objective-c 如何以编程方式打开iOS9中objective c中的WIFI设置
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how to programmatically open the WIFI settings in objective c in iOS9
提问by Helton Fernandes Sampaio
I'm trying to access the WIFI settings through my application using Objective-C. But can not find any way. Could someone help me?
我正在尝试使用 Objective-C 通过我的应用程序访问 WIFI 设置。但是找不到任何办法。有人可以帮助我吗？
Already tested with:
已经测试过：
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=WIFI"]];
Does not work on iOS 9.
不适用于 iOS 9。
采纳答案by Fernando García Corrochano
This is my code
这是我的代码
if (&UIApplicationOpenSettingsURLString != NULL) {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=WIFI"]];
}
Try to add prefs to URL schemes like https://stackoverflow.com/a/31253743/3668465did
尝试向 URL 方案添加首选项，如https://stackoverflow.com/a/31253743/3668465所做的
回答by Karthick Ramesh
As per Apple's New Review standards, we are not supposed to use this way to open Wi-Fi Settings. I have been using this for long time in my app and recently Apple rejected with the below comment.
根据 Apple 的 New Review 标准，我们不应该使用这种方式打开 Wi-Fi 设置。我已经在我的应用程序中使用了很长时间，最近苹果拒绝了以下评论。
Your app uses the "prefs:root=" non-public URL scheme, which is a private entity. The use of non-public APIs is not permitted on the App Store because it can lead to a poor user experience should these APIs change.
您的应用程序使用“prefs:root=”非公共 URL 方案，这是一个私有实体。App Store 不允许使用非公共 API，因为如果这些 API 发生变化，可能会导致糟糕的用户体验。
So you can just navigate to settings of the app by using UIApplicationOpenSettingsURLString.
因此，您可以使用 UIApplicationOpenSettingsURLString 导航到应用程序的设置。
Swift Code:
SWIFT代码：
if let settingsUrl = URL.init(string: UIApplicationOpenSettingsURLString), UIApplication.shared.canOpenURL(settingsUrl) {
UIApplication.shared.openURL(settingsUrl)
}
回答by Shuvo Joseph
This works fine on iOS 10,
这在 iOS 10 上运行良好，
Go to Targets --> (Application) --> Info --> URL Types --> +
转到目标-->（应用程序）--> 信息--> URL 类型--> +
In the URL Schemeswrite
在URL Schemes写
prefs
首选项
Then Call,
然后打电话，
- (void)openWifiSettings
{
if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:@"prefs:root=WIFI"]]) {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"prefs:root=WIFI"]];
} else {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"App-Prefs:root=WIFI"]];
}
}
回答by Vinod Sutar
All conditions:
所有条件：
NSURL * urlCheck1 = [NSURL URLWithString:@"App-Prefs:root=WIFI"];
NSURL * urlCheck2 = [NSURL URLWithString:@"prefs:root=WIFI"];
NSURL * urlCheck3 = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
if ([[UIApplication sharedApplication] canOpenURL:urlCheck1])
{
[[UIApplication sharedApplication] openURL:urlCheck1];
}
else if ([[UIApplication sharedApplication] canOpenURL:urlCheck2])
{
[[UIApplication sharedApplication] openURL:urlCheck2];
}
else if ([[UIApplication sharedApplication] canOpenURL:urlCheck3])
{
[[UIApplication sharedApplication] openURL:urlCheck3];
}
else
{
//Unable to open settings app.
}
回答by Pierre
//Pre iOS 10
NSURL *url = [NSURL URLWithString:@"prefs:root=WIFI"];
if (![[UIApplication sharedApplication] canOpenURL:url])
{ //iOS 10+
url = [NSURL URLWithString:@"App-Prefs:root=WIFI"];
}
[[UIApplication sharedApplication] openURL:url];
回答by Bart
You can't get straight to wifi setting with openURL. All you can do is to open settings for your own app.
您无法使用 openURL 直接进入 wifi 设置。您所能做的就是打开您自己的应用程序的设置。
if (&UIApplicationOpenSettingsURLString != nil) {
NSURL *URL = [NSURL URLWithString:UIApplicationOpenSettingsURLString];
[[UIApplication sharedApplication] openURL:URL];
} else {
...
}
回答by Alessandro Francucci
Swift 4.2, iOS 12
斯威夫特 4.2，iOS 12
This is the function that I'm currently using in my app for it:
这是我目前在我的应用程序中使用的功能：
extension UIApplication {
...
@discardableResult
static func openAppSetting() -> Bool {
guard
let settingsURL = URL(string: UIApplication.openSettingsURLString),
UIApplication.shared.canOpenURL(settingsURL)
else {
return false
}
UIApplication.shared.open(settingsURL)
return true
}
}
Usage: UIApplication.openAppSetting()
用法： UIApplication.openAppSetting()
I also used non-public URL scheme, such as: prefs:root=, but my app was rejected. So if you're trying to do more specific stuff with deeplinking, don't waste your time because at the moment you can't!
我还使用了非公共 URL 方案，例如：prefs:root=，但我的应用程序被拒绝了。因此，如果您想通过深层链接做更具体的事情，请不要浪费时间，因为目前您做不到！
回答by pedro.olimpio
You can use this option:
您可以使用此选项：
iOS >= 4.1 it's possible to obtain SSID of wireless network that device is currenctly connected to.
iOS >= 4.1 可以获取设备当前连接的无线网络的 SSID。
For this you'd use function CNCopyCurrentNetworkInfo
为此，您将使用函数CNCopyCurrentNetworkInfo
Details on implemenation: iPhone get SSID without private library
实现细节：iPhone无需私有库即可获取SSID
