iphone - 使用 webview 通过 safari 打开链接的问题

Question

您好，我在通过 Safari 打开 Web 链接时遇到了问题。这是我的代码

标题

#import <UIKit/UIKit.h>


@interface qrcode_info : UIViewController <UIWebViewDelegate,UIAlertViewDelegate> {

}

@property (nonatomic, retain) IBOutlet UIWebView *Web;



@end



//main file


#import "qrcode_info.h"


@implementation qrcode_info
@synthesize Web;



-(BOOL)Web:(UIWebView *)Web shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType {
    if (navigationType == UIWebViewNavigationTypeLinkClicked) {
        [[UIApplication sharedApplication] openURL:[request URL]];
        return NO;
    }

    return YES;
}

    //[[UIApplication sharedApplication] openURL:[inRequest URL]];
    //return NO;


- (void)viewDidLoad {
    [super viewDidLoad];
    [Web  loadRequest:[NSURLRequest requestWithURL:[NSURL fileURLWithPath:[[NSBundle mainBundle] pathForResource:@"help" ofType:@"html"] isDirectory:NO]]];
    //self.wvTutorial = [[WebViewController alloc] initWithNibName:@”WebView” bundle:[NSBundle mainBundle]];

}

Answer 1

-(BOOL)Web:(UIWebView *)Web shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType {

应该是

- (BOOL)webView:(UIWebView *)webView shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType {

您不能更改委托(delegate)方法的签名并期望它被调用。