date-difference

问题 I have some missing dates in a file. e.g. $cat ifile.txt 20060805 20060807 20060808 20060809 20060810 20060813 20060815 20060829 20060901 20060903 20060904 20060905 20070712 20070713 20070716 20070717 The dates are in the format YYYYMMDD. My intention is fill the missing dates in between the dates if they are missing maximum for 5 day e.g. 20060805 20060806 ---- This was missed 20060807 20060808 20060809 20060810 20060811 ----- This was missed 20060812 ----- This was missed 20060813 20060814

问题 I have some missing dates in a file. e.g. $cat ifile.txt 20060805 20060807 20060808 20060809 20060810 20060813 20060815 20060829 20060901 20060903 20060904 20060905 20070712 20070713 20070716 20070717 The dates are in the format YYYYMMDD. My intention is fill the missing dates in between the dates if they are missing maximum for 5 day e.g. 20060805 20060806 ---- This was missed 20060807 20060808 20060809 20060810 20060811 ----- This was missed 20060812 ----- This was missed 20060813 20060814

问题 I have created one timer application in javascript. Firstly it takes the current UTC date to init timer with some reference. here's the code on_timer: function(e) { var self = this; if ($(e.target).hasClass("pt_timer_start")) { var current_date = this.get_current_UTCDate(); this.project_timesheet_db.set_current_timer_activity({date: current_date}); this.start_interval(); this.initialize_timer(); this.$el.find(".pt_timer_start,.pt_timer_stop").toggleClass("o_hidden"); Now, Once timer is

问题 I need to find the difference between two dates in minutes. Here is the select statement I have been using: select date ('05.04.2017 11:12:00') - date('now'); It is returning -4 mins -21 secs but I want to see just 4 minutes. I could not find the answer in the ref guide, any idea how to show it in the format I need? 回答1: http://ariel.its.unimelb.edu.au/~yuan/Ingres/us_13229.html can you give this a shot... not familiar with ingres select INTERVAL('mins','today'-date('05.04.2017 11:12:00'))

问题 I have this situation, i want to calculate the late of an employee and to get the late, i need to Subtract time to get the minute/hour late of an employee. Suppose to be.. i have this code: $time = strtotime($time_in); $late_time = date("H:i", strtotime('+15 minutes', $time)); if(date("H:i:s",strtotime($time_duty[0]->time)) > $late_time ) { $time_difference = (date("H:i:s",strtotime($time_duty[0]->time)) - $late_time)/60; print_r($time_difference); } then i've encountered error Message: A non

问题 I have this situation, i want to calculate the late of an employee and to get the late, i need to Subtract time to get the minute/hour late of an employee. Suppose to be.. i have this code: $time = strtotime($time_in); $late_time = date("H:i", strtotime('+15 minutes', $time)); if(date("H:i:s",strtotime($time_duty[0]->time)) > $late_time ) { $time_difference = (date("H:i:s",strtotime($time_duty[0]->time)) - $late_time)/60; print_r($time_difference); } then i've encountered error Message: A non

问题 I'm using datatype time to calculate the class taken timings. For calculation I use TIMEDIFF(endtime,starttime) . Query SELECT TIMEDIFF('00:26:08','21:58:18') FROM students_session WHERE id='#' I'm not getting the proper o/p which is 02:27:50 . Instead I get -21:32:10 , which is wrong. How to rectify this? 回答1: The issue is that you know that '00:26:08' is after '21:58:18' (following morning), but MySQL is not aware, thus the result is correct from MySQL point of view. You either need to

问题 I'm using datatype time to calculate the class taken timings. For calculation I use TIMEDIFF(endtime,starttime) . Query SELECT TIMEDIFF('00:26:08','21:58:18') FROM students_session WHERE id='#' I'm not getting the proper o/p which is 02:27:50 . Instead I get -21:32:10 , which is wrong. How to rectify this? 回答1: The issue is that you know that '00:26:08' is after '21:58:18' (following morning), but MySQL is not aware, thus the result is correct from MySQL point of view. You either need to

问题 Here's a sample of my dataset: df=data.frame(id=c("9","9","9","5","5","5","4","4","4","4","4","20","20"), Date=c("11/29/2018","11/29/2018","11/29/2018","5/25/2018","2/13/2019","2/13/2019","6/7/2018", "6/15/2018","6/20/2018","8/17/2018","8/20/2018","12/25/2018","12/25/2018"), Buyer= c("John","John","John","Maria","Maria","Maria","Sandy","Sandy","Sandy","Sandy","Sandy","Paul","Paul")) I need to calculate the difference between dates which I have already done and the dataset then looks like: |

问题 I would like to create a new column in my dataset, which is a difference in years between today and a another column already in the dataset, filled up with dates. the code above: df['diff_years'] = datetime.today() - df['some_date'] df['diff_years'] give me the following output (exemple): 1754 days 11:44:28.971615 and i have to get something like (meaning the output above in years): 4,8 (or 5) I appreciate any help! PS.: i would like to avoid looping the series, path i believe would give me a