I have a string in the format "Fri Jul 09 17:57:44 +0000 2010" which I need to convert to an NSDate.
I have tried a few unsuccessful operations to convert this date, and was wondering if anyone could tell me how I can achieve this.
Regards
I have a string in the format "Fri Jul 09 17:57:44 +0000 2010" which I need to convert to an NSDate.
I have tried a few unsuccessful operations to convert this date, and was wondering if anyone could tell me how I can achieve this.
Regards
Have you tried the brute force method:
This is untested, but should give you some idea of what to do.
// Set up some helper information
NSArray *monthArray = [NSArray arrayWithObjects:@"Jan", @"Feb", @"Mar",
@"Apr", @"May", @"Jun",
@"Jul", @"Aug", @"Sep",
@"Oct", @"Nov", @"Dec", nil];
NSArray *ordinalArray = [NSArray arrayWithObjects:@"01", @"02", @"03",
@"04", @"05", @"06",
@"07", @"08", @"09",
@"10", @"11", @"12", nil];
NSDictionary *monthOrdinalDictionary = [NSDictionary dictionaryWithObjects:ordinalArray
forKeys:monthArray];
// This is the date string to convert.
dateString = @"Fri Jul 09 17:57:44 +0000 2010";
// Split it into it's components
NSArray *dateComponentArray = [dateString componentsSeparatedByString:@" "];
// Create a new date string from the components in a format that NSDate understands
newDateString = [NSString stringWithFormat:@"%@-%@-%@ %@ %@", [dateComponentArray objectAtIndex:6],
[monthOrdinalDictionary objectForKey:[dateComponentArray objectAtIndex:2]],
[dateComponentArray objectAtIndex:3],
[dateComponentArray objectAtIndex:4],
[dateComponentArray objectAtIndex:5]];
// Create the date from this string
NSDate *dateTime = [NSDate dateWithString:newDateString];
Yes, it's ugly but you can refactor most of this out into a helper function. I haven't tested it, but you should have an idea of what to do: Take the date string, convert it into a string that NSDate can use to create a date, use this string to create the NSDate.
At least you'll have code to work on and you can go back and change it if it is a performance bottleneck.