As I said in title I need to convert seconds to hh:mm:ss
I tried this:
ui->label->setText(QDateTime::fromTime_t(10).toString("hh:mm:ss"));
But default value for hours is always 01 but I need it to be 00. As result I should get 00:00:10 but I get 01:00:10.
Your timezone is included in it thats why. Try this:
QDateTime::fromTime_t(10).toUTC().toString("hh:mm:ss");
There is no QTime::fromTime_t; possibly you're using QDateTime::fromTime_t, which accounts for time zones and daylight savings.
Instead you can use QTime().addSecs(10).toString(...).
Other answers work perfect when the number of seconds stays below 24 hours. For most usecases that is good enough. However for our usecase the timer went above 24 hours and reset to 00:00:00. Do note that the following code does not work for times of 100 hours or more. The number 2 in the arg method forces the string to be 2 characters.
int totalNumberOfSeconds = 90242; // nr of seconds more than 1 day.
int seconds = totalNumberOfSeconds % 60;
int minutes = (totalNumberOfSeconds / 60) % 60;
int hours = (totalNumberOfSeconds / 60 / 60);
QString timeString = QString("%1:%2:%3")
.arg(hours, 2, 10, QChar('0'))
.arg(minutes, 2, 10, QChar('0'))
.arg(seconds, 2, 10, QChar('0'));
In Qt 5 you can use QTime::fromMSecsSinceStartOfDay(int msecs).
ui->label->setText(QTime::fromMSecsSinceStartOfDay(10 * 1000).toString("hh:mm:ss"));
Related
I have a start date and an end date. I need to find the difference between these dates and group it under the following categories.
< 1 year, < 2 year and so on till X years.
I'm trying to write a unix C++ program for this problem.
I can easily find the unix time difference between start and end date and compare with the 1 year's time stamp (12 * 30 * 20 * 60 * 60) and so on.
Is there any C++ function that returns the difference in years given the start and end date? Also let's say, the difference is 8 years, I suppose I have to write conditions like this,
if((end_date - start_date) < 12 * 30 * 24 * 60 * 60)
group = " less than 1 year"
...
...
Until what point do I stop at, as I won't know what the maximum difference is between the dates?
Is there any easy way to compute this?
I know i'm confusing here, but i ve put all my efforts to explain the problem here. Thanks in advance.
Also note, this is not a assignment or anything.
Assuming "precise years" (in other words, all years are 365 days long) is not an issue, I would do something like this (counting the number of times each year happens in this case - since the original question doesn't really say WHAT to do with each year)
const int MAX_YEARS = 10;
const int YEAR_IN_SECONDS = 365 * 24 * 60 * 60;
std::array<int, MAX_YEARS+1> bins;
int years = static_cast<int>(difftime(end_date - start_date) / YEAR_IN_SECONDS);
// Outside of range, put it at the end of range...
// We could discard or do something else in this case.
if (years > MAX_YEARS)
{
years = MAX_YEARS;
}
bins[years]++; // Seen one more of "this year".
Obviously, what you do with "bins", and what/how you store data there really depends on what you actually are trying to achieve.
An alternative solution would be to use const double YEAR_IN_SECONDS = 365.25 * 24 * 60 * 60;, which would slightly better cover for leap-years. If you want to be precise about it, you'd have to find out if you are before or after each of the leapday in a particular year that is divisible by 4 (and keep in mind that there are special cases for years divisible by 100 and other rules at 400).
#include <chrono>
using years = std::chrono::duration<std::chrono::system_clock::rep, std::ratio<365 * 24 * 60 * 60, 1>>;
std::chrono::system_clock::time_point end_date = std::chrono::system_clock::now();
std::chrono::system_clock::time_point start_date = end_date - years(2);
years how_many = std::chrono::duration_cast<years>(end_date - start_date);
int how_many_as_int = how_many.count();
std::cout << how_many_as_int << std::endl;
std::unordered_map<int, std::list<whatever>> m;
m[how_many_as_int].push_back(...);
I'm getting radar data as "tracks" and the track data indicates the number of UTC seconds since the last midnight, apparently. This is not the number of seconds since the 1st of jan 1970.
Now I want to convert that to date time, knowing that the clock on the computer could be slightly out of sync with the clock on the radar. I'll assume the radar's seconds are the reference, not the computer's.
I want to convert these seconds to a full date time. Things seem to be a little tricky around
midnight.
Any suggestions? I've got some ideas, but I don't want to miss anything.
I'm working with C++ Qt.
// Function to extend truncated time, given the wall time and period, all
// in units of seconds.
