Question

我尝试使用PropertyListDecoder()解码一个plist但是当我尝试访问密钥时，我收到一个错误，指出它的格式错误。我对自己做错了不知所措。我假设我可以解码Plist文件，就像我可以解码JSON文件一样吗？我不知道，我还是新手。

//struct for PLists
struct AccessControl: Decodable {
    enum AccessControlKeys: String, CodingKey {
         case api
     }

    enum KeySecretKeys: String, CodingKey {
        case apiKey = "KEY"
        case apiSecret = "SECRET"
    }

    var KEYS: [KeySecrets]
//custom decoder
    init(from decoder: Decoder) throws {
        let accessContainer = try decoder.container(keyedBy: AccessControlKeys.self)
        let nestedContainer = try accessContainer.nestedContainer(keyedBy: KeySecretKeys.self, forKey: .api)
        self.KEYS = try nestedContainer([KeySecrets].self, forKey: .apiKey)
        self.KEYS = try nestedContainer.decode([KeySecrets].self, forKey: .apiSecret)
    }
}

struct KeySecrets: Decodable {
    var apiKey: String
    var apiSecret: String
}



func provideAccessKeys(for api: apis = .api, mode: modes = .dev) -> keysForApi? {
    switch api {
    case .api:
        print("Api")
    }
    switch mode {
    case .dev:
        print("mode - developers")
    case .test:
        print("mode - test")
    case .prod:
        print("mode - production")
    }
}

这是我的第一个方法，但它会引发错误说

'无法读取数据，因为格式错误'

if let fileURL = Bundle.main.url(forResource: "Accesscontrol",   withExtension: "plist") {
    do {
        let data = try Data.init(contentsOf: fileURL, options: .mappedIfSafe)
        let decoder = PropertyListDecoder()
        let result = try decoder.decode(AccessControl.self, from: data)

    } catch {
        print(error.localizedDescription)
    }
}

第二种方法，有点放弃Codable一起，仍然无法取出值

guard let fileUrl = Bundle.main.url(forResource: "Accesscontrol", withExtension: "plist") else {return}
let key: String
let secret: String
do {
    let data = try Data.init(contentsOf: fileUrl, options: .mappedIfSafe)
    let plist = try! PropertyListSerialization.propertyList(from:data, options: [], format: nil) as! [Any]
  print(plist)
   let dictionary = plist[api.rawValue]

} catch {
    print(error.localizedDescription)
}

plist文件的结构类似于

<plist version="1.0">
<dict>
    <key>A_GROUP_OF_KEYS</key>
    <array>
        <dict>
            <key>KEY1</key>
            <string>KEY1_STRING</string>
            <key>SECRET1_KEY</key>
            <string>SECRET1_STRING</string>
        </dict>
        <dict>
        <key>KEY2</key>
        <string>KEY2_STRING</string>
        <key>SECRET2_KEY</key>
        <string>SECRET2_VALUE</string>
        </dict>
        <dict>
        <key>KEY</key>
        <string>KEY_STRING</string>
        <key>SECRET_KEY</key>
        <string>SECRET_VALUE</string>
        </dict>
    </array>
<key>ANOTHER_GROUP_OF_KEYS</key>
    <array>
        <dict>
            <key>KEY1</key>
            <string>KEY1_STRING</string>
            <key>SECRET1_KEY</key>
            <string>SECRET1_STRING</string>
        </dict>
        <dict>
        <key>KEY2</key>
        <string>KEY2_STRING</string>
        <key>SECRET2_KEY</key>
        <string>SECRET2_VALUE</string>
        </dict>
        <dict>
        <key>KEY</key>
        <string>KEY_STRING</string>
        <key>SECRET_KEY</key>
        <string>SECRET_VALUE</string>
        </dict>
    </array>
</dict>
</plist>

有什么建议吗？

Answer 1

您的plist文件格式错误，因此无法解码。您不应该使用其他键名称来命名每个键，例如KEY1，KEY2，KEY3等。相反，您应该为键key使用一个名称并放置值字段中的实际名称。 secret也是如此。

这是一个更好的plist文件：

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd">
<plist version="1.0">
<dict>
    <key>A_GROUP_OF_KEYS</key>
    <array>
        <dict>
            <key>key</key>
            <string>KEY1_STRING</string>
            <key>secret</key>
            <string>SECRET1_STRING</string>
        </dict>
        <dict>
            <key>key</key>
            <string>KEY2_STRING</string>
            <key>secret</key>
            <string>SECRET2_VALUE</string>
        </dict>
        <dict>
            <key>key</key>
            <string>KEY3_STRING</string>
            <key>secret</key>
            <string>SECRET3_VALUE</string>
        </dict>
    </array>
    <key>ANOTHER_GROUP_OF_KEYS</key>
    <array>
        <dict>
            <key>key</key>
            <string>KEY1_STRING</string>
            <key>secret</key>
            <string>SECRET1_STRING</string>
        </dict>
        <dict>
            <key>key</key>
            <string>KEY2_STRING</string>
            <key>secret</key>
            <string>SECRET2_VALUE</string>
        </dict>
        <dict>
            <key>key</key>
            <string>KEY3_STRING</string>
            <key>secret</key>
            <string>SECRET3_VALUE</string>
        </dict>
    </array>
</dict>
</plist>

解码这很简单：

struct AccessControl: Decodable {
    struct Key: Decodable {
        var key: String
        var secret: String
    }

    var keyGroup1: [Key]
    var keyGroup2: [Key]

    enum CodingKeys: String, CodingKey {
        case keyGroup1 = "A_GROUP_OF_KEYS"
        case keyGroup2 = "ANOTHER_GROUP_OF_KEYS"
    }
}

使用Swift 4 Codable PropertyListDecoder（）

1 个答案: