I have a birthdate stored as QDate and I would like to kown how many years the person has.
I've tried the daysTo() function but then I cannot convert days in years.
How can I do?
You might want to do this:
int age(const QDate &birthday)
{
const auto today = QDate::currentDate();
auto age = today.year() - birthday.year();
return today.month() >= birthday.month() && today.day() >= birthday.day() ? age : age - 1;
}
Related
TimeOfDay documentation has no comparison operator and primitive comparison does not work. My only solution that I can thinking of right now is to convert TimeOfDay to DateTime and use DateTime's difference method.
Does anyone have a better solution?
Convert it to a double then compare.
double toDouble(TimeOfDay myTime) => myTime.hour + myTime.minute/60.0
extension TimeOfDayExtension on TimeOfDay {
int compareTo(TimeOfDay other) {
if (hour < other.hour) return -1;
if (hour > other.hour) return 1;
if (minute < other.minute) return -1;
if (minute > other.minute) return 1;
return 0;
}
}
Thanks from #Lucas idea, you can calculate hour and minute by
TimeOfDay yourTime ;
TimOfDay nowTime = TimeOfDay.now()
double _doubleYourTime = yourTime.hour.toDouble() +
(yourTime.minute.toDouble() / 60);
double _doubleNowTime = nowTime.hour.toDouble() +
(nowTime.minute.toDouble() / 60);
double _timeDiff = _doubleYourTime - _doubleNowTime;
double _hr = _timeDiff.truncate();
double _minute = (_timeDiff - _timeDiff.truncate()) * 60;
print('Here your Happy $_hr Hour and also $_minute min');
I calculated the difference by turning both values into minute-counts, and comparing those :)
TimeOfDay now = TimeOfDay.now();
int nowInMinutes = now.hour * 60 + now.minute;
TimeOfDay testDate = TimeOfDay(hour: 2, minute: 20);
int testDateInMinutes = testDate.hour * 60 + testDate.minute;
You can use this method. Where you have to provide starttime and endTime in TimesofDay format.
getTime(startTime, endTime) {
bool result = false;
int startTimeInt = (startTime.hour * 60 + startTime.minute) * 60;
int EndTimeInt = (endTime.hour * 60 + endTime.minute) * 60;
int dif = EndTimeInt - startTimeInt;
if (EndTimeInt > startTimeInt) {
result = true;
} else {
result = false;
}
return result;
}
LikeThis
getTime(v1, v2);
TimeOfDay n = TimeOfDay.now();
int nowSec = (n.hour * 60 + n.minute) * 60;
int veiSec = (t.hour * 60 + t.minute) * 60;
int dif = veiSec - nowSec;
Card(
child: ListTile(
onTap: () {
showTimePicker(
context: context,
initialTime: TimeOfDay.now(),
).then((TimeOfDay time) {
double _doubleyourTime =
time.hour.toDouble() + (time.minute.toDouble() /60);
double _doubleNowTime = TimeOfDay.now().hour.toDouble() +
(TimeOfDay.now().minute.toDouble() / 60);`enter code here`
if (_doubleyourTime > _doubleNowTime) {
print('correct format')
});
} else {
print('Sorry You can not set the time')
}
});
},
//dense: true,
leading: Icon(Icons.timer),
title: Text(
'Today On Time',`enter code here`
),
),
),
We can actually use the subtract operator.
Code to make magic happen :
Here I wanted to get the difference in time after the user selects the time (in TimeOfDay format) using showTimePicker()
// current time will be used to find the difference between the time selected by the user.
TimeOfDay _cur_time = TimeOfDay(hour: DateTime.now().hour, minute: DateTime.now().minute);
// scheduled time will be updated as soon as the user inputs a new time using the showTimePicker() function available in Flutter Material librabry.
TimeOfDay _scheduled_time = TimeOfDay(hour: DateTime.now().hour, minute: DateTime.now().minute);
// toDouble Function to convert time to double so that we can compare time and check that the time selected is greater than current time.
double toDouble(TimeOfDay myTime) => myTime.hour + myTime.minute / 60.0;
void _selectTime() async {
Flutter Material widget to select time.
final TimeOfDay? newTime = await showTimePicker(
context: context,
initialTime: _scheduled_time,
);
//Check if the selected time is greater than the cur time
if (toDouble(newTime!) > toDouble(_cur_time)) {
setState(() {
_scheduled_time = newTime;
});
}
}
Function to get the difference between cur time and selected time.
