在 iOS 9 上从 table 视图打开 safari 视图控制器并在 iOS 8 或 7 上在 safari 中打开

Question

您好，如果用户使用 iOS 9 或更高版本，我想从 Safari 视图控制器中的 table 视图单元格打开我的网站。如果用户使用 iOS 7 或 8，网站应该在标准的 Safari 应用程序中打开。

这是我目前使用的打开 safari 的代码。

    case 3: { // Follow us section
        switch (indexPath.row) {
            case 0: { //Website
                NSURL *url = [NSURL URLWithString:@"http://www.scanmarksapp.com"];
                if (![[UIApplication sharedApplication] openURL:url]) {
                    NSLog(@"%@%@",@"Failed to open url:",[url description]);
                }
            }
                break;

            default:
                break;
        }

    }
        break;

我相信这段代码应该可以打开我的网站的 safari 视图控制器。但是我不确定如何组合两组代码。

- (void)openLink:(NSString *)url {

NSURL *URL = [NSURL URLWithString:[NSString stringWithFormat:@"http://www.scanmarksapp.com", url]];
if (URL) {
    SFSafariViewController *sfvc = [[SFSafariViewController alloc] initWithURL:URL];
    sfvc.delegate = self;
    [self presentViewController:sfvc animated:YES completion:nil];
}

#pragma Safari View Controller Delegate

- (void)safariViewControllerDidFinish:(nonnull SFSafariViewController *)controller {
[controller dismissViewControllerAnimated:YES completion:nil];
}

我知道这是用来确定 iOS 版本的代码

if ([[[UIDevice currentDevice] systemVersion] floatValue] < 9.0) {

我听从了你的建议

- (void)openLink:(NSString *)url {

NSURL *URL = [NSURL URLWithString:[NSString stringWithFormat:@"http://www.scanmarksapp.com", url]];
if (URL) {
    SFSafariViewController *sfvc = [[SFSafariViewController alloc] initWithURL:URL];
    sfvc.delegate = self;
    [self presentViewController:sfvc animated:YES completion:nil];
} else {
    // will have a nice alert displaying soon.
}

if ([SFSafariViewController class] != nil) {
    // Use SFSafariViewController
} else {
    NSURL *url = [NSURL URLWithString:@"http://www.scanmarksapp.com"];
    if (![[UIApplication sharedApplication] openURL:url]) {
        NSLog(@"%@%@",@"Failed to open url:",[url description]);
    }
}

然后在我的 table 视图单元格 didSelectRowAtIndexPath

下添加这段代码

        case 3: { // Follow us section
        switch (indexPath.row) {
            case 0: { //Website
                NSURL *URL = [NSURL URLWithString:[NSString stringWithFormat:@"http://www.scanmarksapp.com", url]];
                if (URL) {
                    SFSafariViewController *sfvc = [[SFSafariViewController alloc] initWithURL:URL];
                    sfvc.delegate = self;
                    [self presentViewController:sfvc animated:YES completion:nil];
                } else {
                    // will have a nice alert displaying soon.
                }

                if ([SFSafariViewController class] != nil) {
                    // Use SFSafariViewController
                } else {
                    NSURL *url = [NSURL URLWithString:@"http://www.scanmarksapp.com"];
                    if (![[UIApplication sharedApplication] openURL:url]) {
                        NSLog(@"%@%@",@"Failed to open url:",[url description]);
                    }

                }
            }
                break;

            default:
                break;
        }

    }
        break;

我在这行代码中收到错误 "Use of undeclared identifier url"

NSURL *URL = [NSURL URLWithString:[NSString stringWithFormat:@"http://www.scanmarksapp.com", url]];

删除 NSStringWithFormat 末尾的 url 可使 Safari 视图控制器正常工作。但是在 iOS 低于 9.0 例如8.4 应用程序崩溃。

Answer 1

标准和推荐的方法是检查功能，而不是 OS 版本。在这种情况下，您可以检查 SFSafariViewController class.

是否存在

if ([SFSafariViewController class] != nil) {
    // Use SFSafariViewController
} else {
    // Open in Mobile Safari
}

编辑

您对 openLink: 的实施是错误的。

- (void)openLink:(NSString *)url {
    NSURL *URL = [NSURL URLWithString:url];

    if (URL) {
        if ([SFSafariViewController class] != nil) {
            SFSafariViewController *sfvc = [[SFSafariViewController alloc] initWithURL:URL];
            sfvc.delegate = self;
            [self presentViewController:sfvc animated:YES completion:nil];
        } else {
            if (![[UIApplication sharedApplication] openURL:url]) {
                NSLog(@"%@%@",@"Failed to open url:",[url description]);
            }
        }
    } else {
        // will have a nice alert displaying soon.
    }
}

在 iOS 9 上从 table 视图打开 safari 视图控制器并在 iOS 8 或 7 上在 safari 中打开

Open safari view controller from table view on iOS 9 and open in safari on iOS 8 or 7

iphone

safari

objective-c

ios

sfsafariviewcontroller