У меня следующая ошибка при генерации релиза ios с флаттером, у меня самая последняя версия флаттера, очевидно, проблема в стручках, или simple_permisions, помогите мне, пожалуйста
flutter build ios --release
Running Xcode build...
Xcode build done. 23.4s
Failed to build iOS app
Error output from Xcode build:
↳
** BUILD FAILED **
Xcode's output:
↳
Command CompileSwift failed with a nonzero exit code
/Users/guser/developent/flutter/.pub-cache/hosted/pub.dartlang.org/simple_permissions-0.1.9/ios/Cla
sses/SwiftSimplePermissionsPlugin.swift:51:52: error: 'openSettingsURLString' has been renamed to
'UIApplicationOpenSettingsURLString'
if let url = URL(string: UIApplication.openSettingsURLString) {
^~~~~~~~~~~~~~~~~~~~~
UIApplicationOpenSettingsURLString
UIKit.UIApplication:70:22: note: 'openSettingsURLString' was introduced in Swift 4.2
public class let openSettingsURLString: String
^
/Users/guser/developent/flutter/.pub-cache/hosted/pub.dartlang.org/simple_permissions-0.1.9/ios/Cla
sses/SwiftSimplePermissionsPlugin.swift:54:65: error: cannot convert value of type
'[UIApplication.OpenExternalURLOptionsKey : Any]' (aka 'Dictionary<NSString, Any>') to expected
argument type '[String : Any]'
UIApplication.shared.open(url, options:
convertToUIApplicationOpenExternalURLOptionsKeyDictionary([:]),
completionHandler: nil)
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~
as [String
: Any]
/Users/guser/developent/flutter/.pub-cache/hosted/pub.dartlang.org/simple_permissions-0.1.9/ios/Cla
sses/SwiftSimplePermissionsPlugin.swift:351:108: error: incorrect argument label in call (have
'rawValue:', expected 'string:')
return Dictionary(uniqueKeysWithValues: input.map { key, value in
(UIApplication.OpenExternalURLOptionsKey(rawValue: key), value)})
string
note: Using new build system
note: Planning build
note: Constructing build description
Encountered error while building for device