如何在 Safari 中打开外部链接而不是应用程序的 UIWebView? [英] How can I open an external link in Safari not the app's UIWebView?

问题描述

我有一个 Phonegap (cordova) 应用程序，我想在 phonegap WebView 中加载一些外部网页，我还有其他外部网页，我想在用户激活它们时在 safari 中加载它们.

I have a Phonegap (cordova) application where I want to load some external webpages within the phonegap WebView and I have other external webpages that I want to load in safari when the user activates them.

通常大多数人都会遇到想要在 WebView 中打开外部链接的问题.将 OpenAllWhitelistURLsInWebView 设置为 YES(在 Cordova.plist/Phongap.plist 中)可以解决这个问题.

Typically most people have the problem where they want to open an external link in the WebView. Setting OpenAllWhitelistURLsInWebView to YES (in Cordova.plist/Phongap.plist) solves that problem.

但我不想打开WebView的所有链接，只打开一些.

But I don't want to open all links the the WebView, just some.

我希望我可以调用 window.open('http://someexternalsite') 在 Safari 中打开和 window.parent.location.href = 'http://mysite' 在 WebView 中打开它.

I was hoping I could just call window.open('http://someexternalsite') to open in Safari and window.parent.location.href = 'http://mysite' to open it in the WebView.

知道怎么做吗?

推荐答案

如果你想在 safari 中打开的链接都包含一个公共字符串，你可以使用下一段代码.

If the links you want to open in safari all contain a common string, you can use the next piece of code.

- (BOOL)webView:(UIWebView *)theWebView shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType
{
    NSURL *url = [request URL];

    // Intercept the external http requests and forward to Safari.app
    // Otherwise forward to the PhoneGap WebView
    if ([[url scheme] isEqualToString:@"SCHEME"]) {
        [[UIApplication sharedApplication] openURL:url];
        return NO;
    }
    else {
        return [ super webView:theWebView shouldStartLoadWithRequest:request navigationType:navigationType ];
    }
}

放置在 AppDelegate.m 中的这段代码将在 Safari 中打开所有使用指定 SCHEME 的 URL.

This code placed in the AppDelegate.m will open all URL that use the specified SCHEME in Safari.

恐怕这就是我能想到的全部.

I'm afraid that is all I could come up with.

希望能帮到你

更新:

代码应该放在 MainViewControler 中，至少对于cordova 2.2.0.

The code should be placed in the MainViewControler, at least for cordova 2.2.0.

该方法最初被注释.我不得不用它来重定向谷歌地图链接:

The method is initially commented. I had to use it to redirect Google maps links :

NSRange isGoogleMaps = [[url absoluteString] rangeOfString:@"maps.google.com" options:NSCaseInsensitiveSearch];
NSRange isGoogleTerms = [[url absoluteString] rangeOfString:@"terms_maps.html" options:NSCaseInsensitiveSearch];

if(isGoogleMaps.location != NSNotFound || isGoogleTerms.location != NSNotFound ) {
        [[UIApplication sharedApplication] openURL:url];
       return NO;
}
else 
    return [super webView:theWebView shouldStartLoadWithRequest:request navigationType:navigationType];

这篇关于如何在 Safari 中打开外部链接而不是应用程序的 UIWebView?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！