use*_*386 19 xcode objective-c ios
我想在safari中打开一个url,超出应用程序,而不是在webview中.
我实现了UIWebViewDelegate,但我仍然无法打开网址.基本上我无法点击网址.
以下是代码:
-(void)newView:(NSString *)title Description:(NSString *)desc URL:(NSString *)url{
webView =[[UIWebView alloc]initWithFrame:CGRectMake(15, 17, 190, 190)];
webView.backgroundColor=[UIColor clearColor];
webView.delegate=self;
webView.opaque = NO;
[webView loadHTMLString:[NSString stringWithFormat:@"<html><body p style='color:white' text=\"#FFFFFF\" face=\"Bookman Old Style, Book Antiqua, Garamond\" size=\"5\">%@ %@</body></html>", desc,url] baseURL:nil];
v = [[HUDView alloc] initWithFrame:CGRectMake(60, 70, 220, 220)];
cancelButton = [UIButton buttonWithType:UIButtonTypeCustom];
cancelButton.frame = CGRectMake(0, 0, 30, 30);
[cancelButton setBackgroundImage:[UIImage imageNamed:@"closebox.png"] forState:UIControlStateNormal];
[cancelButton addTarget:self action:@selector(cancelButtonPressed) forControlEvents:UIControlEventTouchUpInside];
[v addSubview:cancelButton];
[v addSubview:webView];
[self.view addSubview:v];
}
-(BOOL) webView:(UIWebView *)inWeb shouldStartLoadWithRequest:(NSURLRequest *)inRequest navigationType:(UIWebViewNavigationType)inType {
if ( inType == UIWebViewNavigationTypeLinkClicked ) {
[[UIApplication sharedApplication] openURL:[inRequest URL]];
return NO;
}
return YES;
}
JTA*_*pps 43
这个答案很容易通过谷歌获得:
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://www.apple.com"]];
只需将其放入按钮按下或任何您想要调用它的事件,然后传递一个URL(替换@"http:/www.apple.com").
Geo*_*rge 39
阅读评论后,我认为这是您正在寻找的:
实现此方法:
- (BOOL)webView:(UIWebView *)webView shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType;
来自UIWebViewDelegate并取决于您应该返回的请求参数TRUE或FALSE.如果您不希望Web视图打开它,您应该致电:
[[UIApplication sharedApplication] openURL:request.URL];
正如其他人提到的那样FALSE.
希望这可以帮助.干杯!
编辑:如果您的网络视图中无法识别链接,请尝试以下操作:
[webView setDataDetectorTypes:UIDataDetectorTypeLink]
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://www.google.com"]];
对于 iOS 10+
// Objective-C
UIApplication *application = [UIApplication sharedApplication];
[application openURL:URL options:@{} completionHandler:nil];
// Swift
UIApplication.shared.open(url, options: [:], completionHandler: nil)
有关更多信息，请参阅这个。
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