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: Get week of year in JavaScript like in PHP 15 answers I'm looking for a tested solid solution for getting current week of the year for specified date. All I can find are the ones that doesn't take in account leap years or just plain wrong. Does anyone have this type of stuff? Or even better a function that says how many weeks does month occupy. It is usually 5, but can be 4 (feb) or 6 (1st is sunday and month has 30-31 days in it) ================= UPDATE: Still not sure about getting week #, but

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 a table in MySQL. What would be the sql statement look like to add say 2 days to the current date value in the table? UPDATE classes SET date = date + 1 where id = 161 this adds one second to the value, i don't want to update the time, i want to add two days? 回答1: Assuming your field is a date type (or similar): SELECT DATE_ADD(`your_field_name`, INTERVAL 2 DAY) FROM `table_name`; With the example you've provided it could look like this: UPDATE classes SET `date` = DATE_ADD(`date` , INTERVAL 2 DAY) WHERE `id` = 161; 回答 2 : UPDATE

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: Given a table with schema like this: id1 id2 day number How can I turn this: a b day1 2 a b day5 4 a b day9 8 c d day2 1 c d day3 2 c d day5 4 c d day8 3 Into this?: a b day1 2 a b day2 2 a b day3 2 a b day4 2 a b day5 4 a b day6 4 a b day7 4 a b day8 4 a b day9 8 c d day2 1 c d day3 2 c d day4 2 c d day5 4 c d day6 4 c d day7 4 c d day8 3 To clarify, for each group of id1 and id2, I need to fill in the missing rows with dates ranging from the minimum date for that grouping to the maximum date. Furthermore, the rows that get filled in must

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 a MultipleChoiceField with a CheckboxSelectMutliple widget: weight_training_days = forms.MultipleChoiceField( help_text=u'(Required) 3 days must be selected', widget=forms.CheckboxSelectMultiple(attrs={ 'inline': True, }), choices=( (0, "Mon"), (1, "Tue"), (2, "Wed"), (3, "Thu"), (4, "Fri"), (5, "Sat"), (6, "Sun"), ), ) What I'm trying to is dynamically set 3 of the 7 checkboxes to "True". Ideally I would do this from the view. def change_challenge_settings_page(request): c = Challenge.objects.get(user__exact = request.user,chal

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 quite a beginner when it comes to Oracle. I am having trouble figuring out how to do something similar to this : SELECT ID, NAME, TO_CHAR(DATEBIRTH, 'DD/MM/YYYY HH24:MI:SS') FROM PEOPLE WHERE DATEBIRTH >= ANOTHERDATE - NDAY To put it short, I want to select everyone who were born N days before a specific date and time but I am not quite sure that this is the way to do it nor that it would give me the results I expect. PS: I am developping under oracle8i. 回答1: Your query looks correct to me. That's how you subtract days from dates 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: A="2002-20-10" B="2003-22-11" How to find the difference in days between two dates? 回答1: If you have GNU date , it allows to print the representation of an arbitrary date ( -d option). In this case convert the dates to seconds since EPOCH, subtract and divide by 24*3600. Or you need a portable way? 回答2: The bash way - convert the dates into %y%m%d format and then you can do this straight from the command line: echo $(( ($(date --date="031122" +%s) - $(date --date="021020" +%s) )/(60*60*24) )) 回答3: And in python $python -c "from datetime

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 would like to have a custom agenda view in my org agenda, that shows only weekend days. For example, when I open [m]onth view (M-x org-agenda a v m), I see the current month. I would like to hide the work days and show weekend days only. How can I do that? 回答1: Here is an interactive function that shows the agenda for the current or upcoming weekend: (defun org-next-weekend-agenda () "Produce an agenda view for the current or upcoming weekend from all files in variable `org-agenda-files'." (interactive) (let* ((day (string-to-number

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 about to create a solution that will list events that will occur in specific days, and was looking for a Calendar control that I can deal with to make only the days with events clickable, and of course receive this click event and handle the rest myself in my Controller. (something like the old asp:Calendar server-side control in Webforms). is there any that match this scenario? Update: What I'm exactly looking for is a Mini Calendar, not a full Calendar like the one in Outlook. this is what I'm exactly looking for: 回答1: jQuery UI can do

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: What's the best way to get the length of time represented by an interval in lubridate , in specified units? All I can figure out is something like the following messy thing: > ival [1] 2011-01-01 03:00:46 -- 2011-10-21 18:33:44 > difftime(attr(ival, "start") + as.numeric(ival), attr(ival, "start"), 'days') Time difference of 293.6479 days (I also added this as a feature request at https://github.com/hadley/lubridate/issues/105 , under the assumption that there's no better way available - but maybe someone here knows of one.) Update -

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 java, I want to get the number of days between two dates, excluding those two dates. For example: If first date = 11 November 2011 and the second date = 13 November 2011 then it should be 1 . This is the code I am using but doesn't work ( secondDate and firstDate are Calendar objects): long diff=secondDate.getTimeInMillis()-firstDate.getTimeInMillis(); float day_count=(float)diff / (24 * 60 * 60 * 1000); daysCount.setText((int)day_count+""); I even tried rounding the results but that didn't help. How do I get the number of days between