You should be using = to assign values and not : You can do like this. import datetime example_timestamp = int(‘1629617204525776950’) timestamp_to_date_time = datetime.datetime.fromtimestamp(example_timestamp/1000000000).strftime(‘%Y-%m-%d %H:%M:%S,%f’) print(timestamp_to_date_time) 2021-08-22 07:26:44,525777 solved Convert Timestamp(Nanoseconds format) to datetime [closed]
On keyup, store the current time in a hidden input. http://jsfiddle.net/trevordixon/hL8YB/ <input type=”text” name=”name” id=”name_input”> <input type=”hidden” name=”name_time” id=”name_time_input” size=”100″> <script> $(‘#name_input’).keyup(function() { var now = new Date(); $(‘#name_time_input’).val(now.toString()); }); </script> 1 solved how to get the time when exiting a textfield? [closed]
I would use date() instead, which is something I’ve just done on my website: <?php echo date(“D M d, Y G:i a”); ?> it’s explained in a little more detail here, as regards formatting. Maybe you want: <?php echo date(“G:i”); ?> the formatting is up to you, but date() is what you want 😉 4 … Read more
Consider this: var date = “11252017”; var arrDate = date.ToArray(); var strDate = arrDate[0] + arrDate[1] + “https://stackoverflow.com/” + arrDate[2] + arrDate[3] + “https://stackoverflow.com/” + arrDate[4] + arrDate[5] + arrDate[6] + arrDate[7]; // 98/25/2017 Notice that: ‘1’ + ‘1’ = 98* ⇒ char + char = int 98 + “https://stackoverflow.com/” = “98/” ⇒ int + … Read more
A possible solution is to use str.lstrip Ex: import datetime Date = datetime.datetime.today().strftime(“%d %B %Y”) print(Date.lstrip(“0”)) Output: 9 May 2018 0 solved Datetime – Strftime and Strptime
You want to use a replace technique (in whatever language/environment you are using) on your substrings by capturing like this: (\d{2})(\d{2})(\d{2}) *note the curly brackets are for improved efficiency. And replace with: $1.$2.$3 Here is a demo link. Here is a SO page discussing the execution of replacements on nano. 1 solved How to add … Read more
Here is a code that may solve your purpose.The concept is simple.Just take the time you want to validate from Calendar class and compare it with the current time and then take an appropiate action. TextView textView = (TextView) findViewById(R.id.text); Calendar cal = Calendar.getInstance(); cal.setTimeInMillis(System.currentTimeMillis()); cal.set(Calendar.HOUR_OF_DAY, 3); cal.set(Calendar.MINUTE, 00); cal.set(Calendar.SECOND, 0); long noon_start = cal.getTimeInMillis();//3.00 … Read more
You can create a timer fileprivate var timer: Timer? var seconds: Int = 0 func runTimer() { timer = Timer.scheduledTimer(timeInterval: 1, target: self, selector: (#selector(ViewController.updateTimer)), userInfo: nil, repeats: true) } func updateTimer() { seconds += 1 } Now its just a matter on when you start the timer. Options : viewDidLoad , viewWillAppear, viewDidAppear Then … Read more
You can change the timezone with the offset at the end of the date string (e.g. -05:00). And then create the date string you want using get functions. You will need to pad the month, day, hour and minutes. However, the .get (eg .getMonth) functions return numbers, so you will also have to convert them … Read more
tl;dr OffsetDateTime .parse( “Wed, 20 Feb 2019 07:14:06 +0100” , DateTimeFormatter.RFC_1123_DATE_TIME ) .toString() 2019-02-20T07:14:06+01:00 Details Your problem has nothing to do with Java 8 versus Java 17. Tip: Before blaming software that is formally specified, is thoroughly tested by enormous test suites, and is used by millions of programmers daily, suspect your own code first. … Read more
With given format, you can use dateutil.parser.parse to handle it. Here’s some code: d = “2017-10-23T03:36:23.337+02:00” time = dateutil.parser.parse(d) print(time.strftime(“%d/%m/%y %H/%M/%S”)) The output is: 23/10/17 03/36/23 4 solved user friendly date format using python
First convert your dates into a date-time class using asPOSIXct df = data.frame(x = c(“2012-03-01T00:05:55+00:00”, “2012-03-01T00:06:23+00:00”, “2012-03-01T00:06:52+00:00”)) df$times = as.POSIXct(df$x, format = “%Y-%m-%dT00:%H:%M+%S”) Then extract just the hour part using format df$hour = format(df$times, ‘%H’) This give you : x times hour 1 2012-03-01T00:05:55+00:00 2012-03-01 05:55:00 05 2 2012-03-01T00:06:23+00:00 2012-03-01 06:23:00 06 3 2012-03-01T00:06:52+00:00 2012-03-01 … Read more
That looks like a UNIX timestamp, if so it’ll be: Int64 timestamp = Convert.ToInt64(“1319556419”); DateTime newDate = new DateTime(1970,1,1).AddSeconds(timestamp); edit: It’s actually seconds instead of milliseconds by the looks of it. This gives the date of October 25, 2011. 3 solved How do I convert this string to a DateTime? [closed]
here is what would help you: date(‘Y-m-d H:i:s’, strtotime(‘Mon, 7 Apr 2014 13:46:23 +0300′)); for setting to date format refer to PHP’s Date function solved Exactly parse string to datetime format [closed]
Introduction This article provides a solution to the problem of parsing a string into a datetime format. It explains the various methods available for parsing a string into a datetime format, and provides examples of how to use each method. It also provides a comparison of the different methods, and discusses the advantages and disadvantages … Read more