days

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: This question already has an answer here: How can I increment a date by one day in Java? 24 answers How do I add x days to a date in Java? For example, my date is (dd/mm/yyyy) = 01/01/2012 Adding 5 days, the output should be 06/01/2012 . 回答1: SimpleDateFormat sdf = new SimpleDateFormat("dd/MM/yyyy"); Calendar c = Calendar.getInstance(); c.setTime(new Date()); // Now use today date. c.add(Calendar.DATE, 5); // Adding 5 days String output = sdf.format(c.getTime()); System.out.println(output); 回答2: java.time With the Java 8 Date and Time API

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I'm using pandas and I'm wondering what's the easiest way to get the business days between a start and end date using pandas? There are a lot of posts out there regarding doing this in Python (for example ), but I would be interested to use directly pandas as I think that pandas can probably handle this quite easy. 回答1: Use BDay() to get the business days in range. from pandas.tseries.offsets import * In [185]: s Out[185]: 2011-01-01 -0.011629 2011-01-02 -0.089666 2011-01-03 -1.314430 2011-01-04 -1.867307 2011-01-05 0.779609 2011-01-06 0

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I tried to create a table with range partitioning. But it shows the following error: A primary must include all columns in the table's partitioning location. This is my SQL statement: CREATE TABLE `tbl_emp_confirmation` ( `fld_id` int(11) NOT NULL AUTO_INCREMENT, `fldemp_id` varchar(100) DEFAULT NULL, `fldempname` varchar(100) DEFAULT NULL, `fldjoindate` varchar(100) DEFAULT NULL, `fldconfirmdate` Date NOT NULL, `fldresigndate` varchar(100) DEFAULT NULL, `fldstatus` varchar(50) DEFAULT NULL, `fldcon_status` varchar(100) DEFAULT NULL, UNIQUE

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I want to use a simple countdown timer starting at 30 seconds from when the function is run and ending at 0. No milliseconds. How can it be coded? 回答1: var count=30; var counter=setInterval(timer, 1000); //1000 will run it every 1 second function timer() { count=count-1; if (count To make the code for the timer appear in a paragraph (or anywhere else on the page), just put the line: where you want the seconds to appear. Then insert the following line in your timer() function, so it looks like this: function timer() { count=count-1; if (count

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: i did some research, but still can't find how to get the days... Here is what I got: int seconds = (int) (milliseconds / 1000) % 60 ; int minutes = (int) ((milliseconds / (1000*60)) % 60); int hours = (int) ((milliseconds / (1000*60*60)) % 24); int days = ????? ; Please help, I suck at math, thank's. 回答1: If you don't have another time interval bigger than days: int days = (int) (milliseconds / (1000*60*60*24)); If you have weeks too: int days = (int) ((milliseconds / (1000*60*60*24)) % 7); int weeks = (int) (milliseconds / (1000*60*60*24*7)

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Somebody could please tell me where I could find the good ol' BDE installer? These links won't work anymore: http://info.borland.com/devsupport/bde/bdeupdate.html 回答1: http://www.jrsoftware.org/download.php/bdeinst.cab Extract the archive and run: regsvr32 BdeInst.dll 回答2: Took months for me to realize there's nothing special about the BDE installation except for copying the files and adding the branches into the registy. I use Actual Installer to build the project but these lines from the .aip file should help you understand just what to

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I saw examples for C#, Java, but for C++ i cant find solution to calculate how many days between two dates. For example between 2012-01-24 and 2013-01-08 Thanks! 回答1: This is one way. #include <iostream> #include <ctime> int main () { struct std :: tm a = { 0 , 0 , 0 , 24 , 5 , 104 }; /* June 24, 2004 */ struct std :: tm b = { 0 , 0 , 0 , 5 , 6 , 104 }; /* July 5, 2004 */ std :: time_t x = std :: mktime (& a ); std :: time_t y = std :: mktime (& b ); if ( x != ( std :: time_t )(- 1 ) && y != ( std :: time_t )(- 1 ) ) { double

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I am trying to calculate the difference between two LocalDateTime . The output needs to be of the format y years m months d days h hours m minutes s seconds . Here is what I have written: import java.time.Duration; import java.time.Instant; import java.time.LocalDateTime; import java.time.Period; import java.time.ZoneId; public class Main { static final int MINUTES_PER_HOUR = 60; static final int SECONDS_PER_MINUTE = 60; static final int SECONDS_PER_HOUR = SECONDS_PER_MINUTE * MINUTES_PER_HOUR; public static void main(String[] args) {

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have start date and end date. I need to find out the day that is Sunday or Monday etc dependent upon user click on check box. How can I find/calculate that in PHP? 回答1: You could create a function that uses strtotime() recursively to count the number of days. Since strtotime("next monday"); works just fine. function daycount($day, $startdate, $counter) { if($startdate >= time()) { return $counter; } else { return daycount($day, strtotime("next ".$day, $startdate), ++$counter); } } echo daycount("monday", strtotime("01.01.2009"), 0);

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: I changed the lifecycle for a bunch of my buckets on Amazon S3 so their storage class was set to Glacier. I did this using the online AWS Console. I now need those files again. I know how to restore them back to S3 per file. But my buckets have thousands of files. I wanted to see if there was a way to restore the entire bucket back to S3, just like there was a way to send the entire bucket to Glacier? I'm guessing there's a way to program a solution. But I wanted to see if there was a way to do it in the Console. Or with another