Question

Dearest stackoverflow brains, I’m having some trouble with iOS again… I am new to this all though.

I’m trying to open a safari with a url, that includes a URL parameter based on the value of the table cell. So if the table cell contained “stack”, I want to go to “http://stack.com?param=stack”. This should be a simple task, but I’m having trouble getting it to work.

It’s failing before I even get to the UIApplication line… so I don’t even know if that will work. I’m getting this error:

2011-10-14 10:18:32.297 NPT[1085:207] *** Terminating app due to uncaught exception
'NSInvalidArgumentException', reason: '*** initialization method -
initWithCharactersNoCopy:length:freeWhenDone: cannot be sent to an abstract object of
class NSCFString: Create a concrete instance!'

When I use this code:

- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath {
    NSString* thisValue = [[[self.searchResults cellForRowAtIndexPath:indexPath] textLabel] text];

    NSString* thisPlateURL = [
                          [NSString new] 
                          initWithFormat:@"http://www.site.co.uk?&regno=%@", thisValue
                          ];

    [[UIApplication sharedApplication] openURL:[NSURL URLWithString: thisPlateURL]];

    [thisPlateURL release];
}

I’m guessing it’s something to do with me not setting thisValue properly, and I’m sure I’ll kick myself if someone spots it, but I just don’t see what the problem is.

Please help!

Answer 1

I think that the problem is in NSString initialization:

try with:

NSString* thisPlateURL = [NSString stringWithFormat: @"http://www.site.co.uk?&regno=%@", thisValue];