Question

This is the a button which when clicked will open up Safari.

-(IBAction)loginClicked:(id)sender{
NSLog(@"loginClicked");

NSLog(@"currentSelectedRow = %i", currentSelectedRow );

loginObj = [appDelegate.loginArray objectAtIndex:currentSelectedRow];
NSLog(@"URL = %@", loginObj.loginURL);

Error-->[[UIApplication sharedApplication] openURL:[NSURL URLWithString: @"%@", loginObj.loginURL]];

}

Error: Too many arguments to method call, expected 1, have 27

If I replace [[UIApplication sharedApplication] with

[[UIApplication sharedApplication] openURL:[NSURL URLWithString: @"www.google.com"]];

Safari can be launched and will go to Google and my debugger shows the following

2012-01-26 16:04:15.546 Login2[197:707] loginClicked
2012-01-26 16:04:15.550 Login2[197:707] currentSelectedRow = 0
2012-01-26 16:04:15.555 Login2[197:707] URL = www.amazon.com

It seems that I have pulled the URL from my array correctly but I can’t implement it to the code to open the URL in Safari.

Answer 1

Error-->[[UIApplication sharedApplication] openURL:[NSURL URLWithString: @"%@", loginObj.loginURL]];  //Edit this line as
Correct-->[[UIApplication sharedApplication] openURL:[NSURL URLWithString:loginObj.loginURL]];