//
// Example: Suppose the truncated period was one hour, and you were
// given a truncated time of 25 minutes after the hour. Then:
//
// o Actual time of 07:40:00 results in 07:25:00 (07:40 + -15)
// o Actual time of 07:10:00 results in 07:25:00 (07:10 + +15)
// o Actual time of 07:56:00 results in 08:25:00 (07:56 + +29)
double extendTruncatedTime(double trunc, double wall, int period) {
return wall + remainder(trunc - wall, period);
}
#define extendTruncatedTime24(t) extendTruncatedTime(t, time(0), 24 * 60 * 60)
Some commentary:
The units of wall are seconds, but its base can be arbitrary. In Unix, it typically starts at 1970.
Leap seconds are not relevant here.
You need #include <math.h> for remainder().
The period in extendTruncatedTime() is almost always twenty-four hours, 24 * 60 * 60, as per the OP's request. That is, given the time of day, it extends it by adding the year, month, and day of month, based on the 'wall' time.
The only exception I know to the previous statement is, since you mention radar, is in the Asterix CAT 1 data item I001/141, where the period is 512 seconds, and for which extendTruncatedTime() as given doesn't quite work.
And there is another important case which extendTruncatedTime() doesn't cover. Suppose you are given a truncated time consisting of the day of month, hour, and minute. How can you fill in the year and the month?
The following code snippet adds the year and month to a time derived from a DDHHMM format:
time_t extendTruncatedTimeDDHHMM(time_t trunc, time_t wall) {
struct tm retval = *gmtime_r(&trunc, &retval);
struct tm now = *gmtime_r(&wall, &now);
retval.tm_year = now.tm_year;
retval.tm_mon = now.tm_mon;
retval.tm_mon += now.tm_mday - retval.tm_mday > 15; // 15 = half-month
retval.tm_mon -= now.tm_mday - retval.tm_mday < -15;
return timegm(&retval);
}
As written, this doesn't handle erroneous inputs. For example, if today is July 4th, then the non-nonsensical 310000 will be quietly converted to July 1st. (This may be a feature, not a bug.)
If you can link against another lib, i'd suggest to use boost::date_time.
It seems you want to take current date in seconds from midnight (epoch) then add the radar time to it, then convert the sum back to a date time, and transform it into a string.
Using boost will help you in:
getting the right local time
calculating the date back
incorporating the drift into the calculation
taking leap seconds into account
since you'll have concept like time intervals and durations at your disposal. You can use something like (from the boost examples):
ptime t4(date(2002,May,31), hours(20)); //4 hours b/f midnight NY time
ptime t5 = us_eastern::local_to_utc(t4);
std::cout << to_simple_string(t4) << " in New York is "
<< to_simple_string(t5) << " UTC time "
<< std::endl;
If you want to calculate the drift by hand you can do time math easily similar to constructs like this:
ptime t2 = t1 - hours(5)- minutes(4)- seconds(2)- millisec(1);
I had the exact same problem but I'm using C#. My implementation is included here if anyone needs the solution in C#. This does not incorporate any clock drift.
DateTime UTCTime = DateTime.UtcNow.Date.AddSeconds(secondSinceMidnightFromRadar);
I have a loop and in every loop I get the current seconds the application has been running for I then want to convert this time into how many, Days, Hours and Seconds that the seconds calculate to but not 'real time' I need to be able to customize how many seconds are in a day, I have tried examples on SO and the web but nothing seems to be out there for this. I have some defines
#define DAY 1200
#define HOUR DAY / 24
#define MINUTE HOUR / 60
#define SECOND MINUTE / 60
So in my define a day would last for 1200 seconds. I have then been trying to convert elapsed seconds into 'my' seconds
seconds_passed = fmodf(SECOND, (float)(GetTicks() / 1000));
Which returns what SECOND equals (0.013889) but then every loop is the same, it never changes I was thinking I would just be able to convert for example: 1real second into 1.25fake seconds then
Minute = (seconds_passed / MINUTE);
seconds_passed = fmodf(seconds_passed, MINUTE);
work out how many (fake)minutes, (fake)hours and (fake)days have elapsed since the application started.
Hope that makes sense, thank you for your time.
Since you want to customise how many seconds are in a day, all you're really doing is changing the ratio of 1 second : 1 second.
For instance, if you did was 1200 seconds in a day your ratio is:
1:72
that is, for every 1 second that passes in your day, it is the equivilent of 72 real seconds.
So yes basically all you need to do in your program is find the ratio of 1 second to 1 second, times your elapsed seconds by that to get the 'fake' seconds, and then use that value...