Duration _getDelayedDuration(){
var hourDelay = _scheduled_time.hour - _cur_time.hour;
print(hourDelay);
var minuteDelay = _scheduled_time.minute - _cur_time.minute;
print(minuteDelay);
return Duration(hours: hourDelay, minutes: minuteDelay);
}
Solution for negative duration calculations
All these answers are pretty good but they didn't help me when I had a user select a range of times in my app. In order to calculate the total duration of the specified time period, I tried all the solutions which work pretty well but fail in certain scenarios. The scenarios are:
When the start time comes after the end time (i.e- when the duration is supposed to be over 12 hours)
When the start time is before 12am at the night and end time is after that
And the code to overcome it is:
String durationFromTimeOfDay(TimeOfDay? start, TimeOfDay? end) {
if (start == null || end == null) return '';
// DateTime(year, month, day, hour, minute)
final startDT = DateTime(9, 9, 9, start.hour, start.minute);
final endDT = DateTime(9, 9, 10, end.hour, end.minute);
final range = DateTimeRange(start: startDT, end: endDT);
final hours = range.duration.inHours % 24;
final minutes = range.duration.inMinutes % 60;
final _onlyHours = minutes == 0;
final _onlyMinutes = hours == 0;
final hourText = _onlyMinutes
? ''
: '$hours${_onlyHours ? hours > 1 ? ' hours' : ' hour' : 'h'}';
final minutesText = _onlyHours
? ''
: '$minutes${_onlyMinutes ? minutes > 1 ? ' mins' : ' min' : 'm'}';
return hourText + minutesText;
}
It is important to note that you need to prefill the DateTime for end TimeOfDay with a day value which is greater than the same in start DateTime. The other parameters (for year and month) can be anything you want.
This outputs a really nicely formatted string that is short, concise, and extremely legible
This, however, doesn't satisfy the requirement that the solution is devoid of conversion to DateTime. But at least it uses a different approach over the difference method. And this makes the correct duration calculation more reliable in a few lines of code comparatively.
I don't think this is possible. You can use .subtract in DateTime as also .difference
I have QTableView with data generated from .csv file. The QTableView has column called Date (QDateTime) that covers over a month. I want to select a date and split the data for that date from the QTableView, so I can summarize and average the values in the other columns.
Similar to SQL Table Queries, or to this c# LINQ syntax:
var result = context.data.Where(n => n.date == date);
You can iterate over your table view data and extract the information you need. As a demonstration please look at the given example:
void findDates(QTableView *table, const QDateTime &date)
{
auto model = table->model();
const int dateColumn = 1; // The column with the dates data
for (int row = 0; row < model->rowCount(); ++row)
{
auto idx = model->index(row, dateColumn);
auto data = idx.data();
auto d = data.toDateTime();
// Comparison of dates
if (d > date)
{
// Do something
}
else
{
// Do something else
}
}
}
I need a method which would give me the number of hours and minutes from any random number.For example if the random number is 500 i would like to have information as 5 hrs and 0 minutes.Another example is 2359 is 23 hrs 59 minutes.The random number is entered by the user and i am only concerned with the hours and minutes(not seconds).I need not even worry about rounding off of minutes .So i wrote this method ,which to me is not efficient.Can any one suggest a better way of doing it ?or is this good enough?
private void calculateDateTime(int someNumber){
if(someNumber<=0){
return;
}
String number = Integer.toString(someNumber);
String hrs ="";
String mins ="00";
if(number.length()>4){
hrs =number.substring(0, 2);
mins = number.substring(2,4);
}
else{
float f =((float)someNumber)/100;
String s = Float.toString(f);
String [] splitArray = s.split("\\.");
if(splitArray.length>1) {
hrs = splitArray[0];
mins = splitArray[1];
}
}
int hr = Integer.valueOf(hrs);
int min = Integer.valueOf(mins);
if(hr>=24||min>=60){
return;
}
Thats how i am getting the hr and mins respectively.Suggest me a better approach if you have one.
Thanks
Pad the input string with zeros until length is 4 and then spilt the string in the middle?
Maybe:
int hours = someNumber / 100;
int minutes = someNumber % 100;
if(hours >= 24 || minutes >= 60) {
//Do whatever
}
If you want your values to be repeatable, I would use the passed in value as the seed into a Random generator.
Random r = new Random(someNumber);
int hr = r.nextInt(24);
int min = r.nextInt(60);
I want to find the day of the week of a particular date in Qt.
e.g.: 1/05/2010 is Sunday.
Is it possible to find the weekday using date?
QDate date;
date.setDate(2010,5,1);
int day = date.dayOfWeek();
QString weekDay = QDate::longDayName(day);
This isn't tested. But hope it will work. Check it out and let know.
int QDate::dayOfWeek () const
Returns the weekday (1 to 7) for this date.
