在XCode中添加一个断点向我展示了一个Swift 1.2 ACAccount对象有一个 _properties 对象,该对象又拥有 @user_id:0123456789 的键/值对。我需要访问这个数值。我已阅读文档,谷歌搜索,SO'ed并盲目地尝试以下所有内容无济于事:
let accountStore = ACAccountStore()
let accountType = accountStore.accountTypeWithAccountTypeIdentifier( ACAccountTypeIdentifierTwitter )
let allACAccounts = accountStore.accountsWithAccountType( accountType )
let someACAccount = allACAccounts[ 0 ]
someACAccount.valueForKey( "properties" )
someACAccount.mutableArrayValueForKey( "properties" )
someACAccount.valueForKeyPath( "properties" )
someACAccount.valueForUndefinedKey("properties")
someACAccount.valueForKey( "user_id" )
someACAccount.mutableArrayValueForKey( "user_id" )
someACAccount.valueForKeyPath( "user_id" )
someACAccount.valueForUndefinedKey("user_id")
someACAccount.properties
someACAccount._properties
someACAccount.user_id
someACAccount.valueForKey(“properties”) - 拼写为“properties”和“_properties” - 生成Builtin.RawPointer的结果。我不知道这是否有用。
答案 0 :(得分:3)
采用更强类型的方法可能更容易,以下适用于全新的iOS项目:
let store = ACAccountStore()
let type = store.accountTypeWithAccountTypeIdentifier(ACAccountTypeIdentifierTwitter)
store.requestAccessToAccountsWithType(type, options: nil) { success, error in
if success, let accounts = store.accountsWithAccountType(type),
account = accounts.first as? ACAccount
{
println(account.username)
if let properties = account.valueForKey("properties") as? [String:String],
user_id = properties["user_id"] {
println(user_id)
}
}
}