用于启动URL的Xcode按钮

Question

我创建了一个简单的应用程序,底部工具栏中有一个按钮,我似乎无法使按钮工作.我试图附加一个动作,它会在按下时打开一个URL,但它不起作用.所以我已经删除了那个动作,我希望有人可以提供帮助.我已经在这个URL上发布了压缩的xcode项目https://www.box.com/s/5d45ce1df7d9dd0fe205

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Answer 1

像这样的东西会起作用.这是纯文本按钮的示例:

UIButton *button = [UIButton buttonWithType:UIButtonTypeCustom];
[button setFrame:CGRectMake(0, 0, 100, 40)];
[button setBackgroundColor:[UIColor clearColor]];
[button setTitle:@"Google" forState:UIControlStateNormal];
[button setTitleColor:[UIColor blueColor] forState:UIControlStateNormal];
[button addTarget:self action:@selector(openGoogleURL) forControlEvents:UIControlEventTouchUpInside];
[self.view addSubview:button];

按钮的选择器:

-(void)openGoogleURL
{
    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"https://www.google.com"]];
}