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!
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