug)->select('name','ar_name','slug_profile','id')->firstOrFail();
$this->validate($request,[
'no_of_seats'=>'required',
'r_date'=>'required',
'r_time'=>'required',
'restaurant_id'=>'required',
]);
$data = [
'no_of_seats'=>$request->no_of_seats,
'r_date'=>$request->r_date,
'r_time'=>$request->r_time,
'restaurant_id'=>$request->restaurant_id
];
$minutes = 1;
Session::put('reservation',json_encode($data),$minutes);
return redirect($restaurant->slug_profile.'/complete-reservation');
}
But it saves the time as 12:00:00 each time.
I do not know how 12:00:00 is generated whenever I save the time. I tried a lot but could not figure out.
Hope you guys will help me solve this.
Thanks in advance.
First remove spaces between time "10 : 52 AM" to "10:52 AM" and then change your line to this:
$this->attributes['r_time'] = date('H:i', strtotime( $value ));
This will surely help, I already tested it.
For example you can remove spaces through:
$a = '10 : 52 PM';
$x = preg_replace('/\s*:\s*/', ':', $a);
date("H:i", strtotime($x));
I am using the codeigniter time helper to echo the TIMESTAMP (CURRENT_TIMESTAMP) row of my mysql database.
The timestamp in my database in raw format is: 2011-11-15 14:40:45
When I echo the time in my view using the helper i get the following: 15/11/2011 02:40
My time now appears to be in AM. Why???
This is the code I use to echo the time:
$the_date = mdate('%d/%m/%Y %h:%i', strtotime($row->date))
echo $the_date
You need to change the format of your data when returning it from the database. Chage the lowercase h (%h) to a capital H (%H) to return 24-hour format rather than the 12 hour you're currently getting.
You're code should look like the below:
$the_date = mdate('%d/%m/%Y %H:%i', strtotime($row->date))
I'm working with a large data frame called exp (file here) in R. In the interests of performance, it was suggested that I check out the idata.frame() function from plyr. But I think I'm using it wrong.
My original call, slow but it works:
df.median<-ddply(exp,
.(groupname,starttime,fPhase,fCycle),
numcolwise(median),
na.rm=TRUE)
With idata.frame, Error: is.data.frame(df) is not TRUE
library(plyr)
df.median<-ddply(idata.frame(exp),
.(groupname,starttime,fPhase,fCycle),
numcolwise(median),
na.rm=TRUE)
So, I thought, perhaps it is my data. So I tried the baseball dataset. The idata.frame example works fine: dlply(idata.frame(baseball), "id", nrow) But if I try something similar to my desired call using baseball, it doesn't work:
bb.median<-ddply(idata.frame(baseball),
.(id,year,team),
numcolwise(median),
na.rm=TRUE)
>Error: is.data.frame(df) is not TRUE
Perhaps my error is in how I'm specifying the groupings? Anyone know how to make my example work?
ETA:
I also tried:
groupVars <- c("groupname","starttime","fPhase","fCycle")
voi<-c('inadist','smldist','lardist')
i<-idata.frame(exp)
ag.median <- aggregate(i[,voi], i[,groupVars], median)
Error in i[, voi] : object of type 'environment' is not subsettable
which uses a faster way of getting the medians, but gives a different error. I don't think I understand how to use idata.frame at all.
Given you are working with 'big' data and looking for perfomance, this seems a perfect fit for data.table.
Specifically the lapply(.SD,FUN) and .SDcols arguments with by
Setup the data.table
library(data.table)
DT <- as.data.table(exp)
iexp <- idata.frame(exp)
Which columns are numeric
numeric_columns <- names(which(unlist(lapply(DT, is.numeric))))
dt.median <- DT[, lapply(.SD, median), by = list(groupname, starttime, fPhase,
fCycle), .SDcols = numeric_columns]
some benchmarking
library(rbenchmark)
benchmark(data.table = DT[, lapply(.SD, median), by = list(groupname, starttime,
fPhase, fCycle), .SDcols = numeric_columns],
plyr = ddply(exp, .(groupname, starttime, fPhase, fCycle), numcolwise(median), na.rm = TRUE),
idataframe = ddply(exp, .(groupname, starttime, fPhase, fCycle), function(x) data.frame(inadist = median(x$inadist),
smldist = median(x$smldist), lardist = median(x$lardist), inadur = median(x$inadur),
smldur = median(x$smldur), lardur = median(x$lardur), emptyct = median(x$emptyct),
entct = median(x$entct), inact = median(x$inact), smlct = median(x$smlct),
larct = median(x$larct), na.rm = TRUE)),
aggregate = aggregate(exp[, numeric_columns],
exp[, c("groupname", "starttime", "fPhase", "fCycle")],
median),
replications = 5)
## test replications elapsed relative user.self
## 4 aggregate 5 5.42 1.789 5.30
## 1 data.table 5 3.03 1.000 3.03
## 3 idataframe 5 11.81 3.898 11.77
## 2 plyr 5 9.47 3.125 9.45
Strange behaviour, but even in the docs it says that idata.frame is experimental. You probably found a bug. Perhaps you could rewrite the check at the top of ddply that tests is.data.frame().
In any case, this cuts about 20% off the time (on my system):
system.time(df.median<-ddply(exp, .(groupname,starttime,fPhase,fCycle), function(x) data.frame(
inadist=median(x$inadist),
smldist=median(x$smldist),
lardist=median(x$lardist),
inadur=median(x$inadur),
smldur=median(x$smldur),
lardur=median(x$lardur),
emptyct=median(x$emptyct),
entct=median(x$entct),
inact=median(x$inact),
smlct=median(x$smlct),
larct=median(x$larct),
na.rm=TRUE))
)
Shane asked you in another post if you could cache the results of your script. I don't really have an idea of your workflow, but it may be best to setup a chron to run this and store the results, daily/hourly whatever.
Is there a built-in function to get the day in last month, same as today? Examples:
2010/05/02 -> 2010/04/02
2010/05/15 -> 2010/04/15
2010/05/31 -> 2010/04/30
Thanks!
You can subtract entire months with <<.
>> d = Date.parse('2010-05-31')
=> #<Date: 4910695/2,0,2299161>
>> d.to_s
=> "2010-05-31"
>> (d<<1).to_s
=> "2010-04-30"
More info
You could do this:
(Date.today - 1.month).strftime("%Y/%m/%d")
You can try using
Time.parse('2010/05/31').months_since(-1).
You could for instance make a time object
old_time = Time.now
Then create a new time object based on that
new_time = Time.local(old_time.year, (old_time.month - 1), old_time.day, old_time.hour, old_time.min, old_time.sec)
However, as deceze pointed out, what is the criterion for 5/31 becoming 4/30?
In irb, 4/31 'overflows' to 5/01.
Ruby's date class allows you to add, subtract days from a particular date.