UIWebView 在 Safari 中打开链接 [英] UIWebView open links in Safari

问题描述

我有一个非常简单的 UIWebView，其中包含来自我的应用程序包的内容.我希望 web 视图中的任何链接在 Safari 中打开，而不是在 web 视图中打开.这可能吗?

I have a very simple UIWebView with content from my application bundle. I would like any links in the web view to open in Safari instead of in the web view. Is this possible?

推荐答案

将此添加到 UIWebView 委托:

Add this to the UIWebView delegate:

(编辑以检查导航类型.您也可以通过 file:// 请求，这些请求将是相对链接)

(edited to check for navigation type. you could also pass through file:// requests which would be relative links)

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

    return YES;
}

Swift 版本:

func webView(webView: UIWebView, shouldStartLoadWithRequest request: NSURLRequest, navigationType: UIWebViewNavigationType) -> Bool {
        if navigationType == UIWebViewNavigationType.LinkClicked {
            UIApplication.sharedApplication().openURL(request.URL!)
            return false
        }
        return true
    }

Swift 3 版本:

Swift 3 version:

func webView(_ webView: UIWebView, shouldStartLoadWith request: URLRequest, navigationType: UIWebViewNavigationType) -> Bool {
    if navigationType == UIWebViewNavigationType.linkClicked {
        UIApplication.shared.openURL(request.url!)
        return false
    }
    return true
}

Swift 4 版本:

Swift 4 version:

func webView(_ webView: UIWebView, shouldStartLoadWith request: URLRequest, navigationType: UIWebView.NavigationType) -> Bool {
    guard let url = request.url, navigationType == .linkClicked else { return true }
    UIApplication.shared.open(url, options: [:], completionHandler: nil)
    return false
}

更新

由于 openURL 在 iOS 10 中已被弃用:

As openURL has been deprecated in iOS 10:

- (BOOL)webView:(UIWebView *)webView shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType {
        if (navigationType == UIWebViewNavigationTypeLinkClicked ) {
            UIApplication *application = [UIApplication sharedApplication];
            [application openURL:[request URL] options:@{} completionHandler:nil];
            return NO;
        }

        return YES;
}

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