将浮点数转换为 NSDate [英] converting an float to NSDate

问题描述

我想将浮点数转换为 NSDate

I want to convert a float to a NSDate

我使用以下方法将 NSDate 转换为浮点数:

I converted a NSDate into a float using this:

// Turn the date into Integers 
NSCalendar *calendar= [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];  
NSCalendarUnit unitFlags = NSHourCalendarUnit | NSMinuteCalendarUnit | NSSecondCalendarUnit;  
NSDateComponents *dateComponents = [calendar components:unitFlags fromDate:nsdate_wakeTime];  
NSInteger hour = [dateComponents hour];  
NSInteger min = [dateComponents minute];  

//Convert the time in 24:60 to x.x format.
float myTime = hour + min/60; 

在我对 mytime 变量做了一些数学运算之后，我得到了许多相同浮点格式的其他时间.

after some math stuff I do on the mytime variable i get a bunch of other times in the same float format.

如何将浮点数转换为 NSDate?

How do I turn a float into a NSDate?

谢谢！

推荐答案

如果您将计算的时间转换为秒(即 mytime * 60)，那么您可以使用 dateWithTimeIntervalSinceReferenceDate:回到 NSDate.根据您正在做的数学计算，此处引用的日期似乎是相关日期的 00:00.不过正如杰森所说，可能有更好的方法来完成您想要完成的任务.

If you convert the time you've computed to seconds (so, mytime * 60), then you can use dateWithTimeIntervalSinceReferenceDate: to get back to an NSDate. From the math you are doing, it looks like the referenced date here would be 00:00 for the day in question. As Jason mentioned though, there's probably a better way to do what you are trying to accomplish.

此外，如果您确实需要分钟，则需要将myTime"计算更改为除以 60.0；您的示例代码将小于 60 的整数值除以整数值 60，该整数值始终为 0.

Also, you need to change your "myTime" computation to dividing by 60.0 if you actually want the minutes; your sample code is dividing an integer value less than 60 by the integer value 60, which will always be 0.

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