Question

Apple提供了用Swift编写的Generic Keychain示例，我想继续使用Objective-C。

我已在两个应用程序和canOpenUrl中启用了钥匙串共享我可以从A调用应用程序B，现在我想要从应用程序A到应用程序B共享用户名和密码。应用程序ID对于这两个应用程序都是相同的。< / p>

我看过各种教程也不想使用任何第三方项目。无法知道如何将参数从app A传递给app B.

Answer 1

启用钥匙串共享：

启用钥匙串共享功能。
选择开发人员团队
将Keychain组名称指定为有意义的名称（例如testKeychainG1）
打开.entitlements文件并将$（AppIdentifierPrefix）替换为您的APP ID（例如AB123CDE45.testKeychainG1）

访问钥匙串（检索共享项目）：

let itemKey = "Item Key"

let keychainAccessGroupName = "AB123CDE45.testKeychainG1"


let query:[String:Any] = [
    kSecClass as String: kSecClassGenericPassword,
    kSecAttrAccount as String: itemKey,
    kSecReturnData as String: kCFBooleanTrue,
    kSecMatchLimit as String: kSecMatchLimitOne,
    kSecAttrAccessGroup as String: keychainAccessGroupName
]

var result: AnyObject?

let resultCodeLoad = withUnsafeMutablePointer(to: &result) {
    SecItemCopyMatching(query as CFDictionary, UnsafeMutablePointer($0))
}

if resultCodeLoad == noErr {
    if let result = result as? Data,
        let keyValue = NSString(data: result,
                                encoding: String.Encoding.utf8.rawValue) as? String {

        // Found successfully
        print(keyValue)
    }
} else {
    print("Error: \(resultCodeLoad)")
}

Answer 2

第1步：
设置网址架构并将 AppA 的网址添加到 AppB 的info.plist之类这样：<key>LSApplicationQueriesSchemes</key> <array> <string>Aapp_Scheme</string> </array>

第2步：
在应用程序A中：

let url = URL.init(string: "B_Scheme://name=Obama&password=Trump");
    if UIApplication.shared.canOpenURL(url) {
        UIApplication.shared.open(url, options: ["":""], completionHandler: nil);
    }

第3步：
在应用B的 AppDelegate.swift 中添加代码：

func application(_ app: UIApplication, open url: URL, options: [UIApplicationOpenURLOptionsKey : Any] = [:]) -> Bool {
    var name  = "", password = "";
    let param = url.absoluteString.replacingOccurrences(of: "B_Scheme://", with: "");
    let paramArray = param.components(separatedBy: "&");
    for temp in paramArray {
        if (temp.range(of: "name=") != nil){
            name = temp.replacingOccurrences(of: "name=", with: "");
        }
        else if (temp.range(of: "password=") != nil){
            password = temp.replacingOccurrences(of: "password=", with: "");
        }
    }
    if name == "Obama" && password == "Trump" {
        print("get param success!");
    }
    return true;
}

两个应用程序之间的钥匙串共享

2 个答案: