How to set JSON output into UILabel in Swift 3.0 ?

Question

I have these JSON data.

(
        {
        email = "b@p.com.my";
        login = ID001;
        pw = 1234;
    },
        {
        email = "p@d.com.my";
        login = ID002;
        pw = 12345;
    }
)

Right now, I can only print myJSON value in x code output.

My question is, how to display each JSON into UILabel _email, _login, _pw?

Someone said that I need to store as variable and set into UILabel but i don't know how to do it. Appreciate if someone can help me on this matters.

ViewController.swift

import UIKit
class ViewController: UIViewController {
    @IBOutlet var _email: UILabel!
    @IBOutlet var _login: UILabel!
    @IBOutlet var _pw: UILabel!

    override func viewDidLoad() {
        super.viewDidLoad()
        let url = URL(string: "http://localhost/get.php")
        let task = URLSession.shared.dataTask(with: url!) {
            (data, response, error) in
            if error != nil {
                print("Error")
                return
            }
            else {
                if let content = data {
                    do {
                        let myJSON = try JSONSerialization.jsonObject(with: content, options: JSONSerialization.ReadingOptions.mutableContainers) as AnyObject
                        print(myJSON)
                    }
                    catch {}   
                }
            }
        }
        task.resume()
    }
}

Thanks.

Store your response in one array and from that array just set a value into label — Himanshu Moradiya, Jan 13 '17 at 04:22
Parsing your `json` response depends on the printed `myJSON` format. — amagain, Jan 13 '17 at 04:23
not sure how to store it as array and set it into label . I'm a beginner .. still struggle to understand swift 3.0 concept. sorry. — Jamilah, Jan 13 '17 at 05:06
In your Json Data are two entries of email/login/password but you have only one uilabel for each. Which entry do you want to show on screen? — jboi, Jan 13 '17 at 06:19
@UmairAfzal . nope . just the simple json output . i want to show each json data for it each label . — Jamilah, Jan 13 '17 at 06:27
Still, I see two email entries in the json data but only one one label to show them. Which one do you want to show. Or do you want both in one label somehow concatenated? Please tell us how the outcome should look like. — jboi, Jan 13 '17 at 06:29
@jboi . _email label = ""b@p.com.my"", _login = "ID001",_pw = "1234" .. no need for the second json object .. — Jamilah, Jan 13 '17 at 06:34

score 0 · Answer 1 · edited May 23 '17 at 11:53

0

EDIT: After the block where you print your myJSON, try this:

for eachItem in myJSON {
if let emailParsed = eachItem["email"] as? String {
    print(emailParsed)
    _email.text = emailParsed 
} 
}

This loop runs between all the dictionaries and This solution could be helpful for you. Please go through it!

edited May 23 '17 at 11:53

Community

1
1

answered Jan 13 '17 at 04:28

amagain

2,042
2
20
36

Hi, i tried your link suggestion . It gives me an error "Could not cast value of type '__NSArrayI' (0x10d055d88) to 'NSDictionary' (0x10d056288)." – Jamilah Jan 13 '17 at 04:47
@Jamilah did you use tableview for set all UILabel _email, _login, _pw value ? – Himanshu Moradiya Jan 13 '17 at 04:59
Nope . just using simple UILabel . Reason i'm doing this is to understand the concept of JSON Parse in Swift 3.0 . :) – Jamilah Jan 13 '17 at 05:01
@Jamilah how many label you declare for display this json response ? – Himanshu Moradiya Jan 13 '17 at 05:11
just three label .. emal, pw and login – Jamilah Jan 13 '17 at 05:11
i tried to add for loop inside do statement . it gives me an error "Type AnyObject does not conform to protocol 'Sequence'" – Jamilah Jan 13 '17 at 05:12
@Jamilah i suggest you do same thing with tableview and print your whole json data in label so you get idea about this. – Himanshu Moradiya Jan 13 '17 at 05:15
i haven't check it yet . Just need to 'walk before i can run' :) .. i'm a beginner .. still struggle to understand this language – Jamilah Jan 13 '17 at 07:00

Pushpendra · Answer 2 · 2017-01-13T06:28:13.413

0

Hi Please try this one

   for i..0 < myJson.count { 
      let object = myJson[i] as AnyObject
      if let emailParsed = myJSON["email"] as? String {
         _email.text = emailParsed 
      }     
   }

EDITED

for i..0 < myJson.count { 
      let object = myJson[i] as [String : AnyObject]
      if let emailParsed = myJSON["email"] as? String {
         _email.text = emailParsed 
      }     
   }

edited Jan 13 '17 at 06:28

answered Jan 13 '17 at 04:34

Pushpendra

1,492
1
17
33

what exactly you are facing – Pushpendra Jan 13 '17 at 05:18
problem is i can't view each of the json data in UILabel . an error occurs "Type AnyObject does not conform to protocol 'Sequence'" – Jamilah Jan 13 '17 at 06:21
Check the edited Answer, their might be some syntactical error, Because this code run in `Swift3` – Pushpendra Jan 13 '17 at 06:29

Dhruv Khatri · Answer 3 · 2017-01-13T06:43:37.477

It's simple because your json object gives an array of dictionary so first you have to take first object from array and after that you can take the string passing the key

override func viewDidLoad() 
{
  super.viewDidLoad()
  let url = URL(string: "http://localhost/get.php")
  let task = URLSession.shared.dataTask(with: url!) 
             {
             (data, response, error) in
                  if error != nil
                     {
                       print("Error")
                       return
                     }
                  else 
                     {
                       if let content = data 
                          {
                             do {
                                  let myJSON = try JSONSerialization.jsonObject(with: content, options:JSONSerialization.ReadingOptions.mutableContainers) as AnyObject
                                 //   print(myJSON)
                                    for data in myJSON
                                    {
                                        let email = data.1["email"] as? String
                                        _email.text = email

                                    }
                                }
                                catch {}   
                          }
                      }
             }
             task.resume()
      }
}

For batter option you can use some predefined libraries like swiftyJSON. To get best result.

somehow i did get an error on this line "for data in myJSON" . it says "Type AnyObject does not conform to protocol 'Sequence'" — Jamilah, Jan 13 '17 at 06:20
error appear on line if let emailParsed. . "Type NSFastEnumerationIterator.Element" (aka 'Any') has no subscript members — Jamilah, Jan 13 '17 at 06:25
what is the suggestions are there if you write "data." (it's 0 or 1). — Dhruv Khatri, Jan 13 '17 at 08:37

How to set JSON output into UILabel in Swift 3.0 ?

3 Answers3