This will provide the Monday of a certain week this year: $thisYear = date(‘Y’); $weekNum = 40; $date = date(‘Y-m-d’, strtotime(“$thisYear-W$weekNum-1”)); // Outputs 2013-09-30 3 solved Get date from monday depending on week number [closed]
The format argument you give should reflect the format that the data is currently in, not the format that you want to convert it to, so you would have to set format = “%Y-%m-%d”. Read the documentation on strptime again and it should make sense. 1 solved strptime returning NA values [closed]
I fixed it by: adding this line df1$yr <- strftime(df1$Close.Date, “%Y”) and then replacing Close.Date with yr in the ggplot line. solved ggplot2: Group histogram data by year
You can format and parse the date from cal.getTime() using df (SimpleDateFormat(“MM/dd/yyyy”)) which will get rid of time and then compare the result with d. solved How to ignore time while checking date using before(Date d)
Change the first format to SimpleDateFormat tempFormat = new SimpleDateFormat(“yyyy-MM-dd”); as DD is the day in the year. 22 is definitely in January 0 solved simple date format giving wrong month [duplicate]
Will be [0-9]{2}\.[0-9]{2}\.[0-9]{4} [0-9]stands for any integer in range 0 to 9, the number in curly brackets ({2} in this case) indicates how many times should the pattern be repeated. You need to escape the dots with a backslash because otherwise they will be interpreted as any character. 0 solved Looking for regex solution [closed]
As you need please use below method. it will give you proper month formated date func monthFormatechange(_ dateString : String) -> String { let dateFormatter = DateFormatter() dateFormatter.dateFormat = “dd-MM-yyyy” let date = dateFormatter.date(from:dateString) dateFormatter.dateFormat = “dd-MMM-yyyy” let actualDate = dateFormatter.string(from: date!) return actualDate } solved Date string convertion wrong month
I think this query will do what you want. It uses (YEAR(CURDATE())*12+MONTH(CURDATE())) – (YEAR(STR_TO_DATE(join_date, ‘%d-%m-%Y’))*12+MONTH(STR_TO_DATE(join_date, ‘%d-%m-%Y’))) – – 1 to get the number of whole months of experience for the user, DAY(LAST_DAY(STR_TO_DATE(join_date, ‘%d-%m-%Y’))) – DAY(STR_TO_DATE(join_date, ‘%d-%m-%Y’)) + 1 to get the number of days in the first month, and DAY(CURDATE()) to get the number of … Read more
Try this: echo date(“M, d”, strtotime($from)); 1 solved How can I change date format? [closed]
LocalDate and ThreeTenABP String dateString = “2019-11-27”; LocalDate date = LocalDate.parse(dateString); int epochDay = (int) date.toEpochDay(); System.out.println(epochDay); This snippet outputs: 18227 The documentation explains: The Epoch Day count is a simple incrementing count of days where day 0 is 1970-01-01 (ISO). So my suggestion is that this number is fine for feeding into your BarEntry … Read more
import re from datetime import datetime, timedelta dateString = “hello hello 2017-08-13, 2017-09-22” dates = re.findall(‘(\d+[-/]\d+[-/]\d+)’, dateString) for d in dates: originalDate = d newDate = datetime.strptime(d, “%Y-%m-%d”) newDate = newDate + timeDelta(days=5) newDate = datetime.strftime(newDate, “%Y-%m-%d”) dateString = dateString.replace(originalDate, newDate) Input: hello hello 2017-08-13, 2017-09-22 Output: hello hello 2017-08-18, 2017-09-27 6 solved Replace YYYY-MM-DD … Read more
Convert string in iso date format to milliseconds. Iso date format includes (“yyyy-MM-dd”) and doesn’t have timezone solved Convert string in iso date format to milliseconds. Iso date format includes (“yyyy-MM-dd”) and doesn’t have timezone
Date objects don’t have a format. Formatting can only be applied when you display them as a strings. A Date object simply stores all the info about the date (day, month, year, time, etc). When displaying that data, it is displayed as a string, and that is when it makes sense to arrange the data … Read more
Your code didn’t compile for me unless I changed the YYYY to yyyy. After that it seems to work fine. 0 solved Calendar add method error when going beyond the year
In .NET you could do use the AddMonths method: DateTime date = new DateTime(2013, 5, 19); DateTime newDate = date.AddMonths(14); As far as parsing a date from a string using a specified format you could use the TryParseExact method: string dateStr = “19/05/2013”; DateTime date; if (DateTime.TryParseExact(dateStr, “dd/MM/yyyy”, CultureInfo.InvariantCulture, DateTimeStyles.None, out date)) { // successfully … Read more