使用 NSDateFormatter 或 mktime 创建格式化的 NSDate
我只是想创建一个 NSDate
对象,字符串如下:2011-02-10 4:30:45
看一下代码:
Approach OBJC :
-(NSDate *)get_date_df:(NSString *)dstr
{
NSLog(@"1> %@",dstr);
NSDateFormatter *df_in = [[[NSDateFormatter alloc] init]autorelease];
[df_in setDateFormat:@"yyyy-MM-dd HH:mm:ss"];
NSDate *d = [df_in dateFromString:dstr];
NSLog(@"2> %@",d);
NSLog(@"3> %@",[df_in stringFromDate:d]);
return d;
}
这是输出:
1> 2011-02-10 4:30:45
2> 2011-02-09 23:00:45 +0000
3> 2011-02-10 04:30:45
那么为什么 2>
打印错误?
假设在2011-02-10 4:20:45
,我尝试设置一个具有该日期的 LocalNotification,它会立即触发,因为处理器假设时间已经过期!
UILocalNotification *lnf = [[UILocalNotification alloc] init];
[lnf setFireDate:[self get_date_df:@"2011-02-10 4:30:45"]];
[lnf setAlertBody:@"WTF"];
[[UIApplication sharedApplication]scheduleLocalNotification:lnf];
[lnf release];
而且,我已经尝试为 en_US、en_FR
设置各种 TimeZones
和 NSLocale
。没有帮助。
方法 C:
仅供参考,现在我正在切换到 C,尝试使用以下内容:
-(NSDate *)get_date_c:(const char *)date_str
{
char date[10][6] = {0};
sscanf(date_str,"%[0-9] - %[0-9] - %[0-9] %[0-9] : %[0-9] : %[0-9]",date[0],date[1],date[2],date[3],date[4],date[5]);
struct tm *t_info = (struct tm *)malloc(sizeof(struct tm));
t_info->tm_sec = atoi(date[5]);
t_info->tm_min = atoi(date[4]);
t_info->tm_hour = atoi(date[3]);
t_info->tm_mday = atoi(date[2]);
t_info->tm_mon = atoi(date[1]);
t_info->tm_year = atoi(date[0]);
NSLog(@"ans: %f",(double)mktime(t_info));
NSDate *d = [NSDate dateWithTimeIntervalSince1970:mktime(t_info)];
free(t_info);
return d;
}
输出: 答:-1.000000
我再次陷入将 time_t
转换为 NSTimeInterval
的困境。
所以,如果您能通过任何方法帮助我,请提前致谢:(
I'm just trying to create a NSDate
object with string as:2011-02-10 4:30:45
Take a look at the code:
Approach OBJC:
-(NSDate *)get_date_df:(NSString *)dstr
{
NSLog(@"1> %@",dstr);
NSDateFormatter *df_in = [[[NSDateFormatter alloc] init]autorelease];
[df_in setDateFormat:@"yyyy-MM-dd HH:mm:ss"];
NSDate *d = [df_in dateFromString:dstr];
NSLog(@"2> %@",d);
NSLog(@"3> %@",[df_in stringFromDate:d]);
return d;
}
And here's the output:
1> 2011-02-10 4:30:45
2> 2011-02-09 23:00:45 +0000
3> 2011-02-10 04:30:45
So why is 2>
getting printed wrong?
Suppose at 2011-02-10 4:20:45
, I try setting up a LocalNotification with that date, it fires instantly because the processor assumes the time is already expired!
UILocalNotification *lnf = [[UILocalNotification alloc] init];
[lnf setFireDate:[self get_date_df:@"2011-02-10 4:30:45"]];
[lnf setAlertBody:@"WTF"];
[[UIApplication sharedApplication]scheduleLocalNotification:lnf];
[lnf release];
And, I've already tried setting up various TimeZones
and NSLocale
for en_US, en_FR
. No help.
Approach C:
And FYI now I'm switching to C, trying something with:
-(NSDate *)get_date_c:(const char *)date_str
{
char date[10][6] = {0};
sscanf(date_str,"%[0-9] - %[0-9] - %[0-9] %[0-9] : %[0-9] : %[0-9]",date[0],date[1],date[2],date[3],date[4],date[5]);
struct tm *t_info = (struct tm *)malloc(sizeof(struct tm));
t_info->tm_sec = atoi(date[5]);
t_info->tm_min = atoi(date[4]);
t_info->tm_hour = atoi(date[3]);
t_info->tm_mday = atoi(date[2]);
t_info->tm_mon = atoi(date[1]);
t_info->tm_year = atoi(date[0]);
NSLog(@"ans: %f",(double)mktime(t_info));
NSDate *d = [NSDate dateWithTimeIntervalSince1970:mktime(t_info)];
free(t_info);
return d;
}
Output:ans: -1.000000
There again I'm stuck with converting time_t
to NSTimeInterval
.
So, thanks in advance if you could help me out with any approach :(
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在您的输入字符串和格式 yyyy-MM-dd HH:mm:ss 中,您没有指定时区。所以默认为UTC。
In your input string and format yyyy-MM-dd HH:mm:ss you didn't specify the timezone. So it defaults to UTC.