I need to convert this string date "2011-06-25T11:00:26+01:00” into a long like value.
I tried this
NSDateFormatter *df = [[[NSDateFormatter alloc] init] autorelease];
[df setTimeZone:[NSTimeZone timeZoneWithName:@"UTC"]];
[df setDateFormat:@"yyyy-MM-ddHH:mm:ssZ"];
NSDate *date = [df dateFromString:[time stringByReplacingOccurrencesOfString:@"T" withString:@""]];
[df setDateFormat:@"eee MMM dd, yyyy hh:mm"];
NSLog(@"time%@", time);
long lgTime = (long)[date timeIntervalSince1970];
but this doesn't work. Please help me.
Thanks in advance.
First of all, I missed this the first time but "2011-06-25T11:00:26+01:00” is cannot be parsed. The correct string would be "2011-06-25T11:00:26+0100”.
Once you've the string in that format, use the date format – "yyyy-MM-dd'T'HH:mm:ssZ".
Example usage
NSString * time = @"2011-06-25T11:00:26+0100";
NSDateFormatter *df = [[[NSDateFormatter alloc] init] autorelease];
[df setDateFormat:@"yyyy-MM-dd'T'HH:mm:ssZ"];
NSDate *date = [df dateFromString:time];
long lgTime = (long)[date timeIntervalSince1970];
NSLog(@"%ld", lgTime);
If you want to convert "2011-06-25T11:00:26+01:00” to "2011-06-25T11:00:26GMT+01:00" which can be parsed
NSMutableString *string=[[NSMutableString alloc]initWithString:@"2011-06-25T11:00:26+01:00"];
NSRange range = [string rangeOfString:@"+" options:NSBackwardsSearch];
[string insertString:@"GMT" atIndex:range.location];
NSLog(@"string is %@",string);
Hope this helps you
