XMLParser.sharedParser.decode() in swift3
Asked Answered
C

2

0

I used this code XMLParser.sharedParser.decode() in swift2. But I migrate swift3 this code didn't work.

My code is here,

func xmppStream(_ sender: XMPPStream!, didReceive message: XMPPMessage!) {


    let result1 = XMLParser.sharedParser.decode(String(message))

    for (key, value) in result1 {

        let title = result1["title"]

    }

   if message.isChatMessage(){

       sms = String(message.body())
       if( sms == "" || message.body() == nil){

       }else{
           self.prefs.set(sms, forKey: "SMS")

       }
   }

}
Coincident answered 21/11, 2016 at 9:50 Comment(3)
this code didn't work This is not very useful. Instead, please explain what are the error(s), what doesn't work, what it should do but doesn't, etc. Thanks.Phenoxide
Type 'XMLParser' has no member 'sharedParser' @EricAyaCoincident
how to write in swift3? @EricAyaCoincident
L
0

If you are using codes from this repo, unfortunately, it hasn't been upgraded to Swift 3.

Latonya answered 25/11, 2016 at 16:52 Comment(0)
C
0
var event : DDXMLElement = message.forName("event", xmlns: "http://jabber.org/protocol/pubsub#event")!

            let items : [DDXMLElement] = event.elements(forName: "items")

            for i: DDXMLElement in items {

                var item  = i.elements(forName: "item")

                for j: DDXMLElement in item {
                    var xml = j.elements(forName: "xml")
                    for x : DDXMLElement in xml {
                        var about = x.elements(forName: "about")

                       debugPrint(about.cs_arrayValue().first?.stringValue)

                    }
                }

            }

After I added the above code, my problem is solved!

Coincident answered 26/5, 2017 at 7:30 Comment(0)

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