The main problem with your format is that Python does not support anything more granular than microseconds, as explained here. In fact, assuming you have 000 as the last three decimals of seconds, you can use the format below. from datetime import datetime datetime_object = datetime.strptime(’24/NOV/18 05:15:00.000000000 AM’, ‘%d/%b/%y %I:%M:%S.%f000 %p’) If you cannot make … Read more
Introduction Converting a 12 hour string into a 24 hour datetime can be a tricky task. Fortunately, there are a few methods that can be used to make this conversion. In this article, we will discuss the different methods that can be used to convert a 12 hour string into a 24 hour datetime. We … Read more
$now = “2014-11-12 20:20:40”; $delayed_time = “2014-11-12 20:21:00”; $diff = strtotime($delayed_time) – strtotime($now); // $diff = 20; solved How to get the difference in seconds in PHP?
There are 1000ms in a second. 1000 / 10 is 100. Divide the number of 10 ms ticks by 100 and you’ll have the number of seconds. 1 solved How to convert 10ms ticks value into seconds using Java?
tick is a channel in Go. If you look at the docs, tick should send something to the channel once each time interval, which is specified by time.Duration(1000/config.Samplerate) * time.Millisecond in your code. <-tick just waits for that time interval to pass. i keeps track of how many seconds pass, so every time it ticks, … Read more
The average case complexity for both cases is O(n). if k is the number of times the first if fails, then the number of comparisons is 2*n – 2 – k. maxmin(a,n,max,min){ max=min=a[1]; for i=2 to n do{ // goes through the loop n-1 times if a[i]>max then max:=a[i]; // out of n-1 times succeeds … Read more
The LocalTime class represents a time of day, without any date component. That seems to be the class you want to use here. You can create LocalTime objects with LocalTime.of, and compare them with isBefore and isAfter. Like this. LocalTime sevenThirty = LocalTime.of(7,30); LocalTime eightTwenty = LocalTime.of(8,20); LocalTime nineOClock = LocalTime.of(9,0); if(eightTwenty.isAfter(sevenThirty) && eightTwenty.isBefore(nineOClock)) { … Read more
Try this: DECLARE @SUM VARCHAR(50)=’33.90′; SELECT CAST(CAST(LEFT(@SUM,CHARINDEX(‘.’,@SUM,0)-1) AS INT)+ CAST(SUBSTRING(@SUM,CHARINDEX(‘.’,@SUM,0)+1,LEN(@SUM)) AS INT)/60 AS VARCHAR(10))+’.’+ CAST(CAST(SUBSTRING(@SUM,CHARINDEX(‘.’,@SUM,0)+1,LEN(@SUM)) AS INT)%60 AS VARCHAR(10)) result: 34.30 1 solved How to convert the Varchar of sum to time using sql server 2008 [closed]
As already answered in the similar question “How can I get a range of dates in php?”, you can make use of the DatePeriod class in PHP to iterate over each 45 minute times’ between a start and end time: Demo: <?php /** * PHP add 45 Minutes to a Time with Loop * * … Read more
Here some tipps to get started: You can parse the String to a Java Date like this: SimpleDateFormat format = new SimpleDateFormat(“HH:mm:ss”); Date startTimer1Date = format.parse(StartTimer1); You can get the current Date and Time like this: Date dateNow = new Date(); And since you are working with time only (and not with date and time), … Read more
Barebones solution: long remainingMillis = countdownEnds.getTime() – System.currentTimeMillis(); long remainingMinutes = TimeUnit.MILLISECONDS.toMinutes(remainingMillis); String countdownEndsString = String.format(“%d minutes”, remainingMinutes); For a nicer solution use java.time, the modern Java date and time API, for the calculation of the minutes: long remainingMinutes = ChronoUnit.MINUTES.between( Instant.now(), DateTimeUtils.toInstant(countdownEnds)); In this case also see if you can get rid of the … Read more
You can get the date with javascript like so: var date = new Date(); var d = date.getDate(); var day = (d < 10) ? ‘0’ + d : d; var m = date.getMonth() + 1; var month = (m < 10) ? ‘0’ + m : m; var yy = date.getYear(); var year = … Read more
i): The outer loop is running n times so you touch every element n times in the outer loop. The inner loop is also running n times, so you touch every element n*n times so its O(n^2) ii) The loop is running n*n times so it simply is O(n^2) solved time complexity for given scenario … Read more
It seems like the obvious structure would be something like this: while (current_time < end_time) { current_number = *next_number++; if (meets_conditions(current_number)) output(current_number); } solved Execution Control In C [closed]
Just maintain a field in database which saves the last file upload date and time and if the uploads reach to the limit of 5 then before uploading file check that last file upload time. If the difference between last file upload time and current time is greater than 2 minutes then allow the file … Read more