get url from String using URL(string: ) constructor

Question

I'm creating a function which makes an API request. The function receives a parameter and put it inside the API call. It is like this:

static func queryCities(cityNameString : String, completion: @escaping (City)->()){
    let urlString = "http://api.geonames.org/searchJSON?q=\(cityNameString)&username=myusername"
    guard let url = URL(string: urlString) else {return}
    print(url)

}

But only when I try to convert my String into a URL the function doesn't return anything. I need to precise that the API call is working well if I paste it in safari. How can I solve this problem?

Your method does not send a request at all... It just parses the string and prints the URL out. — Sweeper
– Sweeper, Commented Mar 9, 2020 at 18:00
Yes I omitted the part where I did stuff because when I print out the url it doesn't print out anything and I want to know why is that. Basically it seems that the URL(string: urlString) constructor is not doing its job — user12422782
– user12422782, Commented Mar 9, 2020 at 18:03
Be aware that URL(string: returns nil if the city name contains a space character. Basically you have to (percent)encode the string — vadian
– vadian, Commented Mar 9, 2020 at 18:04
What cityNameString did you give it? By answering this, you are providing a minimal reproducible example, which helps us see what's wrong with your code. — Sweeper
– Sweeper, Commented Mar 9, 2020 at 18:22

Gereon · Accepted Answer · 2020-03-09 18:53:05Z

1

I would recommend using URLComponents to assemble your URL:

static func queryCities(cityNameString : String, completion: @escaping (City)->()) {
    var components = URLComponents(string: "http://api.geonames.org/searchJSON")
    components?.queryItems = [
        URLQueryItem(name: "q", value: cityNameString),
        URLQueryItem(name: "username", value: "myusername")
    ]

    guard let url = components?.url else { return }
    print(url)

    // do the http request here
}

answered Mar 9, 2020 at 18:53

Gereon

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Comments

Harish · Accepted Answer · 2020-03-09 18:34:00Z

0

Check for space and remove it from String

 let cityName = "  My CityName  "
 let trimmed = cityName.trimmingCharacters(in: .whitespacesAndNewlines)

answered Mar 9, 2020 at 18:34

Harish

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2 Comments

user12422782 Over a year ago

Thanks, I also found the solution for trimming here: stackoverflow.com/questions/28570973/…

vadian Over a year ago

This won't work either because it does not remove the inner space

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