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如何从给定的数字中获取星期几

分类: 技术问答 • 2022-06-02 17:29:20
问题描述：

我想要给定数字的星期几，这是伪代码:

I want to have the day of week name for a given number, here is the pseudo-code :

getDayStringForInt:0 = sunday
getDayStringForInt:1 = monday
getDayStringForInt:2 = tuesday
getDayStringForInt:3 = wenesday
getDayStringForInt:4 = thursday
getDayStringForInt:5 = friday
getDayStringForInt:6 = saturday

我已经尝试使用以下代码，但有些东西不起作用......

I have tried with the follow code, but some thing is not working ...

- (void) setPeriodicityDayOfWeek:(NSNumber *)dayOfWeek{
    gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
    dateFormatter = [[NSDateFormatter alloc] init];
    NSLocale *frLocale = [[NSLocale alloc] initWithLocaleIdentifier:@"en_US"];
    [dateFormatter setLocale:frLocale];
    [gregorian setLocale:frLocale];
    NSDate *today = [NSDate date];
    NSDateComponents *nowComponents = [gregorian components:NSYearCalendarUnit | NSWeekCalendarUnit | NSHourCalendarUnit | NSMinuteCalendarUnit | NSSecondCalendarUnit fromDate:today];

    [nowComponents setWeekday:dayOfWeek];

    NSDate *alertDate = [gregorian dateFromComponents:nowComponents];

    [dateFormatter setDateFormat:@"EEEE"];
    NSLog(@"Day Of Week : %@ - Periodicity : %@", dayOfWeek, [dateFormatter stringFromDate:alertDate]);
    alert.periodicity = [dateFormatter stringFromDate:alertDate];

}

我的日志很奇怪:

Day Of Week : 0 - Periodicity : monday
Day Of Week : 1 - Periodicity : wenesday
Day Of Week : 2 - Periodicity : friday
Day Of Week : 3 - Periodicity : friday
Day Of Week : 4 - Periodicity : tuesday
Day Of Week : 5 - Periodicity : sunday
Day Of Week : 6 - Periodicity : sunday

有什么想法吗?任何更好的解决方案...

Any idea ? any better solution ...

答

既然这已成为公认的答案，我也会在这里发布正确"的解决方案.归功于 Rob 的回答.

Since this has become the accepted answer, I'll post the "right" solution here too. Credits to Rob's answer.

使用NSDateFormatter的[shortWeekdaySymbols][1]方法可以简单地完成整个事情，所以完整的解决方案归结为

The whole thing can simply be achieved using the [shortWeekdaySymbols][1] method of NSDateFormatter, so the full solution boils down to

- (NSString *)stringFromWeekday:(NSInteger)weekday {
    NSDateFormatter * dateFormatter = [NSDateFormatter new];
    dateFormatter.locale = [[NSLocale alloc] initWithLocaleIdentifier:@"en_US"];
    return dateFormatter.shortWeekdaySymbols[weekday];
}

原答案

请注意，您正在将指向 NSNumber 的指针传递给需要 NSInteger 的方法.编译器不会警告您，因为指针确实是一个整数，只是不是您所期望的.考虑这个简单的测试:

Original answer

Beware, you're passing a pointer to NSNumber to a method that requires a NSInteger. The compiler is not warning you since a pointer is indeed an integer, just not the one you would expect. Consider this simple test:

- (void)foo:(NSInteger)a {
    NSLog(@"%i", a);
}

- (void)yourMethod {
    [self foo:@1]; // @1 is the boxed expression for [NSNumber numberWithInt:1]
}

这会打印类似于 185035664 的内容，这是指针值，即当转换为 NSInteger 时为 NSNumber *.

This prints something like 185035664, which is the pointer value, i.e. NSNumber * when cast to NSInteger.

您应该使用 [dayOfWeek integerValue] 或直接将 dayOfWeek 转换为方法签名中的 NSInteger.

You should either use [dayOfWeek integerValue] or directly turn dayOfWeek into a NSInteger in your method signature.

此外，我认为您还有其他错误:来自 setWeekday:

Also I think you're getting something else wrong: from the doc of setWeekday:

设置接收器的工作日单位数.工作日单位是数字 1 到 n，其中 n 是一周中的天数.例如，在公历中，n 为 7，星期日为用 1 表示.

Sets the number of weekday units for the receiver. Weekday units are the numbers 1 through n, where n is the number of days in the week. For example, in the Gregorian calendar, n is 7 and Sunday is represented by 1.

星期日是 1，所以你最好也检查一下与你的代表的对应关系.

Sunday is 1, so you'd better check the correspondence with your representation too.

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