怎样让uiwebview里 点击的网页链接 支持uinavigationbar的返回操作？

Question

在webView里面设置一个返回按钮 就可以呀  

Answer 1

你是想点击链接实现页面的返回[self.navigationController popViewControllerAnimated:YES];还是想点击上面的返回键，回到网页的上一页
-(id)initWithNibName:(NSString *)nibNameOrNil bundle:(NSBundle *)nibBundleOrNil {
if (self = [super initWithNibName:nibNameOrNil bundle:nibBundleOrNil]) {
// 下一个界面的返回按钮
    UIBarButtonItem *temporaryBarButtonItem = [[UIBarButtonItem alloc] init];
    temporaryBarButtonItem.title = @"back";
    temporaryBarButtonItem.target = self;
    temporaryBarButtonItem.action = @selector(back:);
    self.navigationItem.backBarButtonItem = temporaryBarButtonItem;}return self;}
-(void) back
{
    if ([webView canGoBack]) {
        [webView goBack];    
    }
   else
    {
        [self.navigationController popViewControllerAnimated:YES];    }
}

Answer 2

补充上面的，你要想点击链接返回
- (void)webViewDidFinishLoad:(UIWebView *)webView {        if ([webView.request.URL isEqual: @"你的链接地址"]) {
           [self.navigationController popViewControllerAnimated:YES];
        }
         
}