Opening Link in Safari Problem

[mikelowe]

I am trying to open a link in safari and I know how to do it, but I am running into some problems.

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"%@", link]];

this seems like it should work, but for some reason i get the following error:

Too many arguments to function 'URLWithString:'

help is appreciated

[JasonR]

You don't need the @"%@" before the link. You only use that when you are using a format string.

[bobbypage]

Why don't you just do it like this?

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://www.apple.com"]];

[mikelowe]

Originally Posted by bobbypage

Why don't you just do it like this?

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://www.apple.com"]];

Because the eventually the user will be defining the link so it cant be one definite url.

[SkyTrix]

URLWithString only takes one argument, you gave it two. You're trying to format the string, so this is what you need:

[[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"%@", link]]];

[WillSDev]

or you could do

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

i think