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: How can I calculate the number of work days between two dates from table (from the 1st row to the end) in SQL Server 2008? I tried something like this, but it does not work DECLARE @StartDate as DATETIME, @EndDate as DATETIME Select @StartDate = date2 from testtable ; select @EndDate = date1 from testtable ; SELECT (DATEDIFF(dd, @StartDate, @EndDate) + 1) -(DATEDIFF(wk, @StartDate, @EndDate) * 2) -(CASE WHEN DATENAME(dw, @StartDate) = 'Sunday' THEN 1 ELSE 0 END) -(CASE WHEN DATENAME(dw, @EndDate) = 'Saturday' THEN 1 ELSE 0 END) 回答1: I would

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: JSLint Validation error "combine this with the previous var statement" How do I combine this so I don't get JSLint Validation error? I get the validation error on the lines of code in getClassName function. $(document).ready(function () { 'use strict'; // This function is used to calculate the date function dateString(dateToDisplay) { var monthNames = ['January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December'], dayNames = ['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday',

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 is my first time using netCDF and I'm trying to wrap my head around working with it. I have multiple version 3 netcdf files (NOAA NARR air.2m daily averages for an entire year). Each file spans a year between 1979 - 2012. They are 349 x 277 grids with approximately a 32km resolution. Data was downloaded from here . The dimension is time (hours since 1/1/1800) and my variable of interest is air. I need to calculate accumulated days with a temperature Day 1 = +4 degrees, accumulated days = 0 Day 2 = -1 degrees, accumulated days = 1 Day 3

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: Can somebody give good logic for set repeat days of week alarm? I have done weekly Alarm by using alarmCalendar.set(Calendar.HOUR, AlarmHrsInInt); alarmCalendar.set(Calendar.MINUTE, AlarmMinsInInt); alarmCalendar.set(Calendar.SECOND, 0); alarmCalendar.set(Calendar.AM_PM, amorpm); Long alarmTime = alarmCalendar.getTimeInMillis(); Intent intent = new Intent(Alarm.this, AlarmReciever.class); intent.putExtra("keyValue", key); PendingIntent pi = PendingIntent.getBroadcast(Alarm.this, key, intent, PendingIntent.FLAG_UPDATE_CURRENT); am

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: Say part of my dataframe df[(df['person_num'] == 1) | (df['person_num'] == 2) ] looks like this: person_num Days IS_TRUE 1 1 1 1 4 1 1 5 0 1 9 1 2 1 1 2 4 1 2 5 0 2 9 1 And for each person_num , I want to count something like " how many IS_TRUE=1 happens within seven days before a certain day ". So for Day 9, I count the number of IS_TRUE=1 s from Day 2 to Day 8, and add the count to a new column IS_TRUE_7day_WINDOW . The result would be: person_num Days IS_TRUE IS_TRUE_7day_WINDOW 1 1 1 0 1 4 1 1 1 5 0 2 1 9 1 1 2 1 1 0 2 4 1 1 2 5 0 2 2 9

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 using pandas/python and I have two date time series s1 and s2, that have been generated using the 'to_datetime' function on a field of the df containing dates/times. When I subtract s1 from s2 s3 = s2 - s1 I get a series, s3, of type timedelta64[ns] 0 385 days, 04:10:36 1 57 days, 22:54:00 2 642 days, 21:15:23 3 615 days, 00:55:44 4 160 days, 22:13:35 5 196 days, 23:06:49 6 23 days, 22:57:17 7 2 days, 22:17:31 8 622 days, 01:29:25 9 79 days, 20:15:14 10 23 days, 22:46:51 11 268 days, 19:23:04 12 NaT 13 NaT 14 583 days, 03:40:39 How do I

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've been searching for 4 hours now, and have not found a solution to get the difference between two dates in years, months, and days in JavaScript, like: 10th of April 2010 was 3 years, x month and y days ago. There are lots of solutions, but they only offer the difference in the format of either days OR months OR years, or they are not correct (meaning not taking care of actual number of days in a month or leap years, etc). Is it really that difficult to do that? I've had a look at: http://momentjs.com/ -> can only output the difference in

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: From October 1st to March 31 the fee is $1 (season 1). From April 1st to September 30 the fee is $2 (season 2). How can I calculate the total fee of a given date range (user input) depending on how many days of this date range fall into season 1 and season 2? $user_input_start_date = getdate( $a ); $user_input_end_date = getdate( $b ); $start_date_new = mktime( 12, 0, 0, $user_input_start_date['mon'], $user_input_start_date['mday'], $user_input_start_date['year'] ); $end_date_new = mktime( 12, 0, 0, $user_input_end_date['mon'], $user_input

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: StackOverflow to the rescue!, I need to find the medians for five columns at once, in one query call. The median calculations below work for single columns, but when combined, multiple uses of "rownum" throws the query off. How can I update this to work for multiple columns? THANK YOU. It's to create a web tool where nonprofits can compare their financial metrics to user-defined peer groups. SELECT t1_wages.totalwages_pctoftotexp AS median_totalwages_pctoftotexp FROM ( SELECT @rownum := @rownum +1 AS `row_number` , d_wages.totalwages

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: In MySQL, how would I get a timestamp from, say 30 days ago? Something like: select now() - 30 The result should return a timestamp. 回答1: DATE_SUB will do part of it depending on what you want mysql> SELECT DATE_SUB(NOW(), INTERVAL 30 day); 2009-06-07 21:55:09 mysql> SELECT TIMESTAMP(DATE_SUB(NOW(), INTERVAL 30 day)); 2009-06-07 21:55:09 mysql> SELECT UNIX_TIMESTAMP(DATE_SUB(NOW(), INTERVAL 30 day)); 1244433347 回答2: I think you are after DATE_SUB . 回答3: You could use: SELECT unix_timestamp(now()) - unix_timestamp(maketime(_,_,_)); For unix