I am converting an NSString like #"12:25 AM May 27 2011" into NSDate.But as I convert it using the following code
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat:#"hh:mm a MMM dd YYYY"];
NSDate *date = [[dateFormatter dateFromString:#"12:25 AM May 27 2011"] copy];
[dateFormatter release];
I get the wrong Date as 2010-12-25 19:25:00 +0000
Can anyone please help me around here
It's not a wrong date, the time is displayed GMT hance (+0000) if you set the correct timezone very thing will be oke.
Related
I want to display date like Today, 25 Mar 2014 15:18, if the date is today's date.
Also expecting Tomorrow, 26 Mar 2014 15:18, if it is tomorrow's date and Wednesday, 27 Mar 2014 15:18, if it is day after tomorrow's date.
Anyone please help me.
Sample from developer.apple.com
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setTimeStyle:NSDateFormatterNoStyle];
[dateFormatter setDateStyle:NSDateFormatterMediumStyle];
NSLocale *frLocale = [[NSLocale alloc] initWithLocaleIdentifier:#"en_US"];
[dateFormatter setLocale:frLocale];
[dateFormatter setDoesRelativeDateFormatting:YES];
NSDate *date = [NSDate dateWithTimeIntervalSinceNow:60*60*24*1];
NSString *dateString = [dateFormatter stringFromDate:date];
NSLog(#"dateString: %#", dateString);
NSDateFormatter *detailFormatter = [[NSDateFormatter alloc] init];
[detailFormatter setLocale:frLocale];
[detailFormatter setDateFormat:#"dd MMM yyyy HH:mm"];
NSString *detailString = [detailFormatter stringFromDate:date];
NSLog(#"detailString: %#", detailString);
NSString *finalString = [NSString stringWithFormat:#"%#, %#",dateString, detailString];
NSLog(finalString);
Apple doesn't directly support using relative date combined with absolute date. What you'd have to do is to create two formatters, one with relative enabled and one without. Then compare the output strings (e.g. "Today" and "Mar 25") and if they are not equal, combine them: "Today, Mar 25", but if they are equal, just use one.
And as an aside, actually on "après-après-demain", I think you may have found a bug (either in docs or code). Do you actually set "après-après-demain"? I tried Apple's sample code:
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setTimeStyle:NSDateFormatterNoStyle];
[dateFormatter setDateStyle:NSDateFormatterMediumStyle];
NSLocale *frLocale = [[NSLocale alloc] initWithLocaleIdentifier:#"fr_FR"];
[dateFormatter setLocale:frLocale];
[dateFormatter setDoesRelativeDateFormatting:YES];
NSDate *date = [NSDate dateWithTimeIntervalSinceNow:60*60*24*3];
NSString *dateString = [dateFormatter stringFromDate:date];
NSLog(#"dateString: %#", dateString);
Which Apple says produces output of dateString: après-après-demain no matter what your language is set to (as it switches the date formatter to French).
But when I ran the code on both iOS and OS X, I get the actual date: Mar 28 (and if I change to *2, I get après-demain.
So, if you're seeing the same, I'd suggest filing a bug report with Apple.
I am not able to parse the following string into date
2012-12-28T00:01:51Z ,
I need date in format 28 Dec 2012
Please help me .
Thanks in advance
Abhishek
Try this way:
NSString *dateString=#"2012-12-28T00:01:51Z";
NSDateFormatter *dateFormatter=[NSDateFormatter new];
[dateFormatter setDateFormat:#"YYYY-MM-dd'T'hh:mm:ss'Z'"];
NSLog(#"=> time %#",[dateFormatter dateFromString:dateString]);
NSDate *formattedDate=[dateFormatter dateFromString:dateString];
[dateFormatter setDateFormat:#"dd MMM YYYY"];
NSLog(#"Formatted date : %#",[dateFormatter stringFromDate:formattedDate]);
Why dont u pass the string simply as it is and then convert it into any format u want ?
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat:#"yyyy-MM-dd"];
I am trying this since last two hours and finaly left it out for you guys :P
I need to convert this String Mon, 14 May 2012, 12:00:55 +0200
into Date dd/mm/yyyy hh:ss format..
Any kind of help towards the goal will be really appreciated.
What I have tried
I have tried using NSDateFormatter but I am unable to figure out the exact format of the above date.
This [dateFormatter setDateFormat:#"EEEddMM,yyyy HH:mm:ss"]; is how I have tried and many other formats too
Eg:
NSString *finalDate = #"Tue, 29 May 2012, 14:24:56 +0200";
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat:#"EEEddMM,yyyy HH:mm:ss"];
NSDate *date = [dateFormatter dateFromString:finalDate];
Here the date alwasys comes as nil
The most important part of date formatting is often forgotten, tell the NSDateFormatter the input language:
NSDateFormatter *dateFormatter = [NSDateFormatter new];
[dateFormatter setDateFormat:#"EEE, dd MMMM yyyy, HH:mm:ss Z"];
dateFormatter.locale = [[NSLocale alloc] initWithLocaleIdentifier:#"EN"];
NSDate *date = [dateFormatter dateFromString:#"Mon, 14 May 2012, 12:00:55 +0200"];
NSLog(#"date: %#", date);
I've checked the output: date: 2012-05-14 10:00:55 +0000
Be aware that the HH in the date formatter is for 24hr.
Now I would suggest not to use a fixed output scheme, but use one of the NSDateFormatterStyle:
[dateFormatter setDateStyle:NSDateFormatterShortStyle];
[dateFormatter setTimeStyle:NSDateFormatterShortStyle];
NSString *dateString = [dateFormatter stringFromDate:date];
NSLog(#"date: %#", dateString);
Which in american english will output: 5/14/12 12:00 PM
This code is valid for ARC if you are not using ARC add autorelease to the NSLocale and release the NSDateFormatter after you are done with it.
