在iOS9中以编程方式打开手机设置 [英] Open phone settings programmatically in iOS9

问题描述

我最近有一个应用程序登陆应用程序商店，他们拒绝批准我使用的以下代码手机设置：

I recently have an app summited to the app store and they refused to approve the following code I used open phone settings:

let url:NSURL! = NSURL(string : "prefs:root=")
    UIApplication.sharedApplication().openURL(url)

所以我得到了以下代码（如何在点击按钮时打开手机设置并获得批准：

So I got the following code (How do i open phone settings when a button is clicked ios) and got it approved:

UIApplication.sharedApplication().openURL(NSURL(string:UIApplicationOpenSettingsURLString)!)

不幸的是它不做我真正需要的是，它有时会打开我的应用程序设置，有时会打开手机设置。

Unfortunately it does not do what I really need as it opens my application settings sometimes and other times it opens the phone settings.

我需要打开一些东西 just&只改为 手机设置。

And I need something to open just & only the phone settings instead.

推荐答案

在iOS 10中，prefs必须替换为App-Prefs。
然后您可以使用此代码测试所有可能的网址。

In iOS 10, "prefs" must be replaced with "App-Prefs". You can then use this code which tests over all the possible urls.

NSArray* urlStrings = @[@"prefs:root=<your_path>", @"App-Prefs:root=<your_path>"];
for(NSString* urlString in urlStrings){
    NSURL* url = [NSURL URLWithString:urlString];
    if([[UIApplication sharedApplication] canOpenURL:url]){
        [[UIApplication sharedApplication] openURL:url];
        break;
    }
}

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