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

时间:2019-10-18 02:30:56

标签: ios swift firebase firebase-remote-config

我使用Firebase Remote Config比较应用程序的版本以进行更新,但是我无法获得Remote Config的值。

如何快速获取数组json之类的值?

这是我的价值

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

这是我的代码

获取价值:

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 {

        }
    }
}

我可以使用“版本”:“ 1.0.4”,但不能获得“文本”。 我该怎么办?

2 个答案:

答案 0 :(得分:0)

当您使用JSON对象时,建议您在此处使用SwiftyJSON。

在项目中安装SwiftyJSON

然后在文件顶部导入SwiftyJSON

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"])
            
        }
    }
        
}

答案 1 :(得分:0)

此解决方案对我有用,无需使用 SwiftyJSON

func getSomeValue() -> SomeValue? {
    let configValue = remoteConfig?.configValue(forKey: "some_key")
    if let data = configValue?.dataValue {       
        do {
            let result = try JSONDecoder().decode(SomeValue.self, from: data)
            return result
        } catch {
            return nil
        }
    }
    return nil
}