NSDateFormatter *inputFormatter = [[NSDateFormatter alloc] init];
[inputFormatter setDateFormat:#"dd/mm/yyyy hh:ss"];
NSString *formatterDate = [inputFormatter stringFromDate:inDate];
[inputFormatter release];
get inforamtion about all Date Formate
my Blog Link is http://parasjoshi3.blogspot.in/2012/01/date-formate-info-for-iphone-sdk.html
and also small function is bellow.....
-(NSString *) dateInFormat:(NSString*) stringFormat {
char buffer[80];
const char *format = [stringFormat UTF8String];
time_t rawtime;
struct tm * timeinfo;
time(&rawtime);
timeinfo = localtime(&rawtime);
strftime(buffer, 80, format, timeinfo);
return [NSString stringWithCString:buffer encoding:NSUTF8StringEncoding];
}
////////Like.......
NSString *mydate = [self dateInFormat:#"%Y%m%d-%H%M%S"];
hope,this help you....
I have stuck in issue in which i have to convert date format is Thu, 28 Jul 2011 22:33:57 +0000 into yyyy-mm-dd hh:mm:ss
please give me some idea how to do this?
What you want is to use the NSDateFormatter. Something like this:
NSDateFormatter* f = [[[NSDateFormatter alloc] init] autorelease];
[f setDateFormat:#"yyyy-MM-dd hh:mm:ss"];
NSString* dateString = [f stringFromDate:date];
Do note that the hours will still follow the users selected locale. Use kk:mm:ss to enforce a 24-hour time.
what have you tried? it's difficult to answer questions like this...
first you have to parse the date into an NSDate, use an NSDateFormatter, the incoming format looks like POSIX date format so should be easy.
then you want to output to the format you specify with another NSDateFormatter
You need to use an NSDateformatter to convert the first date to a string with the following syntax.
NSDateFormatter * dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat: #"yyyy-MM-dd hh:mm:ss"];
NSString *dateString = [dateFormatter stringFromDate: [NSDate date]]; //Put whatever date you want to convert
Then if you want the date as an NSDate and you have the string generated above just put the following code.
NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init];
[dateFormatter setDateFormat: #"yyyy-MM-dd hh:mm:ss"];
NSDate * date = [dateFormatter dateFromString: dateString]; //String generated above
NSDate *My_StartDate,*My_EndDate ;
NSDateFormatter * df= [[NSDateFormatter alloc]init];
[df setDateFormat:#"dd/MM/yyyy hh:mm:ss"];
My_StartDate = [df dateFromString:#"01/05/2010 10:15:33"];
My_EndDate = [df dateFromString:#"01/05/2010 10:45:33"];
NSLog(#"%#",My_StartDate);
NSLog(#"%#",My_EndDate);
In the log i get something like this for the my_startdate as 2010-05-01 04:45:33 +0000 and end date as 2010-05-01 05:15:33 +0000 instead i should have got value as for start date as 2010-05-01 10:15:33 +0000 and end date as 2010-05-01 10:45:33 +0000
Try with below function:
-(NSString *)getDateStringFromDate :(NSDate *)dateValue{
NSDateFormatter *dateFormat = [[NSDateFormatter alloc] init];
[dateFormat setDateFormat:#"yyyy-MM-dd"];
NSDateFormatter *timeFormat = [[NSDateFormatter alloc] init];
[timeFormat setTimeStyle:NSDateFormatterShortStyle];
[timeFormat setDateFormat:#"HH:mm:ss"];
//[timeFormat setDateFormat:#"HH:mm"];
//[timeFormat setDateFormat:#"HH:mm a"];
////
NSString *theDate = [dateFormat stringFromDate:dateValue];
NSString *theTime = [timeFormat stringFromDate:dateValue];
NSLog(#"\n"
"theDate: |%#| \n"
"theTime: |%#| \n"
, theDate, theTime);
return theDate;
}
Change Format of data as per your need.
Let me know in case of any difficulty.
Cheers.
This shows date which follow American standard time string but by this reason you don't get any problem in making your logic.Also
[df setDateFormat:#"dd/MM/yyyy hh:mm:ss"];
this format using 12 hour format (means 2:03 pm and 2:03 am) and date object never use am and pm for showing date object value but when you convert it correctly then it gives you right date and time.
If you feel you get any problem then use different locale for that.
It is displaying asper the GMT+4.30 time.It displays like that only.When you are converting that date to string using the DateFormatter it gives the same date(Whichever you want like start date as 01/05/2010 10:15:33 and end date as 01/05/2010 10:45:33).
NSDateFormatter * dateformatter= [[NSDateFormatter alloc]init];
[dateformatter setDateFormat:#"dd/MM/yyyy hh:mm:ss"];
NSString *dat = [dateformatter stringfromDate:My_StartDate];
then you will get the output as 01/05/2010 10:15:33
You might want to set the time zone of the date formatter to GMT here. Do it using
[df setTimeZone:[NSTimeZone timeZoneWithName:#"GMT"]];
before you do dateFromString: calls. This will give you what you want.
Just need to update here in your code:
I might be like that your time would be in 24 hours format, so at that time you need to use this ....other than that you need to set the timezone.
Follow this link for All zone : http://unicode.org/reports/tr35/tr35-6.html#Date%5FFormat%5FPatterns
[df setDateFormat:#"dd/MM/yyyy hh:mm:ss"];
to
[df setDateFormat:#"dd/MM/yyyy HH:mm:ss"];
You are Done ;)