将字符串转换为日期并查找时差

Question

我想将字符串转换为日期时间对象，然后减去2个日期之间的差异。但是我一直收到一个无效的指针警告。日期格式为20131209 02:34。我的代码在

之下

NSDateFormatter *df =[[NSDateFormatter alloc]init];
[df setDateFormat:@"YYYYMMDD HH:MM"];
NSDate *currTime = [df dateFromString:self.currentTime];
NSDate *predicTime =[df dateFromString:self.predictedTime];
NSTimeInterval newtime=[predicTime timeIntervalSinceDate:currTime];

Answer 1

您的日期格式不正确[df setDateFormat:@"YYYYMMDD HH:MM"];我认为您的字符串有效！

尝试：

NSDateFormatter *df =[[NSDateFormatter alloc]init];
[df setDateFormat:@"yyyyMMdd HH:mm"];  // 24hr format, for 12hr format use 'hh'
NSDate *currTime = [df dateFromString:@"20131209 02:34"];
NSDate *predicTime =[df dateFromString:@"20131209 04:34"];
NSTimeInterval newtime=[predicTime timeIntervalSinceDate:currTime];

Answer 2

试试这段代码

 NSDateFormatter *df =[[NSDateFormatter alloc]init];
 [df setDateFormat:@"yyyyMMdd hh:mm"];
 NSDate *currTime = [df dateFromString:self.currentTime];
 NSDate *predicTime =[df dateFromString:self.predictedTime];
 NSTimeInterval newtime=[predicTime timeIntervalSinceDate:currTime];