Question

当用户点击网页视图上网页显示的网址链接时，我使用下面的代码启动了Safari：

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

它适用于iOS 4和iOS 5。

然而，在iOS4上，它启动了safari，但是当我关闭浏览器并返回应用程序时，网页视图仍然继续转到我发送到safari的网址。

如何避免这种情况？

Answer 1

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

Answer 2

删除后尝试！ from if（！[[UIApplication sharedApplication] openURL：[request URL]]），如下所示 -

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

它可能对你有帮助....

强制应用程序启动Safari

2 个答案: