URLRequest in Swift5 - ios

Question

I'm using the OpenWeather Current Weather Data Api, and trying to make a url request to get json data from the api in Swift5. I need to print the json data. Here is some code I found on the internet that I have been trying to use, but has not been working.
Note: I do NOT want to use any external libraries. like alamofire.
let url = URL(string: "https://api.openweathermap.org/data/2.5/weather?lat=35&lon=139&appid={APIKEY}")!
var request = URLRequest(url: url)
let session = URLSession.shared
let task = session.dataTask(with: request) { (data, response, error) in
if let error = error {
print(error)
} else if let data = data {
print(data)
} else {
print("nope")
}
}
task.resume()

Answer 1

The Openweathermap API documentation is a bit misleading, the expression {API key} indicates the API key without the braces.
Insert the key with String Interpolation
let url = URL(string: "https://api.openweathermap.org/data/2.5/weather?lat=35&lon=139&appid=\(APIKEY)")!
The URLRequest is not needed and dataTask returns either valid data or an error
let task = URLSession.shared.dataTask(with: url) { (data, _, error) in
if let error = error { print(error); return }
print(String(data: data!, encoding: .utf8)!)
}
task.resume()
To display the data create an appropriate model and decode the data with JSONDecoder

Answer 2

So, at first you should be aware that you are registered and already have your own API Key. The main reason that can occur here for not opening link is that You are using a Free subscription and try requesting data available in other subscriptions . And for future if you want to do just get request you don't need to do session.dataTask(with: request), the session.dataTask(with: url) will be OK.)
Here is simpler way of your code.
guard let url = URL(string: "https://api.openweathermap.org/data/2.5/weather?lat=35&lon=139&appid={APIKEY}") else {return}
let session = URLSession.shared
let task = session.dataTask(with: url) { (data, response, error) in
if let error = error {
print(error)
} else if let data = data {
print(data)
} else {
print("nope")
}
}
task.resume()

Answer 3

Not all APIs work with just URL
So if the API has a header in request, try this code.
Note: The header are dependent on your API.
let semaphore = DispatchSemaphore (value: 0)
let param = [
"language": "english",
"serviceRequestId": 1,
"location": ["latitude": "12.34","longitude": "12.34"]
] as [String : Any]
var request = URLRequest(url: URL(string: "UrlHere")!,timeoutInterval: Double.infinity)
request.addValue("tokenHere", forHTTPHeaderField: "Authorization")
do{
let i = try JSONSerialization.data(withJSONObject: param, options: .prettyPrinted)
// print("\(type(of: i))")
print(String(data: i,
encoding: .ascii) ?? "nil")
request.httpMethod = "POST"
request.httpBody = i
let task = URLSession.shared.dataTask(with: request) { data, response, error in
guard let data = data else {
print(String(describing: error))
semaphore.signal()
return
}
semaphore.signal()
do{
let postData = try JSONDecoder().decode(ModelRootClassHere.self, from: data)
print(postData)
MyData = postData
completion()
}
catch{
print(error)
print("error............")
}
}
task.resume()
semaphore.wait()
}catch{
print(error)
}