iPhone:UIWebview 在 safari 中打开超链接

在我的应用程序中，我使用了 UIwebview，我需要在其中超链接一些电话号码、电子邮件和 URL。我做了以下操作:

 - (void)viewDidLoad
    {
        credits.delegate=self;
    }



-(void)loadScreen
{
scroll=[[UIScrollView alloc]init];
 credits=[[UIWebView alloc]init];

NSString *Address = @"www.***.com/php/getjson.php?method=";


    jmax = [NSURL URLWithString:Address];

    NSData *Data =[ NSData dataWithContentsOfURL:jmax];


    String = [[NSString alloc] initWithDataata encoding:NSUTF8StringEncoding];

    NSString * creditsHtml = [NSString stringWithFormat" <font face=\"Arial\"><H3 align=\"center\"> Credits </H3> %@ <br>hone%@</font>", String,phone];

    credits.userInteractionEnabled = YES;
    credits.frame=CGRectMake(10, (frame1.size.height + 564), 300, 100);
    credits.dataDetectorTypes=UIDataDetectorTypeAll;
    credits loadHTMLString:creditsHtml baseURL:nil];

[scroll addSubview:credits];
}



   -(BOOL)webViewUIWebView*)webView shouldStartLoadWithRequestNSURLRequest*)request navigationTypeUIWebViewNavigationType)navigationType{


        NSString *urls=  jmaxString;

        NSURL *requestURL =[NSURL URLWithString: urls]; 
        if ( ( [ [ requestURL scheme ] isEqualToString: @"http" ] || [ [ requestURL scheme ] isEqualToString: @"https" ] || [ [ requestURL scheme ] isEqualToString: @"mailto" ] ) 
            && ( navigationType == UIWebViewNavigationTypeLinkClicked ) ) { 
            return ![ [ UIApplication sharedApplication ] openURL: requestURL ]; 
        } 

        return YES; 
    }

编辑 shouldStartLoadWithRequest: 方法

-(BOOL)webViewUIWebView*)webView shouldStartLoadWithRequestNSURLRequest*)request navigationTypeUIWebViewNavigationType)navigationType{

    NSURL *requestURL =[request URL];

    if ( ( [ [ requestURL scheme ] isEqualToString: @"http" ] || [ [ requestURL scheme ] isEqualToString: @"https" ] || [ [ requestURL scheme ] isEqualToString: @"mailto" ] ) 
        && ( navigationType == UIWebViewNavigationTypeLinkClicked ) ) { 
        return ![ [ UIApplication sharedApplication ] openURL: requestURL ]; 
    } 
    NSLog(@"URL %@",requestURL);


    return YES; 
}

NSURL *requestURL =[request URL]; 出现错误，Expected ;在末尾 & Extraneous ')' before ';'

---

Best Answer-推荐答案

您在 loadScreen 函数中分配/初始化 UIWebView 的一个新对象，而没有设置其委托(delegate)，因为您假设您在 viewDidLoad 中这样做了。您必须在定义实例“信用”后设置委托(delegate)。

关于iPhone:UIWebview 在 safari 中打开超链接，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/9703175/