ios - 如何从 Firebase remote-config 获取数组 json 值？(Swift)

Question

我使用 Firebase 远程配置来比较要更新的应用程序版本，但无法获得远程配置值。

如何快速获得数组 json 之类的值？

这是我的值(value)

[
{
"version": "1.0.4",
"status": "2",
"text": "Fix Error"
}
]

这是我的代码

获取值(value):

class ConstraintValues {
static let shared = ConstraintValues()
init() {
    RemoteConfig.remoteConfig().setDefaults(["a":"b" as NSObject])
    let debugSettings = RemoteConfigSettings()
    RemoteConfig.remoteConfig().configSettings = debugSettings
}
func fetch(completetion:@escaping ( _ stringarray:String?)->()) {
    RemoteConfig.remoteConfig().fetch(withExpirationDuration: 0) { (status, error) in
        if let error = error {
            print(error)
            completetion(nil)
            return
        }
        RemoteConfig.remoteConfig().activate()
        let rc = RemoteConfig.remoteConfig().configValue(forKey: "VERSION")

        do {
            let json = try JSONSerialization.jsonObject(with: rc.dataValue, options: .allowFragments) as! [String:String]

            guard let dictionary1 = json["version"] else {return }
            guard let dictionary2 = json["text"] else {return }
            guard let dictionary3 = json["status"] else {return }
            completetion(dictionary1)
            completetion(dictionary2)
            completetion(dictionary3)
        }
        catch let error{
            print(error)
            completetion(nil)
        }

    }
}}

检查版本:

func checkversion(controller:UIViewController){
    ConstraintValues.shared.fetch { (version) in
        let cversion = Bundle.main.infoDictionary?["CFBundleShortVersionString"] as! NSString//localversion
         let url = URL(string: "https://www.google.com/")
        let appStore = version//newversion
        //Compare
        let versionCompare = cversion.compare(appStore! as String, options: .numeric)
        if versionCompare == .orderedSame {
            print("same version")
        } else if versionCompare == .orderedAscending {

            let alertc = UIAlertController(title: "UPDATE!!", message: "New Version: \(version!)", preferredStyle: .alert)
            let alertaction = UIAlertAction(title: "Update", style: .default, handler: { (ok) in
                if #available(iOS 10.0, *) {
                    UIApplication.shared.open(url!, options: [:], completionHandler: nil)
                } else {
                    UIApplication.shared.openURL(url!)
                }
            })
            alertc.addAction(alertaction)
            controller.present(alertc, animated: true, completion: nil)

        } else if versionCompare == .orderedDescending {

        }
    }
}

我可以取"version": "1.0.4"，但是取不到"text"。
这个方法我该怎么办？

Answer 1

当您使用 JSON 对象时，我建议您在此处使用 SwiftyJSON。
安装 SwiftyJSON在你的项目中
然后导入 SwiftyJSON在文件顶部
您的代码应如下所示

import Firebase
import SwiftyJSON

class MyViewController: UIViewController {
        
    var remoteConfig: RemoteConfig!
    
    override func viewDidLoad() {
        super.viewDidLoad()
        
        remoteConfig = RemoteConfig.remoteConfig()
        let settings = RemoteConfigSettings()
        settings.minimumFetchInterval = 0
        remoteConfig.configSettings = settings
        
        var remoteConfigDefaults = [String: NSObject]()        
        let testArr: NSArray = []
        remoteConfigDefaults["VERSION"] = testStr
        remoteConfig.setDefaults(remoteConfigDefaults)        
        
        remoteConfig.fetch() { (status, error) -> Void in
            if status == .success {
                print("Config fetched!")
                self.remoteConfig.activate() { (error) in
                    // ...
                }
            } else {
                print("Config not fetched")
                print("Error: \(error?.localizedDescription ?? "No error available.")")
            }
          
            let version = self.remoteConfig.configValue(forKey: "VERSION").jsonValue            
            let json = JSON(version)            
            print(json[0]["version"], json[0]["status"], json[0]["text"])
            
        }
    }
        
}