Question

我有4个标签，所有标签都有UIWebView。我希望第一个标签应该立即加载和显示，而不必等待其他人加载。我在UITabBarController课程中做了这个：

for(UIViewController * viewController in  self.viewControllers){
        if(![[NSUserDefaults standardUserDefaults]boolForKey:@"isSessionExpired"])
        {
            if((int)[self.viewControllers indexOfObject:viewController] != 4)
            {
                viewController.tabBarItem.tag = (int)[[self viewControllers] indexOfObject:viewController];
                [viewController view];
            }
        }
    }

但主线程等待加载所有标签我使用dispatch_async

尝试使用 GCD

dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_HIGH, 0), ^{ // 1
        dispatch_async(dispatch_get_main_queue(), ^{
        UIViewController *firstContrl = [self.viewControllers objectAtIndex:0];
        firstContrl.tabBarItem.tag = (int)[[self viewControllers] indexOfObject:firstContrl];
        [firstContrl view];
        dispatch_async(dispatch_get_main_queue(), ^{ // 2
            UIViewController *secContrl = [self.viewControllers objectAtIndex:1];
            secContrl.tabBarItem.tag = (int)[[self viewControllers] indexOfObject:secContrl];
            [secContrl view];
            dispatch_async(dispatch_get_main_queue(), ^{ // 3
                UIViewController *thirdContrl = [self.viewControllers objectAtIndex:2];
                thirdContrl.tabBarItem.tag = (int)[[self viewControllers] indexOfObject:thirdContrl];
                [thirdContrl view];
                dispatch_async(dispatch_get_main_queue(), ^{ // 4
                    UIViewController *fourthContrl = [self.viewControllers objectAtIndex:3];
                    fourthContrl.tabBarItem.tag = (int)[[self viewControllers] indexOfObject:fourthContrl];
                    [fourthContrl view];
                });
            });
        });
    });
});

但这也不起作用。显示第一个标签需要相同的时间如何解决这个问题？

Answer 1

Ehm .. GCD 和 UIWebView＆＃39> <{strong> -loadRequest:并不像您想象的那样有效。上面的所有代码都非常相同：

dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_HIGH, 0), ^{ dispatch_async(dispatch_get_main_queue(), ^{ UIViewController *firstContrl = [self.viewControllers objectAtIndex:0]; firstContrl.tabBarItem.tag = (int)[[self viewControllers] indexOfObject:firstContrl]; [firstContrl view]; UIViewController *secContrl = [self.viewControllers objectAtIndex:1]; secContrl.tabBarItem.tag = (int)[[self viewControllers] indexOfObject:secContrl]; [secContrl view]; UIViewController *thirdContrl = [self.viewControllers objectAtIndex:2]; thirdContrl.tabBarItem.tag = (int)[[self viewControllers] indexOfObject:thirdContrl]; [thirdContrl view]; UIViewController *fourthContrl = [self.viewControllers objectAtIndex:3]; fourthContrl.tabBarItem.tag = (int)[[self viewControllers] indexOfObject:fourthContrl]; [fourthContrl view]; }); });

要解决您的问题，您需要实现队列取决于- (void)webViewDidFinishLoad:(UIWebView *)webView。这个想法是：

第一个视图控制器的webView开始加载请求异步;

当加载（或失败）请求时，webView将通知您委托;

使用webView通知所有其他视图控制器，让它们启动加载他们的请求;

如果我是你，我会以下一个方式实施它：

// FirstViewController.h extern NSString * const FirstViewControllerDidLoadWebView; // FirstViewController.m NSString * const FirstViewControllerDidLoadWebView=@"FirstViewControllerDidLoadWebView"; - (void)webViewDidFinishLoad:(UIWebView *)webView { // notify other controllers, to let them load their web views: [[NSNotificationCenter defaultCenter] postNotificationName:FirstViewControllerDidLoadWebView object:nil]; } - (void)webView:(UIWebView *)webView didFailLoadWithError:(NSError *)error { // loading failed. Try to load it few more times but anyway notify other controllers after: [[NSNotificationCenter defaultCenter] postNotificationName:FirstViewControllerDidLoadWebView object:nil]; }

之后，＆＃39;订阅＆＃39;所有其他（仅选项卡）查看FirstViewControllerDidLoadWebView通知的控制器，并在其到达后加载其WebView。

有关详细信息，请查看NSNotificationCenter和UIWebViewDelegate

使用GCD连续显示多个视图

1 个答案: