如何将NSString转换为NSTimeInterval

Question

我有NSString有价值" 22/04/2013 05:56",根据要求我只想计算这个时间和日期,并显示一些基于条件的消息.

第一个条件: If(字符串日期=当前日期和字符串时间=当前时间)||(字符串日期=当前日期和字符串时间<当前时间1分钟)

第二个条件: If(String date =当前日期和stringtime>当前时间由多少分钟或小时)

第三个条件:要知道字符串日期是昨天.

第四个条件:To Know是字符串日期是前天.

我从服务器收到这个字符串.如何通过这个"22/04/2013 05:56"字符串实现上述目标.

Answer 1

你需要采取两步:

1. 将字符串转换为 NSDate
2. 将日期转换为timeStamp

如下:

- (void) dateConverter{
    NSString *string = @"22/04/2013 05:56";
    NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
    // this is imporant - we set our input date format to match our input string
    // if format doesn't match you'll get nil from your string, so be careful
    [dateFormatter setDateFormat:@"dd/MM/yyyy hh:mm"];
    NSDate *date = [[NSDate alloc] init];
    // voila!
    date = [dateFormatter dateFromString:string];
    NSLog(@"dateFromString = %@", date);

    //date to timestamp
    NSTimeInterval timeInterval = [date timeIntervalSince1970];
}

然后实现像以前一样的方法下面的方法将有所帮助,虽然它不完全适合你,但我相信你可以修改它来帮助你!

- (NSString *) timeAgoFor : (NSDate *) date {
    double ti = [date timeIntervalSinceDate:[NSDate date]];
    ti = ti * -1;

    if (ti < 86400) {//86400 = seconds in one day
        return[NSString stringWithFormat:@"Today"];
    } else if (ti < 86400 * 2) {
        return[NSString stringWithFormat:@"Yesterday"];
    }else if (ti < 86400 * 7) {
        int diff = round(ti / 60 / 60 / 24);
        return[NSString stringWithFormat:@"%d days ago", diff];
    }else {
        int diff = round(ti / (86400 * 7));
        return[NSString stringWithFormat:@"%d wks ago", diff];
    }
}