For example,
QDate date;
date.setDate(2010, 5, 1);
switch (date.dayOfWeek()) {
case 1:
// Monday
break;
// etc...
}
I think you want the QDate class and the dayOfWeek function.
int QDate::dayOfWeek () const
Returns the weekday (1 to 7) for this date.
So, I am trying to calculate the time between two dates that fits certain criteria (here: work / non-work) and I'm confused about the results as I can't find out why it's wrong.
But first, some code;
**Input Date A:** 2009-01-01 2:00 pm
**Input Date B:** 2009-01-02 9:00 am
So, as you can see, the total timespan (calculated e.g. by DateB.Substract(DateA)) is 19 hours.
I now want to calculate how many hours in this timespan are "non-work" hours, based on an average work day from 8am to 5pm - Result should be 15 hours (So, 19 - 15 = 4 hours total work time) (2:00 pm to 5:00 pm plus 8:00 am to 9:00 am)
But, following my code
DateTime _Temp = new DateTime(2009, 1, 1, 14, 0, 0);
DateTime _End = new DateTime(2009, 1, 2, 9, 0, 0);
int _WorkDayStart = 8;
int _WorkDayEnd = 17;
int _Return = 0;
while (_End > _Temp)
{
if (_Temp.Hour <= _WorkDayStart || _Temp.Hour >= _WorkDayEnd)
_Return++;
_Temp = _Temp.AddHours(1);
}
the result is 16 hours (19 - 16 = 3 hours total work time) - I don't see where the mistake is, so there is one hour missing?! I refactored it on paper and it should work as intended... but doesn't :-/
Anyone sees the mistake?
You're counting both ends as non-work, instead of just one. This line:
if (_Temp.Hour <= _WorkDayStart || _Temp.Hour >= _WorkDayEnd)
should probably be:
if (_Temp.Hour < _WorkDayStart || _Temp.Hour >= _WorkDayEnd)
You're effectively stepping through "start hours". So 8am itself should count as a work hour, because it's the start of a work hour, but 5pm won't because it's the start of a non-work hour.
I would also strongly advise you to either put braces around the body of the if statement or at least indent the following line. (I'd further advise you to use camelCase for local variables, but that's just a convention thing - I've never seen C# code written with that convention for local variables before now. You may want to read Microsoft's naming conventions document - it doesn't specify local variables, but they;re generally in camel case.)
Finally, I personally find it easier to read conditions where the "thing that's changing" is on the left - so I'd change your loop condition to:
while (_Temp < _End)
An alternative is to change it into a for loop. With all of these changes, the code would be:
DateTime start = new DateTime(2009, 1, 1, 14, 0, 0);
DateTime end = new DateTime(2009, 1, 2, 9, 0, 0);
int workDayStart = 8;
int workDayEnd = 17;
int nonWorkHours = 0;
for (DateTime time = start; time < end; time = time.AddHours(1))
{
if (time.Hour < workDayStart || time.Hour >= workDayEnd)
{
nonWorkHours++;
}
}
Finally, extract this into a method:
public static int CountNonWorkHours(DateTime start, DateTime end,
int workDayStart, int workDayEnd)
{
int nonWorkHours = 0;
for (DateTime time = start; time < end; time = time.AddHours(1))
{
if (time.Hour < workDayStart || time.Hour >= workDayEnd)
{
nonWorkHours++;
}
}
return nonWorkHours;
}
EDIT: Regarding konamiman's suggestion... yes, looping over each hour is very inefficient. However, it's relatively easy to get right. Unless you're going to be doing this a lot with long time periods, I'd use this fairly simple code. It would be very easy to end up with off-by-one errors in various situations if you tried to do a per-day version. While I don't like inefficient code, I don't mind it if it's not hurting me :)
If you plan to reuse this code, I would refactor it to avoid the loop. You could just multiply the number of whole days by the labour hours per day, then treat the first and the last day of the interval as special cases.
You could also use this to avoid a loop
DateTime startDate = new DateTime(2009, 1, 1, 14, 0, 0);
DateTime endDate = new DateTime(2009, 1, 2, 9, 0, 0);
int startTime = 8;
int endTime = 17;
int ret = ((endDate.Subtract(startDate).Days - 1) * 8)
+ (startDate.Hour >= startTime && startDate.Hour < endTime ? endTime - startDate.Hour : 0)
+ (endDate.Hour > startTime && endDate.Hour <= endTime ? endDate.Hour - startTime : 0);