time

问题 I need to store and compare dates (without times) in my app, without caring about time zones. I can see three solutions to this: (date1.year == date2.year && date1.month == date2.month && date1.day == date2.day) This is what I'm doing now, but it's horrible verbose. date1.format("YYYYMMDD") == date2.format("YYYYMMDD") This is still rather verbose (though not as bad), but just seems inefficient to me... Create a new Date class myself, perhaps storing the date as a "YYYYMMDD" string, or number

问题 I'm getting mad with the fact that a client could change cookie values, because I realized that everything on my website was pretty unsecure. Before I just setted a cookie with an expire time, and I was sure that a given user would have kept a particular property (such as "logged in" or "he has this privilege") for that exact amount of time (unless he cleared the cache). Now that I have to switch everything to sessions, which doesn't have an expire time. So, while for the login I implemented

问题 i am using the following to set the date/time to something more readable: set day=%date:~4,2% set mth=%date:~7,2% set yr=%date:~10,4% set hur=%time:~0,2% set min=%time:~3,2% set bdate=[%day%-%mth%-%yr%]-[%hur%-%min%] this works well and outputs something like : [02-06-2020]-[22-59] How ever after midnight the time format changes from HH:MM:SS:MS to H:MM:SS:MS and messes up the format and when it includes the : in the time, since the characters are shifted over by one, when i make the log file

问题 i am using the following to set the date/time to something more readable: set day=%date:~4,2% set mth=%date:~7,2% set yr=%date:~10,4% set hur=%time:~0,2% set min=%time:~3,2% set bdate=[%day%-%mth%-%yr%]-[%hur%-%min%] this works well and outputs something like : [02-06-2020]-[22-59] How ever after midnight the time format changes from HH:MM:SS:MS to H:MM:SS:MS and messes up the format and when it includes the : in the time, since the characters are shifted over by one, when i make the log file

问题 I found that this gives a wrong date. but how i can not solve it. please someone help me. I am new in android Development. Thanks in advance; String timestamp = "1538970640"; SimpleDateFormat formatter = new SimpleDateFormat("dd MMM 'at' hh:mm a z" ); String dateString = formatter.format(new Date(Long.parseLong(timestamp))); This returns: 19 Jan at 01:29 AM GMT+06:oo But it should be: 8 Oct at 9:50 AM GMT+06:00 回答1: The java.util.Date constructor accepts milliseconds since the Epoch, not

问题 I found that this gives a wrong date. but how i can not solve it. please someone help me. I am new in android Development. Thanks in advance; String timestamp = "1538970640"; SimpleDateFormat formatter = new SimpleDateFormat("dd MMM 'at' hh:mm a z" ); String dateString = formatter.format(new Date(Long.parseLong(timestamp))); This returns: 19 Jan at 01:29 AM GMT+06:oo But it should be: 8 Oct at 9:50 AM GMT+06:00 回答1: The java.util.Date constructor accepts milliseconds since the Epoch, not

问题 I'd like a rule to fire when it's 8am OR the sun is up. How to go about this? rule X when ( "it's 8am" and ... and ...) or (Sun( up ) and ... and ...) then // do something end A timer acts like a prerequisite. So I guess it's not useful in this case. An extra time fact would have to be updated every second, which would cause the rule to refire every second. I guess I could split the time fact into hour , minute , and second facts, but that wouldn't really solve the problem, but only make it

问题 I'd like a rule to fire when it's 8am OR the sun is up. How to go about this? rule X when ( "it's 8am" and ... and ...) or (Sun( up ) and ... and ...) then // do something end A timer acts like a prerequisite. So I guess it's not useful in this case. An extra time fact would have to be updated every second, which would cause the rule to refire every second. I guess I could split the time fact into hour , minute , and second facts, but that wouldn't really solve the problem, but only make it

问题 The thing is, I have to somehow get current time of day in milliseconds in convenient format. Example of desired output: 21 h 04 min 12 s 512 ms I know how to get this format in seconds, but I have no idea how to get my hands on milliseconds? 回答1: Using the portable std::chrono auto now = std::chrono::system_clock::now(); auto time = std::chrono::system_clock::to_time_t(now); auto ms = std::chrono::duration_cast<std::chrono::milliseconds>(now.time_since_epoch()) - std::chrono::duration_cast

问题 How can I format the time elapsed from seconds to hours, mins, seconds? My code: start = time.time() ... do something elapsed = (time.time() - start) Actual Output: 0.232999801636 Desired/Expected output: 00:00:00.23 回答1: If you want to include times like 0.232999801636 as in your input: import time start = time.time() end = time.time() hours, rem = divmod(end-start, 3600) minutes, seconds = divmod(rem, 60) print("{:0>2}:{:0>2}:{:05.2f}".format(int(hours),int(minutes),seconds)) Example: In