如何在iOS 10中以编程方式打开键盘设置屏幕？

Question

如何在iOS 10中以编程方式打开键盘设置屏幕？

此代码在iOS 10中不起作用

NSURL *keyboardSettingsURL = [NSURL URLWithString: @"prefs:root=General&path=Keyboard/KEYBOARDS"];
    [[UIApplication sharedApplication] openURL:keyboardSettingsURL];

并添加了URL Scheme

Answer 1

看起来iOS 10中已禁用此功能。我相信https://developer.apple.com/library/content/qa/qa1924/_index.html已不再有效。我检查了几个顶级键盘应用程序，它们已更新为不再具有指向首选项的直接链接，或者具有非工作按钮。

Answer 2

适用于iOS 10+

NSURL *keboardURL = [NSURL URLWithString: @"App-Prefs:root=General&path=Keyboard/KEYBOARDS"];
[[UIApplication sharedApplication] openURL:keyboardURL];

关键点：

@“的App-偏好设置：根=通用及路径=键盘/键盘”

Answer 3

如果您尚未将URL方案添加到项目中，则可能需要添加URL方案。在这个Apple Q&A Reference可以找到说明和更多细节。希望这可以帮助！

Answer 4

在iOS 10中，需要新的URL。尝试使用此代码测试两个网址：

NSArray* urlStrings = @[@"prefs:root=General&path=Keyboard/KEYBOARDS", @"App-Prefs:root=General&path=Keyboard/KEYBOARDS"];
for(NSString* urlString in urlStrings){
    NSURL* url = [NSURL URLWithString:urlString];
    if([[UIApplication sharedApplication] canOpenURL:url]){
        [[UIApplication sharedApplication] openURL:url];
        break;
    }
}

Answer 5

For IOS-10 or higher,openURL is deprecated. use with completion handler like below.

NSString *settingsUrl= @"App-Prefs:root=General&path=Keyboard";
if ([[[UIDevice currentDevice] systemVersion] floatValue] >= 10.0){
    if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:settingsUrl]]) {
    [[UIApplication sharedApplication] openURL:[NSURL URLWithString:settingsUrl] options:@{} completionHandler:nil];
     }
 }