swiftyjson - Call can throw, but it is marked with 'try' and the error is not handled

Question

I am trying to use swiftyjson and I am getting an Error:

Call can throw, but it is marked with 'try' and the error is not handled.

I have validated that my source JSON is good. I've been searching and cannot find a solution to this problem

import Foundation


class lenderDetails
{

func loadLender()
{

    let lenders = ""

    let url = URL(string: lenders)!
    let session =  URLSession.shared.dataTask(with: url)
    {
        (data, response, error) in


        guard let data = data else
        {
            print ("data was nil?")
            return
        }

        let json = JSON(data: data)
        print(json)
    }

    session.resume()
}
}

Thank you for all the help!

score 59 · Answer 1 · answered Jun 08 '17 at 17:04

59

The SwiftyJSON initializer throws, the declaration is

public init(data: Data, options opt: JSONSerialization.ReadingOptions = []) throws

You have three options:

Use a do - catch block and handle the error (the recommended one).

do {
   let json = try JSON(data: data)
   print(json)
} catch {
   print(error)
   // or display a dialog
}

Ignore the error and optional bind the result (useful if the error does not matter).
```
if let json = try? JSON(data: data) {
   print(json)
}
```
Force unwrap the result
```
let json = try! JSON(data: data)
print(json)
```
Use this option only if it's guaranteed that the attempt will never fail (not in this case!). Try! can be used for example in FileManager if a directory is one of the default directories the framework creates anyway.

For more information please read Swift Language Guide - Error Handling

answered Jun 08 '17 at 17:04

vadian

274,689
30
353
361

3

This sould be the selected answer! – Ahmadreza Nov 09 '17 at 16:24
Thanks no 3 work for me.. don't know why but no 1 and 2 didnt work =-=" any idea why? *just curious why – dmh Mar 17 '18 at 10:42
@dmh All three ways are supposed to work. What error message do you get? – vadian Mar 17 '18 at 11:43
"Call can throw, but it is marked with 'try' and the error is not handled". got eror... even after trying to rebuild the project and closing the xcode workspace,, – dmh Mar 17 '18 at 12:02
@dmh Check the balance of the braces of the `do - catch` block – vadian Mar 17 '18 at 12:17

score 4 · Accepted Answer · answered Jun 08 '17 at 16:51

You should wrap it into a do-catch block. In your case:

do {
    let session =  URLSession.shared.dataTask(with: url) {
        (data, response, error) in
            guard let data = data else {
            print ("data was nil?")
            return
        }

        let json = JSON(data: data)
        print(json)
    }
} catch let error as NSError {
    // error
}

score 0 · Answer 3 · answered Jun 01 '17 at 04:11

0

Probably you need to implement do{} catch{} block. Inside do block you have to call throwable function with try.

answered Jun 01 '17 at 04:11

noob

545
6
23

swiftyjson - Call can throw, but it is marked with 'try' and the error is not handled

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