date

问题 Scenario: I am creating an event registration module in which a user would choose a date from a calendar for booking, fill some input fields and pay up to book the date. One of the requirements is that, there can be only one booking per date. This arises a case where two people want to book the same date on the same time. So, we need to show the user who came later to book the date a message that a booking is already in progress for that specific date. Please check back later . For this to

问题 I am unable to apply transformation using below code. getting error The year function doesn't support dt_wstr. The expression iam using is: (DT_I4)((DT_WSTR,4)YEAR(fisc_wk_end_dt) + RIGHT("0" + (DT_WSTR,2)MONTH(fisc_wk_end_dt),2) + RIGHT("0" + (DT_WSTR,2)DAY(fisc_wk_end_dt),2)) 回答1: Problem From the expression you mentioned, it looks like fisc_wk_end_dt column data type is string while YEAR, MONTH, DAY function's parameters must be of type date. From the official documentation: Syntax YEAR

问题 I want merge data from a filtered set into a timeline I created with the help of the timeline package. df1 looks like Date Label Freq 2011-03-12 1 18 2011-03-14 1 16 2011-03-18 1 5 time line produces a vector with dates from a specific starting date until a specified end date. What I want to achieve is a timeline with all days in a certain period. Then I want to merge df1 into timeline. Using left_join from dplyr I first get Error in UseMethod("left_join") : not applicable for 'left_join' for

问题 This question already has answers here : Java: Date from unix timestamp (10 answers) Converting Long to Date in Java returns 1970 (11 answers) android timestamp parsing gone wrong(always in 1970) (4 answers) Closed 1 year ago . I want to convert 1574348400 value to date format using code: public class Main { public Main() { long value = 1574348400; String dateString = new SimpleDateFormat("EEEE dd MMMM, yyyy").format(new Date(value)); System.out.println("Formated time: " + dateString); }

问题 objective: compare the date in the string with today's date Issue: get this error:ValueError: unconverted data remains: 12:00:00 question: how do a fix that error how do i remove the time element of the string? Code from datetime import date from datetime import time from datetime import datetime dateTimeStart = "2014-01-01 12:00:00" def main(): dateTime1 = datetime.strptime(dateTimeStart, "%Y-%m-%d") today = date.today() if dateTime1 > "today": print "No" if __name__ == "__main__": main();

问题 Is it possible merge text and photo taken from camera? I would like to stamp date and time on photo but I don't find anything on Google. 回答1: Use below code to achieve what you required. Bitmap src = BitmapFactory.decodeResource(getResources(), R.drawable.cuty); // the original file is cuty.jpg i added in resources Bitmap dest = Bitmap.createBitmap(src.getWidth(), src.getHeight(), Bitmap.Config.ARGB_8888); SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); String dateTime =

问题 Is it possible merge text and photo taken from camera? I would like to stamp date and time on photo but I don't find anything on Google. 回答1: Use below code to achieve what you required. Bitmap src = BitmapFactory.decodeResource(getResources(), R.drawable.cuty); // the original file is cuty.jpg i added in resources Bitmap dest = Bitmap.createBitmap(src.getWidth(), src.getHeight(), Bitmap.Config.ARGB_8888); SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); String dateTime =

问题 Is it possible merge text and photo taken from camera? I would like to stamp date and time on photo but I don't find anything on Google. 回答1: Use below code to achieve what you required. Bitmap src = BitmapFactory.decodeResource(getResources(), R.drawable.cuty); // the original file is cuty.jpg i added in resources Bitmap dest = Bitmap.createBitmap(src.getWidth(), src.getHeight(), Bitmap.Config.ARGB_8888); SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); String dateTime =

问题 I have data that looks like this Date Fruit 2017-01-01 Orange 2017-01-01 Apple 2017-01-08 Orange 2017-01-09 Orange 2017-01-09 Apple I want to plot Number of Oranges, Number of Apples by Date in a single plot. How would I do that? I grouped them by Date and I see the result. df.groupby(['Date','Fruit']).size() Date Fruit 2017-01-01 Orange 1 Apple 1 2017-01-08 Orange 1 2017-01-09 Orange 1 Apple 1 I tried this but it gives a bar plot having two categories but not against the dates. sns.catplot(x

问题 I have a column in a dataframe that contains the day number ( 0 through 6, 0=Sunday, 1=Monday, etc) and I need to convert that to the day name. How can I do this? Sample data: df <- data.frame(day_number=0:6) 回答1: Simple way with dplyr. library(dplyr) df <- data.frame(day_number=0:6) df$day_number <- recode(df$day_number, "0"="Sunday", "1"="Monday", "2"="Tuesday", "3"="Wednesday", "4"="Thursday", "5"="Friday", "6"="Saturday") 回答2: Treat the column as a factor with day name as label: x <- data