The code may look something like this:
// get the ratio second:fake_second
#define REAL_DAY_SECONDS 86400
int ratio = REAL_DAY_SECONDS / DAY;
fake_to_real = fake_second*ratio;
real_to_fake = real_second/ratio;
You can make your own time durations with one line in chrono:
using fake_seconds = std::chrono::duration<float, std::ratio<72,1>>;
Some sample code
#include <iostream>
#include <chrono>
using namespace std::chrono_literals;
using fake_seconds = std::chrono::duration<float, std::ratio<72,1>>;
int main()
{
auto f_x = fake_seconds(350s);
std::cout << "350 real seconds are:\n" << f_x.count() << " fake_seconds\n";
}
https://godbolt.org/z/f5G86avxr
using Qt 4.8 how can I print the time in the format DD HH SS? I have the seconds and I want to get back a string in that format.
QDateTime::fromTime_t(seconds).toString("ss hh DD");
see http://qt-project.org/doc/qt-5.0/qdatetime.html#toString
If you want a duration ( your question was really unclear) try something like :
QString seconds_to_DHMS(quint32 duration)
{
QString res;
int seconds = (int) (duration % 60);
duration /= 60;
int minutes = (int) (duration % 60);
duration /= 60;
int hours = (int) (duration % 24);
int days = (int) (duration / 24);
if((hours == 0)&&(days == 0))
return res.sprintf("%02d:%02d", minutes, seconds);
if (days == 0)
return res.sprintf("%02d:%02d:%02d", hours, minutes, seconds);
return res.sprintf("%dd%02d:%02d:%02d", days, hours, minutes, seconds);
}
Since you have the server uptime as seconds, you can use the QDateTime class.
QDateTime::fromTime_t(duration).toUTC().toString("dd hh ss");
Notice the toUTC, that's to set the beginning hour to 0. Since you will only be taking the date, hour and seconds, it doesn't really matter if the seconds are not since that date since the year won't be displayed.
You can use QDateTime::fromTime_t :
Returns a datetime whose date and time are the number of seconds that have passed since 1970-01-01T00:00:00, Coordinated Universal Time (Qt::UTC).
What you want to print is a duration of time...not a "moment" in clock time. QDateTime doesn't do much with durations except computing secsTo (and daysTo), and you pretty much have to roll your own printing.
Good news is the math isn't that hard:
Convert seconds to Days, Minutes and Seconds
Although your internationalization of words like seconds / days / years might be a nuisance. :(
The math is incredibly hard. Days are not 24 hours, they are usually 24 hours but sometimes 23 or 25 (daylight savings time changes) or 24 hours and a second or two (leap seconds). The same problem goes for months (obviously, since differently-sized months are common) years (leap day) and really anything that inherits day's problem by being defined in terms of days (weeks).
I have some number of miles and a speed in MPH that I have converted into the number of hours it takes to travel that distance at that speed. Now I need to convert this decimal number into hours, minutes, and seconds. How do I do this? My best guess right now is:
double time = distance / speed;
int hours = time; // double to integer conversion chops off decimal
int minutes = (time - hours) * 60;
int seconds = (((time - hours) * 60) - minutes) * 60;
Is this right? Is there a better way to do this? Thanks!
I don't know c++ functions off the top of my head, however this "psuedocode" should work.
double time = distance / speed;
int hours = time;
double minutesRemainder = (time - hours) * 60;
int minutes = minutesRemainder;
double secondsRemainder = (minutesRemainder - minutes) * 60;
int seconds = secondsRemainder;
Corrected not needing floor.
As far as the comment about it not working for negative times, you can't have a negative distance in physics. I'd say that would be user input error, not coder error!
I don't know if this is better...actually I don't know for sure that it is right as I haven't tested it, but I would first convert the hours to the total number of seconds, then convert that back into hours/minutes/seconds. It would look something like:
int totalseconds = time * 3600.0;
// divide by number of seconds in an hour, then round down by casting to an integer.
int hours = totalseconds/3600;
// divide by 60 to get minutes, then mod by 60 to get the number minutes that aren't full hours
int minutes = (totalseconds/60) % 60;
// use mod 60 to to get number of seconds that aren't full minutes
int seconds = totalseconds % 60;
I'd say you've got it right. :-)
Though I should add, if a method like that Aequitarium Custos has presented is more readable or preferable to you, by all means use that method. Sometimes it's easier to calculate data elements one at a time, and calculate the next one from the one you just calculated, rather than using an absolute formula always beginning from your first datum.
In the end, as long as your math is correct (and I think it is), it's up to you how you want to write the code.
a. Check if speed is not 0
b. it is bad programming to put a double into an int IMHO.
use floor (assuming time is positive...)
c. if speed and distance are int - the result in time will be wrong...
d. #Aequitarum Custos got it right after the edit...